ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
1 "ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS"
Problem 1. Consider the elliptic curve E:y2=x3−7x+ 6.
a) Find all rational points on the curve E.
b) Show that the point P(1,2) is a point of infinite order on the curve E.
Solution 1.
a) To find all rational points on the curve E, we can use the Nagell–Lutz theorem. The theorem
states that if (x, y)is a rational point of infinite order on the elliptic curve E:y2=x3+Ax +B, then
xand yare both integers, and y2=x3+Ax +B.
In our case, A=−7and B= 6. The rational points on Eare the points at infinity and the
points where y= 0 which corresponds to the point (3,0). Thus, the rational points on Eare
(3,0),(3,0),(1,2),(1,−2).
b) To show that the point P(1,2) is a point of infinite order on the curve E, we can use the slope
formula for the tangent line at P. Let P1= (1,2) and P2=P1∗2be the sum of P1+P1, then the
slope λof the line through P1and P2is given by:
λ=3x2
1+A
2y1
=3+(−7)
4=−1.
Now, in order for P(1,2) to be a point of infinite order, the line through P1and P2should not
be parallel to the y-axis. Since the slope λ=−1is not equal to infinity (as required for parallel to
y-axis), P(1,2) is a point of infinite order on the curve E.
2 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve given by the equation y2=x3+ 5x+ 7 over the field of
rational numbers Q.
a) Find the point of order 2 on this elliptic curve.
b) Determine the sum of all rational points on this elliptic curve.
Solution 1.
a) To find the point of order 2, we look for a point P= (x, y)such that 2P is the point at infinity
in the group law. The point at infinity is the identity element of the group. Let’s denote the point at
infinity as O. We have:
2P=O
This means that Pis its own inverse. To compute this, we need to utilize the group law on
elliptic curves. For a point P= (x, y), the tangent at this point intersects the curve at some other
point Q= (x, −y). Then we draw a line passing through Pand Q, and where the line intersects
the curve for the third time will be 2P.
Given the elliptic curve equation y2=x3+ 5x+ 7, we can compute the tangent line at point
P= (x, y), find the other intersection point Q, and then compute 2Pas the third intersection with
the curve.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
b) To determine the sum of all rational points on this elliptic curve, we need to find all integer
solutions (x, y)that satisfy the equation y2=x3+ 5x+ 7. The rational points are the points on the
curve with both xand ybeing rational numbers. By systematically checking integer points on the
curve and verifying which satisfy the equation, we can determine all rational points on the curve.
This approach involves a thorough search for integer solutions to the given equation, followed
by verifying which of these also have rational y-coordinates.
3 THE INTERSECTION OF ELLIPTIC CURVES AND NUMBER THEORY
Problem 3. Consider the elliptic curve E:y2=x3+ 4x+ 1 over the integers Z.
a) Find all integer solutions (x, y)such that y2=x3+ 4x+ 1.
b) Show that the group of rational points E(Q)on Eis infinite.
c) Determine the torsion subgroup of E(Q).
Solution 3.
a) To find the integer solutions to y2=x3+4x+1, we can rewrite the equation as y2−1 = x3+4x.
Notice that the left-hand side can be factored as (y+ 1)(y−1), so we have:
(y+ 1)(y−1) = x(x2+ 4)
Since both y+ 1 and y−1have the same parity, for the product to be even, both terms on the
left must be even. This implies yis odd, so let y= 2m+ 1 for some integer m. Substituting this
back into the equation and rearranging gives:
4m(m+ 1) = x(x2+ 4)
It is now clear that xmust also be even, so let x= 2nfor some integer n. Substituting this back
in gives:
4m(m+ 1) = 8n(4n2+ 1)
Simplifying and dividing by 4 gives:
m(m+ 1) = 2n(4n2+ 1)
One possible solution is m= 2 and n= 2, so y= 5 and x= 4. Thus, one integer solution to
the equation is (x, y) = (4,5).
b) To show that the group of rational points E(Q)is infinite, we can use the Nagell-Lutz theorem
which states that for an elliptic curve in the form y2=x3+ax +b, if there is a rational point (x, y)
with both coordinates in Z, then the point is part of a finite cyclic subgroup.
Since we found one integer solution (4,5) in part a), the Nagell-Lutz theorem guarantees that
the group of rational points E(Q)is infinite.
c) The torsion subgroup of E(Q)consists of all points on the curve Ewith finite order. To
determine this subgroup, we can analyze the curve locally at the point at infinity.
Since our curve Ehas the form y2=x3+ 4x+ 1, we consider the projective closure ¯
E:Y2Z=
X3+ 4XZ2+Z3. At the point at infinity, the curve can be written as Y2=X3, which is isomorphic
to the simple group Z/2Z.
Therefore, the torsion subgroup of E(Q)consists only of the point at infinity, and is isomorphic
to Z/2Z.
4 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 4. Consider the elliptic curve given by the Weierstrass form E:y2=x3+ 5x+ 4.
Determine the rational points on the curve E.
Solution 4. In order to find the rational points on the curve E, we will use the group law on
elliptic curves.
a) Begin by finding the discriminant of the elliptic curve E: The discriminant of an elliptic curve
in Weierstrass form y2=x3+ax +bis given by ∆ = −16(4a3+ 27b2).
For E:y2=x3+ 5x+ 4, we have a= 5 and b= 4. So, the discriminant ∆ = −16(4(5)3+
27(4)2) = −16(500 + 432) = −16(932) = −14912.
b) Next, we find the rational points on the curve Eby adding the point at infinity Oto itself
repeatedly until we cover all the rational points. The formulas for adding points on an elliptic curve
are given by: - If P= (x1, y1)and Q= (x2, y2)are two points on the curve, then their sum
P+Q= (x3, y3)is given by:
x3=m2−x1−x2
y3=m(x1−x3)−y1
where
m=y2−y1
x2−x1
c) Using these formulas, perform the point addition calculations to find the rational points on
the curve E. The point at infinity denoted by Oacts as the identity element in the group law for
elliptic curves.
After performing the necessary calculations, the rational points on the curve Eare: O= (0 : 1 :
0),(1 : 2 : 1),(−4 : 0 : 1), and (−2 : 3 : 1). Thus, the rational points on the curve Eare the points
at infinity, (1,2),(−4,0), and (−2,3).
5 "ELLIPTIC CURVE POINT COUNTING IN CRYPTOGRAPHY"
Problem 5. Consider the elliptic curve E:y2=x3+ 5x+ 7 defined over the finite field F11.
a) Determine the number of F11-rational points on the curve E.
b) Find the order of the group of points on E(F11).
c) Determine the number of points on the elliptic curve Eover the finite field extension F112.
Solution 5.
a) To find the number of F11-rational points, we need to count the solutions to the equation
y2≡x3+ 5x+ 7 (mod 11).
•Let’s substitute x= 0,1,...,10 into the equation:
x y2
0 7
1 9
2 6
3 3
4 9
5 6
6 3
7 2
8 6
9 9
10 10
•We see that for x= 7 and x= 10, there are no corresponding yvalues. For all other values of
x, there are either 0, 1, or 2 solutions for y. Therefore, the elliptic curve Ehas 7 F11-rational
points.
b) The number of points on an elliptic curve over a finite field Fqis equal to the order of the
group of points modulo q. In this case, the number of points is 7.
c) To determine the number of points on the curve over F112, we recall that the number of points
over a finite field extension Fqkis given by qk+ 1 −t, where tis the trace of Frobenius. For our
curve Eover F112, we have q= 11 and k= 2.
The trace of Frobenius is given by t=p+ 1 −#E(Fp)for a prime field Fp. In this case, p= 11
and we already found that #E(F11)=7. Therefore, the trace of Frobenius is t= 11 + 1 −7=5.
So, the number of points on the elliptic curve Eover F112is 112+ 1 −5 = 121 + 1 −5 = 117.
I. Problem:
Let E:y2=x3+ 5x+ 7 be an elliptic curve defined over the rational numbers Q.
a) Find the generator point P= (xp, yp)with integer coordinates on E.
b) Compute the order of the generator point P.
c) Determine the group structure of the points on E(Q)generated by P.
II. Solution:
a) We need to find a point P= (xp, yp)on Esuch that both xpand ypare integers. Set x= 0
to solve for y:
y2= 03+ 5(0) + 7
y2= 7
y=±√7
Since √7is not an integer, we consider the point P= (0,√7). It has integer coordinates on E.
b) To determine the order of P, we compute nP for increasing values of nuntil nP =O, the
point at infinity. Let’s calculate 2P:
x(2P) = s2−2x
2, s x−s2−2x
2−y
=(−5)2−2(0)
2,(−5) 0−(−5)2−2(0)
2−√7
= (12,−52)
Thus, 2P= (12,−52). Calculating further, we find that 3P= (37/9,−124/27) and 4P= (−71/25,158/125).
Finally, 5P= (253/144,−21019/1728) and 6P=O. So, |P|= 6.
c) The group structure of the points on E(Q)generated by Pis isomorphic to the cyclic group
Z/6Z.
6 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Let C1and C2be two circles in the plane, given by the equations x2+y2−6x−
4y+ 9 = 0 and x2+y2−8x−6y+ 16 = 0, respectively. Determine the number of intersection
points of C1and C2.
Solution 1. To find the intersection points, we can set the two equations equal to each other:
x2+y2−6x−4y+ 9 = x2+y2−8x−6y+ 16
−6x−4y+ 9 = −8x−6y+ 16
2x+ 2y= 7
x+y=7
2
Substitute y=7
2−xinto one of the equations to solve for x:
x2+7
2−x2
−6x−47
2−x+ 9 = 0
x2+49
4−7x+x2−6−4x+ 9 = 0
2x2−11x+5
4= 0
(8x−1)(x−5
4) = 0
Thus, we have x=1
8or x=5
4. Substituting these back to find yvalues gives solutions (1
8,3
8)
and (5
4,3
4). Therefore, there are 2intersection points.
Problem 2. Consider the system of equations:
2x2+ 5y2= 39
3x−2y= 1
Find all integer solutions (x, y)to the system.
Solution 2. Rearranging the second equation:
3x= 1 + 2y
x=1+2y
3
Substitute xinto the first equation:
21+2y
32
+ 5y2= 39
4(1 + 2y)2
9+ 5y2= 39
4(1 + 4y+ 4y2)
9+ 5y2= 39
4 + 16y+ 16y2+ 45y2
9= 39
4 + 16y+ 16y2+ 45y2= 351
61y2+ 16y−347 = 0
Solving this quadratic equation gives irrational roots. Thus, there are no integer solutions to the
system.
7 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 8. Consider the elliptic curve E:y2=x3−2x+ 2 over the rational numbers Q.
a) Find the rational points on the elliptic curve E.
b) Determine the rank of the elliptic curve Eover Q.
c) Find the torsion subgroup of E(Q).
Solution 8.
a) To find the rational points on the elliptic curve E, we can start by substituting in y=mx +b
into y2=x3−2x+ 2 and then solving for xin terms of mand b. This will give us a parametrization
of rational points on the curve. Let’s denote y=mx +band substitute into the curve equation:
(mx +b)2=x3−2x+ 2
m2x2+ 2mbx +b2=x3−2x+ 2
(m2−3)x2+ (2mb + 2)x+b2+ 2 = 0
For this equation to be satisfied for all x, the coefficients of x2and xmust both be zero. Solving
these equations gives m2= 3 and b=±√3.
Thus, the rational points on the elliptic curve Eare the points where y=±√3±√3xand x
satisfies the curve equation.
b) To determine the rank of the elliptic curve E, one can use the 2-descent method, which
involves computing the group structure of the curve and finding generators of the group. Alterna-
tively, one can use the Lutz-Nagell theorem for rank calculations. For this specific curve, the rank
can be computationally intensive to determine.
c) The torsion subgroup of E(Q)consists of points of finite order on the curve E. To find the
torsion points, one can calculate the divisors of the curve equation and find the integer solutions
for the points. In this case, it involves solving the equation y2=x3−2x+ 2 over Qfor points where
2P=Ofor P∈E(Q).
8 "TORSION POINTS ON ELLIPTIC CURVES OVER NUMBER FIELDS"
Problem 9. Consider an elliptic curve E:y2=x3+ 7x+ 10 defined over the number field
Q(√2). Determine the torsion subgroup of E(Q(√2)), i.e., find all points P∈E(Q(√2)) such that
nP =Ofor some positive integer n.
Solution 9.
a) To find the torsion points on the curve E, we use the elliptic curve’s group law. The elements
of the torsion subgroup are those points P= (x, y)satisfying [n]P=Ofor some n > 0. Since Eis
defined over Q(√2), torsion points must have coordinates that are algebraic numbers in this field.
First, we compute the discriminant of the cubic polynomial in the Weierstrass form of Eto
determine the possible torsion values. The discriminant is ∆ = −16(4 ·73+ 27 ·102) = −25·52·72.
The possible torsion values are divisors of ∆, i.e., ±1,±2,±5,±7,±10,±14,±35.
b) Next, we calculate the points of finite order. Denote the point at infinity as O. For each
potential torsion value m, we solve the equation [m]P=O.
For m= 1, or equivalently, P= [1]P=O, we have the point at infinity as the only possible
torsion point.
c) Therefore, the torsion subgroup of E(Q(√2)) consists of only the point at infinity, denoted by
O.
9 ANOMALOUS ELLIPTIC CURVES AND RATIONAL POINTS
Problem 10. Consider the elliptic curve E:y2=x3+ 5x−4over the field of rational numbers
Q.
a) Find the discriminant of the elliptic curve.
b) Determine whether the point P= (2,3) lies on the elliptic curve E.
c) Find all rational points on the elliptic curve E.
Solution 10.
a) The discriminant of the elliptic curve E:y2=x3+5x−4is given by ∆ = −16(4·53+27·42) =
−16(500 + 432) = −16(932) = −14912.
b) To check if the point P= (2,3) lies on the curve, substitute x= 2 and y= 3 into the curve
equation:
32= 23+ 5 ·2−4
9 = 8 + 10 −4
9 = 14
Since 9= 14, the point P= (2,3) does not lie on the curve E.
c) To find all rational points on the elliptic curve E:y2=x3+5x−4, we use geometrical methods
in combination with algebra. By drawing a graph of the curve, we can see that it intersects the x-axis
at 1, -2, and -2. Therefore, the rational points on the curve are (1,0),(−2,0), and (−2,0).
Therefore, the rational points on the elliptic curve Eare (1,0),(−2,0), and (−2,0).
10 THE INTERSECTION OF ALGEBRAIC VARIETIES AND RATIONAL POINTS
Problem 11. Consider the elliptic curve E:y2=x3−2x+ 5 defined over the rational numbers.
a) Find the rational points on the curve E.
b) Show that Ehas infinitely many rational points.
c) Find the torsion subgroup of E(Q).
Solution 11.
a) To find the rational points on the curve E, we solve the equation by substitution of xvalues:
Substitute y= 0 in the curve equation: 0 = x3−2x+ 5
This simplifies to the cubic equation x3−2x+ 5 = 0. By inspection, one rational root is x= 1.
We can now perform polynomial long division:
x3−2x+ 5 = (x−1)(x2+x−5)
The quadratic factor x2+x−5=0has discriminant 12−4(−5) = 21 which is not a perfect
square. This means the quadratic has no rational roots.
Hence, the only rational point on the curve Eis (1,0).
b) To show that Ehas infinitely many rational points, we can use the fact that the curve is an
elliptic curve. By Siegel’s theorem, we know that any nonsingular cubic curve has infinitely many
rational points.
c) To find the torsion subgroup of E(Q), we can compute the torsion subgroup by solving the
Mordell-Weil equation. Since the only rational point is (1,0), the torsion subgroup consists only of
the point at infinity: E(Q)tors ={O} where Odenotes the point at infinity.
11 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−4x+ 2 over the rational numbers.
a) Find all rational points on the curve E.
b) Determine the torsion subgroup of E(Q).
c) Find the rank of the elliptic curve E.
Solution 1.
a) To find all rational points on the curve E, we first consider points at infinity. Set x= 1/t2,
which transforms the curve into y2= (1/t2)3−4(1/t2) + 2, or y2= 1 −4t2+ 2t4. This gives us the
point at infinity (0,1).
Now, to find non-trivial rational points on E, consider the line y=mx +band subtitute it into
the equation of E:
m2x2+ 2mxb +b2=x3−4x+ 2
By equating coefficients of x2,x, and constant terms, we get a system of equations in mand
b. Solving this system will give us all rational points on the curve. This process yields the rational
points (1,−1),(1,1),(−1,1),(−1,−1) in addition to the point at infinity.
b) The torsion subgroup of E(Q)consists of all points of finite order. To find these points, we
use the Nagell-Lutz theorem which states that a point (x, y)with y= 0 is in the torsion subgroup
if and only if xand yare integers. For the curve E, we have already found that the only points
with integer coordinates are the points at infinity and (1,−1),(1,1),(−1,1),(−1,−1). Therefore,
the torsion subgroup of Eis isomorphic to Z/4Z.
c) To determine the rank of the elliptic curve E, we can use Mordell-Weil theorem or advanced
techniques such as the 2-descent. It turns out that the rank of Eis 0 for this curve. This can be
proven by showing that the generator of the free part of E(Q)has infinite order by computing its
height or a similar measure, which is a bit involved and goes beyond the scope of this problem.
12 "ELLIPTIC CURVES AND RATIONAL POINTS ON SURFACES"
Problem 13. Consider the elliptic curve given by the equation y2=x3−4x.
a) Find the rational points on this elliptic curve.
b) Show that the point (−2,0) is a point on this elliptic curve.
c) Determine the group of rational points on this elliptic curve.
Solution 13.
a) To find the rational points on the elliptic curve y2=x3−4x, we can substitute y= 0 and
solve for xto find the points on the x-axis.
Setting y= 0 in the equation y2=x3−4x, we get 0 = x3−4x.
Factoring out x, we get x(x2−4) = x(x+ 2)(x−2) = 0.
Thus, the rational points on the curve are (0,0),(−2,0), and (2,0).
b) To show that (−2,0) is a point on the curve, substitute x=−2into the equation y2=x3−4x.
This gives us y2= (−2)3−4(−2), which simplifies to y2=−8+8, so y2= 0.
Therefore, y= 0, and hence (−2,0) is a point on the curve.
c) To determine the group of rational points on the elliptic curve, we use the group law for elliptic
curves.
The group of rational points on this curve is infinite cyclic and is generated by the point (−2,0).
Any rational point on the curve can be obtained by performing repeated "tangent and chord"
operations starting from (−2,0).
13 "BIRATIONAL GEOMETRY AND RATIONAL POINTS ON CURVES"
Problem 14. Consider the curve defined by the equation y2=x3−3x+ 3.
a) Find the point(s) on the curve with rational coordinates.
b) Show that the curve is birationally equivalent to the line y2=x.
c) Find the rational points on the curve that map to the point at infinity under the birational
equivalence of part b.
Solution 14.
a) To find the rational points on the curve, we need to solve the equation y2=x3−3x+ 3 with
both xand ybeing rational numbers. Let’s denote x=p/q and y=r/s, where p, q, r, s are integers
and q, s are not zero.
Substitute x=p/q and y=r/s into the equation:
r
s2=p
q3−3p
q+ 3
r2
s2=p3
q3−3p
q+ 3
After clearing the denominators, we get:
r2q3s2=p3s2−3pqs2+ 3q3s2
This equation shows that rational solutions will exist when s= 1 and ris a rational root of
x3−3x+ 3 = 0.
The rational points on the curve are then the points where ycan be rational, which correspond
to the rational solutions of the equation x3−3x+ 3 = 0.
b) To show that the curve is birationally equivalent to the line y2=x, we construct a rational
parametrization.
Let x=t2+ 1 and y=t(t2+ 1). Substituting these into y2=x3−3x+ 3, we get:
t2(t2+ 1)2= (t2+ 1)3−3(t2+ 1) + 3
Simplifying, we get: 0=0, which confirms that the parametrization lies on the curve.
Thus, the curve y2=x3−3x+ 3 is birationally equivalent to the line y2=xthrough the rational
parametrization described above.
c) The point at infinity on the line y2=xis the point (0 : 1 : 0). Under the birational equivalence,
this point corresponds to the point at infinity on the curve. The point at infinity on the curve is the
limit of the rational points as xor ytends to infinity, hence it does not have rational coordinates.
I. Problem 1. Consider the elliptic curve E:y2=x3+ 7 over the rational numbers.
a) Find all rational points on the curve E.
b) Show that the point (1,2) is a rational point on E.
Solution 1.
a) To find all rational points on the curve E, we first need to find the x-coordinates of the points.
Then, we can substitute these x-coordinates back into the equation of the curve to find the corre-
sponding y-coordinates.
The curve Eis given by y2=x3+ 7. To find rational points, we let y=m
nand x=p
q, where
m, n, p, q are integers and n, q = 0. Substituting these into the equation of the curve gives:
m
n2=p
q3
+ 7 ⇒m2q2=p3n3+ 7n3q3
This equation needs to be satisfied for some integers m, n, p, q. We can see that the point at
infinity (0 : 1 : 0) is a rational point on E. Additionally, using the fact that the curve Ehas no
repeated roots, we can confirm that there are no other rational points on E.
b) We need to verify if the point (1,2) is a rational point on the curve E:y2=x3+ 7. Plugging
in these coordinates:
22= 13+ 7
4 = 8
Since 4= 8, the point (1,2) does not lie on the curve, hence, it is not a rational point on E.
14 THE SUBTOPIC: "ELLIPTIC CURVES AND RATIONAL POINTS ON CURVES"
Problem 16. Consider the elliptic curve E:y2=x3+ 2x+ 2 defined over the rational numbers
Q.
a) Find all rational points (x, y)on the curve E.
b) Show that the point (1,3) is a rational point on E.
c) Find the group structure of the rational points on E.
Solution 16.
a) To find all rational points on the curve E, we substitute y=mx +ninto the equation of the
curve and solve for x:
(mx +n)2=x3+ 2x+ 2
m2x2+ 2mnx +n2=x3+ 2x+ 2
Now, equating coefficients of x2,x, and the constant term, we get the following system of equations:
m2= 1,2mn = 0, n2= 2
From m2= 1, we have m=±1. Since 2mn = 0, we have n= 0 when m= 1 and vice versa. Thus,
the rational points on Eare (1,±1),(−1,±1).
b) The point (1,3) is a rational point on Ebecause it satisfies the equation of the curve:
32= 13+ 2 ·1+2
9=3
c) The group structure of the rational points on Eis isomorphic to Z/2Z, with the point at infinity
as the identity element. The rational points on Eare {(1,1),(1,−1),(−1,1),(−1,−1), O}where O
is the point at infinity.
15 SESHADRI CONSTANTS AND HEIGHTS IN ARITHMETIC GEOMETRY
Problem 17. Consider the elliptic curve E:y2=x3−5x+ 4 defined over the rational numbers.
a) Determine the Seshadri constant at the point P= (1,2) on the curve.
b) Compute the canonical height of the point Q= (3,−4) on the curve.
c) Find all rational points on the curve E.
Solution 17.
a) To find the Seshadri constant at the point P= (1,2), we recall that the Seshadri constant
sE(P)at a point Pon an elliptic curve Eis defined as
sE(P) = lim
n→∞
h(np)
n2
where his the canonical height and np denotes the point Padded to itself ntimes.
In our case, the point P= (1,2), and we want to find sE(P). Let’s start by computing some
multiples of Pand calculating the canonical heights:
2P= (1,−2) ⊕(1,2) = (0,1)
3P= (0,1) ⊕(1,2) = (1,−2)
4P= (1,−2) ⊕(1,2) = (0,1)
Now, we calculate the canonical heights:
h(2P) = log+(max(|x|,|y|)) = log+(1) = 0
h(3P) = log+(1) = 0
h(4P) = log+(1) = 0
Therefore, the Seshadri constant at P= (1,2) is given by:
sE(P) = lim
n→∞
h(np)
n2= lim
n→∞
0
n2= 0
b) To compute the canonical height of the point Q= (3,−4), we use the formula for the canonical
height:
h(Q) = log+(max(|3|,| − 4|)) = log+(4) = log(4)
Therefore, the canonical height of Q= (3,−4) is log(4).
c) To find all rational points on the curve E, we can list the finite rational points using the group
law operation on elliptic curves. In this case, aside from the point at infinity O, there are no other
rational solutions to the equation.
E(Q) = {(1,2),(1,−2),(0,1),(0,−1),O}
16 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 19. Consider the elliptic curve E:y2=x3−4x+ 1 defined over the rational numbers
Q.
a) Find the set of rational points on the elliptic curve E.
b) Determine if the point P= (−1,0) lies on the elliptic curve E.
c) Verify if the elliptic curve Esatisfies the Hasse–Minkowski theorem.
Solution 19.
a) To find the rational points on the elliptic curve E, we need to solve the equation y2=x3−4x+1
with x, y ∈Q.
The rational points on Eare: P1= (1,0) and P2= (1,−2).
b) For the point P= (−1,0) to lie on the elliptic curve E, the coordinates xand yshould satisfy
the equation y2=x3−4x+1. Plugging in the coordinates of P:02= (−1)3−4(−1)+1 0 = −1+4+1
0=4.
Since 0= 4, the point P= (−1,0) does not lie on the elliptic curve E.
c) The Hasse–Minkowski theorem states that an equation has a rational solution if and only if it
has a solution over the real numbers Rand at every completion of the rationals Qpfor each prime
p.
For the elliptic curve E, we need to check if the equation y2=x3−4x+ 1 has real solutions
and solutions modulo pfor each prime p. Since Ehas rational points, real solutions, and solutions
modulo pfor all primes, the Hasse–Minkowski theorem is satisfied for the elliptic curve E:y2=
x3−4x+ 1.
17 "ELLIPTIC CURVES AND THE BIRCH AND SWINNERTON-DYER CONJECTURE"
Problem 19. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers Q.
a) Find all rational points on the curve E.
b) Calculate the rank of E(Q), the group of rational points on E.
c) Determine the order of the Tate-Shafarevich group of E.
Solution 19.
a) To find all rational points on the curve E:y2=x3−5x+ 4, we can use the method of
intersecting the curve with a line. We start with the general rational point (x, y) = p
q,r
son the
curve where p, q, r, s are integers with q= 0 and s= 0. Substituting these into the equation of E
we have:
r
s2=p
q3
−5p
q+ 4
r2q3=p3s2−5pqs2+ 4q3
This gives us a parametric representation of the rational points on Eas:
x=5s2+ 4q2
s(5s−4q), y =2r
(5s−4q)
Therefore, all rational solutions to the curve Eare given by the parametric equations above.
b) To calculate the rank of E(Q), we need to find a basis for the group of rational points on E.
By computing the Mordell-Weil group structure, we find that E(Q)has rank 1, meaning it has a
generator with infinite order.
c) The Tate-Shafarevich group is denoted by Sha(E/Q). It is equal to 0in this case since the
rank of E(Q)is 1.
Thus, the solutions to the given problem are as follows:
a) All rational points on the curve Eare parametrically represented by x=5s2+4q2
s(5s−4q)and y=
2r
(5s−4q).
b) The rank of E(Q)is 1.
c) The Tate-Shafarevich group of Ehas order 0.
18 ARITHMETIC GEOMETRY AND DIOPHANTINE EQUATIONS
Problem 1. Consider the elliptic curve E:y2=x3−5x+ 4 over the rational numbers.
a) Find the points on E(Q).
b) Show that the group of Q-rational points E(Q)is isomorphic to Z/3Z.
Solution 1.
a) To find the rational points on the elliptic curve E, we can use the Chabauty-Coleman method.
First, let’s compute the Mordell-Weil rank of the curve:
The curve Ehas discriminant ∆ = −3889 = 0, so has rank r≤1. To find the generator(s) of
E(Q), we will use a computer algebra system (e.g., SageMath).
Let’s denote the generator(s) as P= (xP, yP). We find that P= (2,0) and the negative of this
point −P= (2,0) are the only two rational points on the curve. Therefore, E(Q) = {O, P, −P}.
b) We know that the group of Q-rational points E(Q)is isomorphic to Z/rZ. Since we have found
two distinct rational points Pand −P, the rank of Eis r= 1, and therefore E(Q)is isomorphic to
Z/3Z.
