Master's Thesis: On Recognizing Congruent Primes

Master's Thesis: On Recognizing Congruent Primes
On Recognizing Congruent Primes
Brett Hemenway
B.Sc., Brown University 2004
A THESIS SUBMITTED I N PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
MASTEROF SCIENCE
I N THE DEPARTMENT
OF
MATHEMATICS
@ Brett Hemenway 2006
SIMON FRASER UNIVERSITY
Fall 2006
All rights reserved. This work may not be
reproduced in whole or in part, by photocopy
or other means, without the permission of the author
APPROVAL
Name:
Brett Hemenway
Degree:
Master of Science
Title of thesis:
On Recognizing Congruent Primes
Examining Committee: Dr. Jason Bell
Chair
Dr. Nils Bruin
Senior Supervisor
Dr. Peter Borwein
Supervisor
Dr. Stephen Choi
Supervisor
Dr. Imin Chen
Internal/External Examiner
Date Approved:
September 12th, 2006
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Abstract
An integer n is called "congruent" if it corresponds to the area of a right triangle with
three rational sides. The problem of classifying congruent numbers has an extensive
history, and is as yet unresolved. The most promising approach to this problem utilizes elliptic curves. In this thesis we explicitly lay out the correspondence between
the congruence of a number n and the rank of the elliptic curve y2 = x3 - n 2 x . By
performing two-descents on this curve and isogenous curves for n = p a prime, we
-
are able to obtain a simple and unified proof of the majority of the known results
concerning the congruence of primes. Finally, by calculating the equations for homogeneous spaces associated to the curve when p
for future analysis.
1 mod 8, we position the problem
Acknowledgments
I would like to thank my entire thesis committee. In particular, thanks to Peter
Borwein for welcon~ingme into t,he SFU mathematical community, for making my
time in Canada possible, and for his continued guidance throughout my time at SFU.
Thanks also to Nils Bruin for introducing me to the study of elliptic curves, for
helping me navigate this project, and for reading many many draftasof this thesis. His
constant encouragement, feedback and support has made him a wonderful advisor.
Dedication
To Emily.
Contents
..
Approval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
111
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vi
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Congruent Numbers . . . . . . . . . . . . . . . . . . . . . . .
Known Results . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1.1
1.2
Genocchi's 1855 Argument . . . . . . . . . . . . . . . . . . .
Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3
2
2.1
2.2
2.3
2.4
2.5
2.6
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Group Law . . . . . . . . . . . . . . . . . . . . . . . . . .
The Structure of the Group . . . . . . . . . . . . . . . . . . .
Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
...
2
4
9
9
9
10
11
14
14
15
2.8
Torsion in the Congruent Number Curve . . . . . . . . . . . .
Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.9
2-Isogenies . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
The Curve y2 = x3 - n2x . . . . . . . . . . . . . . . . . . . . .
TwoDescent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.1
Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.2
The map p
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
2.7
2.10
3
11
18
24
The Two-descent . . . . . . . . . . . . . . . . . . . . . . . . .
Twedescent applied t o congruent number curves . . . . . . . . . . .
4.1
General Bounds . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3
4
4.2
4.3
5
Specific Bounds . . . . . . . . . . . . . . . . . . . . . . . . . .
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
=1
30
32
33
35
mod 8 . . . . . . . . . . . . . . . . . . . . . .
36
36
5.2
A Specific Two-Isogeny . . . . . . . . . . . . . . . . . . . . . .
Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3
5.4
General Bounds . . . . . . . . . . . . . . . . . . . . . . . . . .
p-adics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
38
39
5.5
The Quadratic Character of 1 + i . . . . . . . . . . .
2-adics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Case When p
5.1
5.4.1
6
28
Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.1
Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
6.3
Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Follow up . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Bibliography
vii
37
40
43
43
45
48
Chapter 1
Introduction
1.1 Congruent Numbers
A positive rational number n E Q is said to be a Congruent Number if it is the area
of a right triangle with rational sides, i.e. if there are rational numbers a, b, c such
that
If n is a congruent number, then nr2 is also congruent for any r E Q , since nr2 is
the area of the triangle with sides ra, rb, re. Thus whether a number is congruent
is determined solely by its residue class in the group Q*/(Q*)2.With this in mind,
when searching for congruent numbers we restrict our attention to square free integers. The question of determining which numbers are congruent is known as the
Congruent Number Problem, and while the statement is completely elementary, this
problem has been investigated by mathematicians for over one thousand years (see
[lo]) without leading to a complete solution. By performing elementary transforma-
tions on the equations (1.1),the problem can be formulated in many different ways.
The earliest known reference the Congruent Number Problem is from A.D. 972 (see
[lo]), and states the problem in another form. A rational number n is said to be
congruent if there exists a rational number x such that x2
+ n and x2 - n are both
CHAPTER 1. INTRODUCTION
2
squares. This is also the formulation that Leonardo Pisano (Fibonacci) used to show
that 5 is a congruent number in 1220 [lo]. This formulation gives some insight into
+
the name, since the three squares, x2 - n, x 2 ,x2 n are congruent modulo n. The two
definitions of a Congruent Number are easily seen to be equivalent by the maps
For most of this thesis we will be using a third characterization of congruent
numbers, one relating t o rational solutions to a cubic equation.
A square-free natural number n is a congruent number if we can simultaneously
solve two equations over the rationals
which means
( a i b)2
4
Then setting x =
=
(f) 2 i n .
(i)2 and y = (a2- b 2 ) i gives a solution t o the cubic equation
We will c,arefully examine the curve give by equation 1.2 in the case when n is
prime to determine whether n is congruent.
1.2
Known Results
We now give a brief review of what is currently known about the Congruent Number
Problem. Assuming the Birch and Swinnerton-Dyer Conjecture [4], the problem was
essentially solved by Jerold B. Tunnel1 in 1983.
C H A P T E R 1. INTRODUCTION
Tunnell's Theorem. Define
Suppose n is congruent, if n is even then A, = Bn and if n is odd, then 2Cn = D,. If
the Birch and Swinnerton-Dyer Conjecture holds for curves of the form y2 = x3 - n2x
then, conversely, these equalities imply n is a congruent number.
Proof. See [34], or [20] Chapter IV section 4.
0
Currently this provides a method for showing certain numbers are not congruent,
and assuming the Birch and Swinnerton-Dyer Conjecture this would provide a fairly
efficient method for determining whether a given number is congruent, since counting
solutions to these equations can be done easily. We will not go into details about
the Birch and Swinnerton-Dyer conjecture, but it should be noted that it is widely
believed to be true, and is one of the Clay Mathematics Institute's Millenium Prize
Problems.
Much study has gone into Congruent Number Problem, not assuming the Birch
and Swinnerton-Dyer Conjecture, and we will list the known results here. Let p, and
q, denote distinct primes with p,
= q, = i
mod 8 Then the following results are known
p3 is not a congruent number
Genocchi, 1855 [13]
and Tunnell, 1983 [34]
p3q3,2p5,p5q5 are not congruent numbers
Genocchi [13] and Tunnel1 [34]
p5,p7 are congruent numbers
Monsky, 1990 [24]
1 2p7,2p3_p3q7,2p3q5,2p5q7 are congruent numbers
Monsky, 1990 1241
Overviews of progress made on the Congruent Number Problem can be found in
[lo], [141, PI, [81, [281 and [271.
In this thesis, we obtain upper bounds for the rank of the elliptic curve y2 =
x3-n2x using the method of 2-descent. In this way, we show that pa is not a congruent,
C H A P T E R 1. INTRODUCTION
4
number, p5:p7 are congruent numbers assuming the conjecture that UT(E/K)[2] is
+
+
finite, and for pl, if we decompose pl = a2 b2, then pl is not congruent if (a b)2 $ 1
mod 16. This last result was stated by Bastien in 1915 [2], but a proof does not seem
to have appeared in the literature until Tunnell's proof in [34]. Tunnell's proof comes
-
as a consequence of Tunnell's Theorem, and hence his method is very different from
the one presented here.
The case of when p
3 mod 8 was the first to be resolved. In 1855 Angelo
Genocchi showed that p3 is not a congruent number. His paper predates much of the
general machinery of elliptic curves, and his argument is fairly elementary. Nevertheless, Genocchi's method bears many similarities to the method of descent used in this
thesis. To show that p3 is not congruent, Genocchi shows that if p3 were congruent,
this would lead to an integral point on a quartic, then he shows that because
2 are not squares in Z/p&
- 1 and
these quartics have no rational points. To illustrate his
technique, we give an overview of his original argument in 51.3.
1.3
Genocchi's 1855 Argument
We now make a brief digression to give Genocchi's argument [13] that if p is a prime
with p EE 3 mod 8 then p is not a congruent number. Genocchi's paper is often cited,
as it is one of the earliest demonstrations that an entire class of numbers is not congruent. Unfortunately, his paper has become very difficult to obtain. For its historical
significance, as well as its ingenuity, we give a detailed account of his argument. The
terminology has been updated, but the content of the proof remains the same.
Fibonacci gave t,he following characterization of congruent numbers. A number
is congruent if it is one of the four numbers a , b, a
+ b, a - b and the remaining three
numbers are square. This is equivalent to the statement that n is congruent if there
is a rational number x such that x2 f n are both square. Genocchi begins with
Fibonacci's characterization, and considers four cases separately.
Throughout the proof, we will make use of the fact that all Pythagorean Triples
CHAPTER 1. INTRODUCTION
can be paramatrized as
( r 2- s2,~
T Sr2
,
+ s2).
Suppose n is a congruent number then one of the following four cases holds.
Case 1:
~ = n f ~ , 2 ,ba +
= b~= h 2 , a - b = k
2
Thus we have
a2 - b2 = h2k2,a2 = b2 + ( f ~ k ) ~
So b, h k , a is a Pythagorean Triple. Then parametrizing we have a = r2
+ s2
and b = r2 - s2 or b = 2 ~ s This
.
gives
In the first case, since r , s are relatively prime, we must have r
r
- s = P2.
+s
= a2,
This gives
In the second case, we get
Case 2:
a = f 2 , b = g 2 , a + b = h2 , a - b = n k .2
Thus we have
f 2 +g2
=
h2.
Thus ( f , g, h ) is a pythagorean triple, so ( f , g) = ( r 2 - s2,2rs) or ( f , g)
(2rs,r2 - s 2 ) . From the equation f 2
Expanding gives
- g2 = n k 2 , we
then obtain
=
C H A P T E R 1. INTRODUCTION
Case 3:
a = f 2 , b = g 2 , a + b = n h 2 , a - b = k .2
Thus we have
k2 + g2 = f2.
So parametrizing this pythagorean triple, we have f = r2
+ s2 and g = 2rs or
g = r2 - s2, plugging into the equation
gives
+
+
( r 2 s2)2 ( 2 ~ s=) nh2
~
or
Thus
Case 4:
a = f 2,b=ng 2: a + b = h 2 , a - b = k 2 .
Here, we are forced to use a different method since
and
f 2 - ng2
= k2
tell us only that n is a congruent number. Going back t o the definition of a
congruent number, if n is congruent, we can find x l , x2, X Q E Q such that,
CHAPTER 1. INTRODUCTION
7
Clearing denominators we can find q such that xi =
where pi, q E Z. Here,
we assume that q is the smallest integer such that pl,p2,p3 E Z. Thus we have
the equations
-
nq2 =p:,pi
+ nq
2
= p 32 .
Since
2
p2
3 - pl
we conclude that pl
= p3
= 2nq
2
,
mod 2, thus we can define integers
This gives
pi = rf
+ r,2,
which is a Pythagorean Triple, so we can parametrize this as
Now, we also have ng2 = 2rlr3, so substituting our parametrization for r l , r3 we
get
ng2 = 4ab(a - b) (a
+ b).
Since n is congruent, we also know that we can find integers a , b such that three
of the four integers a, b, a + b, a - b are square and n divides fourth. Assuming
we are not in one of the three previous cases, if n is prime we must have nlb, so
Then, these equations give
So ng2 < ng2, but this contradicts the minimality of q. Thus we conclude that
this case is impossible for n prime.
CHAPTER 1. INTRODUCTION
8
It remains to show that the first three cases are impossible. Genocchi does this
by examining congruence conditions mod p. Gathering equations (1.3) - (1.7), and
unifying notation, we have
If p
=3
mod 8 and pin, then taking remainders modulo p shows that Equations
(1.8),(1.9), (1.12) are impossible since - 1 is not a square mod p. To deal with equation
(1.l o ) , we notice that
and 2 is not a square mod p. Similarly for equation (1.11),we observe
and it suffices to notice that either -1 or 2 is not a square mod p.
Thus if p
=3
mod 8, then p is not a congruent number.
Chapter 2
Elliptic Curves
2.1
Introduction
We will be looking for rational solutions (x,y ) t o the equation
E
:
y2 = x3 - n2x.
This is an example of an elliptic curve. The theory of elliptic curves is well-developed,
and before we begin analyzing the curve
E,we review some of the general properties
of elliptic curves that we will use.
2.2
Definition
Let P2 denote the projective plane. An elliptic curve is the locus of points in P2
E : y 2 Z + a l X Y Z + a 3 Y Z 2= x3+ a 2 x 2 2+ a 4 x Z 2+ a 6 z 3 .
It is usually convenient t o de-homogenize, i.e. change variables and let x = X / Z and
y = Y / Z . Then we have the equation
When the characteristic of the base field is not equal t o 2 or 3, then we can make a
change of variables resulting in the simpler form
CHAPTER 2. ELLIPTIC CURVES
10
We denote by 0 the point [0 : 1 : 0] on E. Since this is the only point with Z = 0
on E , for convenience, we will often denote a point P on E with P
#
0 simply as
P = (x, y), where this is shorthand for the point P = (x : y : 1).
Recall that a function f (x) = x3+ax+b has a double root if and only if 4a3+27b2 =
0. A curve given by the equation y2 = f (x) is called singular if f (x) has a double
root.
2.3
Finite Fields
An elliptic curve E : y2
=
x3
+ ax + b, is said to be defined over Q if a , b E Q.
A
point on E is called rational if its coordinates are rational numbers. When studying
the rational points, it can be useful to examine the points in IF, for q = p f , where IF,
denotes the finite field with q elements. For the curve to remain nonsingular in IF,,
we need -16(4a3
+ 27b"
#
0 E IF, which means p f 2 and p f 4a3
+ 27b2. Assuming
this is the case, we denote the natural map
for any rational point ( X : Y : 2) on E, we can choose X: IT
Z such
, that X , Y, Z E Z
and gcd(X, Y, Z ) = 1. Then we have a map
2) =
,1, we cannot have )? = Y = 2 = 0, thus P is a point in P2(IF,).
Since gcd(X, IT
We can sometimes (as in the proof of Theorem 1) use the finiteness of P"IF,) to great
advantage.
CHAPTER 2. ELLIPTIC CURVES
The Group Law
The set of rational points on an elliptic curve can be made into an abelian group,
with 0 acting as the identity, and it is towards this group that we direct our further
attentions.
For a more in depth discussion of the group law, see [29] 111.2, [5] Chapter 7, or
[18] Chapter 3.
We describe the group law geometrically. First, we need a lemma.
Lemma 1. If P = (XI : Yl : Z1) and Q = (X2 : Y2 : Z2) are two rational points on
the curve E , then the line L through P and Q intersects E in a third rational point,
R.
Proof. This is a direct consequence of Bkzout's Theorem ([I51 Theorem 18.3 or [16]
Corollary 1.7.8).
0
This means that the line at infinity intersects E with multiplicity 3 at 0. The line
through the point 0 and R intersects the curve E in a third point by lemma 1, and
we call this third point of intersection P
+ Q (see figure 2.1). This operation makes
the rational points on E into an abelian group.
It is straightforward to check that)
Associativity can be checked directly, but the calculations are long. A verification
using MAGMA can be found at [30]. We give a short argument for the associativity
given in [5] Chapter 7. First, we note that three points P, Q, R are collinear if and
only if there exists a linear form L1 such that L1 has zeros at P, Q, R. Thus if 0, R, S
are collinear, then there is a linear form La with zeros at 0, R, S.
CHAPTER 2. ELLIPTIC CURVES
Figure 2.1: The addition law on the curve y2 = 2 -412x.
Suppose the points P, Q, R are collinear and 0, R, S are collinear, i.e. P + Q = S.
Since P, Q, R are collinear there exists a linear form L1, such that L1 has zeros at
P, Q, R. Similarly there exists a linear form L2 with zeros at 0, R, S.
Thus the rational function
is defined on the curve E and has zeros at P, Q and
poles at 0, S. In fact, the converse holds as well: if there exists a rational function
f with zeros at P, Q and poles at 0, S, then P + Q = S. With this characterization
2
of the addition law, it becomes straightforward to see that associativity holds. The
equation
X = (P+Q)+T,
is thus equivalent to the existence of a function with simple poles at P, Q, T a double
CHAPTER 2. ELLIPTIC CURVES
13
zero at 0 , and a simple zero at X. But this is exactly the same as the function
corresponding to the equation
so we conclude that ( P
+ Q) + T = P + ( Q + T). This characterization of will also
be useful in our study of isogenies.
The group law can be given explicitly as functions on the coordinates of the
points. As we will not have occasion to use the group la,w in its full generality, we will
calculate only a few special cases here. We will be interested in curves of the form
Let P = (xo,yo) be a point on E. This curve E is now symmetric about the x-axis.
The line through P and 0 is vertical, so it intersects the curve again a t the point
(50, -yo). Thus ( s o ,-yo)
+ (xo,yo) = 0 , because the line through 0 and 0 intersects
at 0 with multiplicity 3. Thus we have
-
As is the case with any abelian group, we can consider the group E as a Zmodule
under the action
[ n ] P =P + . . . + P
n times
Let us calculate 2P for P = (so,yo) on the curve E :
tangent at the point (xo,yo) has slope
So the equation of the tangent line becomes
This intersects the curve at the point (xl, yl) where
y2
= x"
ax
+ b.
The
CHAPTER 2. ELLIPTIC CURVES
2.5
The Structure of the Group
The group of rational points on an elliptic curve is clearly abelian, but we can say
much more.
Mordell-Weil Theorem. The group of rational points on an elliptic curve is finitely
generated.
Proof. See [29]Theorem 4.1, [5]Theorem 13.1 and [18]Theorem 7.4.
0
Fundamental Theorem of Finitely Generated Abelian Groups. Every finitely
generated abelian group is the direct product of a finite torsion group and a number of
copies of infinite cyclic groups (i.e. Z ) .
Proof. See [26]Theorem 10.20.
0
We now know that the group E ( Q ) can be decomposed as
where the integer r is known as the rank of the curve. Thus to fully describe the
group E ( Q ) ,we only need to calculate E(Q)torsand r.
2.6
Torsion
The torsion subgroup of an elliptic curve is well-understood.
Mazur's Theorem. For an elliptic curve E over Q the torsion subgroup E ( Q t o r sis
one of the following 15 groups
Z / N Z for 1 5 N 5 10 or N = 12
Z / 2 Z x Z / 2 N Z for 1
<N
54
This was originally proven by Barry Mazur in [21]and [22].It is also stated without proof as [29] Theorem VIII.7.5, [18] Chapter 1 Theorem 5.3, and [27] Theorem
2.2.
CHAPTER 2. ELLIPTIC CURVES
15
For our purposes we will only consider the two-torsion of a curve, i.e. the points of
order dividing two. Since the group law gives us - ( x , y) = ( x ,- y ) , it is easy to see
that a nonzero point is a two-torsion point if and only if its second coordinate is zero.
We will not be overly concerned with the computation of the torsion group of E ( Q ) .
It should be noted, however, that the problem of calculating torsion points over the
rationals has been solved, and a general algorithm for calculating the torsion group
can be found in [9], Section 3.3.
Torsion in the Congruent Number Curve
On the curve, E : y2 = x3 - n 2 x , we can see that there are four two-torsion points
( 0 , (0,0 ) ,(f
n , 0)). Since 0, ztn are the only roots of x3 - n2x,we conclude that these
are the only two-torsion points on E. We now show that these are the only torsion
points on E following the method in [20].
We begin by calculating the number of points on E over the finite field IFp.
Lemma 2. Let p be a prime, with p { n and p
p + 1 points on E over IFp.
=3
mod 4. Then there are exactly
Proof. The curve E always has the four points ( 0 , (0;0 ) ,( f n , 0)). Notice that these
remain distinct over IF, since p 1 n, and p odd. If p = 3, then these are the only four
points because 0, fn are the only three possible values of the x-coordinate of a point
on E. Let us now examine the points where x
# 0, fn.
There are p - 3 such values
for x. We can group them in to pairs {fx ) . Now, we have a point on our curve E
exactly when "x
not a square in IF,.
n2x is a square in IFp. Since p
=3
mod 4, we know that -1 is
Since the squares form a subgroup of index 2 in
(IF,)*,
we can
see that every pair { f x ) in (IFp)* contains exactly one square. But we also have that
x3 - n2x = - ( ( - x ) ~- n 2 ( - x ) ) ,SO for every pair {fx ) , exactly one will lead to a point
on our curve in IFp. So we have (p - 3)/2 distinct x values. Since each x value leads
to exactly two solutions f y , we have p
-3
additional points on E over Fp. Adding in
the four two-torsion points, we find there are exactly p + 1 points on E over Fp.
CHAPTER 2. ELLIPTIC CURVES
If PI, P2 are points on the curve
condition that
16
E(Q),we
now give a necessary and sufficient
PI = P2.
Lemma 3. Let PI, P2 E P2(FP), i.e.
gcd(Xi, Y,, Zi) = 1. Then
Pl
=
Pi
fi ifl p
= (Xi : Y , :
Zi) with Xi, Y,, Zi E 7,and
divides Y1Z2 - Y2Z1,X2Z1 - X1Z2 and
X1Y2 - X2Yl.
Proof. Notice that these are the con~ponentsof the cross-product of Pl and P2 considered as vectors in
R3.
If p divides the cross-product, then we consider two cases.
-
(1) If p divides X I , then p divides X2Zl and X2Y1. Since p cannot divide both
Yland Z1, we conclude that p divides X2. NOW,we also know YlZ2
mod p, so we have
Y2Zl
(2) If p does not divide X I , then
For the converse, suppose
Pl
= P2. We know that p does not divide all three of
XI,Yl, Z1, so suppose p { XI. The other two cases will proceed in exactly the same
CHAPTER 2. ELLIPTIC CURVES
way. Since
PI = P2we have p { X2, thus
Since the first components of these points are the same, we must have Y1X2 F XlY2
mod p and Z1X2 = X1Z2 mod p. So it only remains to show that Y1Z2 r Y2Z1
mod p. If p divides both Yl and Z1 this is clear, otherwise replacing X1, X2 by Yl, Y2
or Z1, Z2 in the above argument gives the result.
0
Now we are ready to characterize the torsion points of the curve E : y"
x3-n2x.
Theorem 1. I E(Q)torsl= 4.
Proof. We know there are exactly four two-torsion points on E ( Q ) , (0,(0, 0), (fn, 0)).
Suppose there is another torsion point on E(Q). Since this point is not a two-torsion
point it must have order greater than two. Thus the group E(Q),,,, has a subgroup
H of order nL where either m is odd, or m
=
8. In fact, h4azur's Theorem (2.6))
lists all possible torsion groups, but we do not need such heavy machinery here. Let
{PI,.. . , P,). We now examine for which p the reduction map P H P is injective
on H. If we consider P I , . . . , P, as vectors in iR3, since they are distinct in lP2(Q) no
H
=
two are multiples of each other, so the cross product Pi x Pj
# 6.If we let
nij denote
the greatest common divisor of the components of the vector Pi x Pj, by Lemma 3,
Pi #
P' if and only if p { nij.
map P
H
P is
So if we let N = max(nij), we have that the reduction
injective on H for all primes p > N. Thus m divides the order of
the group E(Fp) for all such p. By Lemma 2, if p F 3 mod 4, then IE(Fp)I = p
thus mlp
+ 1, or p G -1
+ 1,
mod m. Thus we have shown that for all but finitely many
primes p with p = 3 mod 4, we have p
= -1
mod m. Now recall Dirichlet's famous
theorem that for any a , m with gcd(a, m ) = 1 there are infinitely many primes p with
p r a mod m. (See [19] Chapter 16 Theorem 1). If m = 8, we have shown that there
are only finitely many primes p
=3
mod 8, which contradicts Dirichlet's Theorem.
If m is odd, then for all but finitely many primes p
=3
mod 4 we also have p
= -1
mod m which together give p $ 3 mod 4m which is a contradiction to Dirichlet's
Theorem if 3 { m. On the other hand, if 31m, then we have if p r 3 mod 4 then for
CHAPTER 2. ELLIPTIC CURVES
all but finitely many primes p
= -1
18
mod 3 k , so there are only finitely many primes
7 mod 12k which again contradicts Dirichlet's Theorem.
of p
0
Calculating the rank of an elliptic curve is significantly more difficult, and currently no general algorithm is known. The bulk of this thesis will be devoted to finding
the rank of certain "congruent number curves".
Isogenies
2.8
An isogeny, 4 : El
4(OE1)
+ E2,is
= 0E2. In fact
a morphisni between elliptic curves El and E2 such that
4
induces a group homomorphism from the group E1(K) to
E 2 ( K ) ,this is [29] Chapter 111 Theorem 4.8. We prove a special case of this theorem
in 52.9. The map
4 also induces an injection of function fields by the pull-back,
4 we define the degree of 4 to be the degree of the field K ( E 1 ) as
an extension of the field 4"K ( E 2 ) . So degree(4) = [I<(El): 4" K (E2)],where K (El)
For non-constant
is the rational function field of K over E l .
If K is a number field, then the Mordell-Weil theorem holds, and we can talk
about the rank of the curves El and E2. In this case we have the property that
isogenous curves have equal rank. A proof is sketched below.
We have already seen the multiplication-by-m map, denoted [m]. It is easy to see
that this is in fact an isogeny. This isogeny is particularly important as it exists for
all curves E and all positive integers m. For a more thorough discussion of isogenies
see [29] Chapter 111, Section 4. The degree of the multiplication-by-nz. map is m2.
4 : El + E2 is an isogeny of degree m , then there is a unique isogeny of degree
m , 4 : E2 + El such that 4 o 4 = [m]. The isogeny 4 is called the dual isogeny to 4.
If
The existence and uniqueness of the dual isogeny is proven in [29] Theorem III.G.l.
Since
4 is a homomorphism, 4 takes elements of finite order in E 2 ( K ) to elements of
finite order in E1(K). The map [m] takes elements of infinite order to elements of
CHAPTER 2. ELLIPTIC CURVES
4
19
q5 = [m], so we must have that q5 takes elements of infinite order
elements of infinite order. This implies that
infinite order and
o
Rank(E1( K ) ) 5 Rank(E2 ( K ) )
Then, applying the same argument t,o
4,we have that
Rank(E1( K ) ) = Rank(E2(K)).
The fact that isogenous curves have equal rank is useful to us, as we can bound
the rank of E1(K) by bounding the rank of E 2 ( K ) .
We now examine in more detail a type of isogeny that will be of use to us. Specifically,
we show how to create a degree two-isogeny from any two-torsion point on a curve.
We follow the method outlined in [5] Chapter 14.
While it is common, given an elliptic curve, to change variables and write the
curve in the form
y2
= x3
+ ax + b.
In this section, to simplify calculations, we write
our elliptic curve in the form E : y2
=
x(z2 + ax
+ b).
This change of variables has
the effect of putting a two-torsion point at (0,O). Consider a map
Notice that since (0,O) is a two-torsion point, we have that +(+(x, y)) = (x, y). The
map
+ : (z,y)
H
(zl,yl) induces an automorphism on the function field K ( E ) . Let
us ~alculat~e
the fixed field. The line through (0,O) and (z,y) intersects the curve at
a third point, (zl, -yl). Solving gives
CHAPTER 2. ELLIPTIC CURVES
20
Since (z: y) and ( z l , -yl) are on the same line through the origin, we must have
Y = -,
-y1 so $ is invariant under $. Using the fact that y2 = x(x2
ax b), we have
5
+ +
51
To find another fixed function, it suffices to notice that $(x, y) = ( X I , yl) and $(XI,yl) =
(x, y) we have that y
+ yl is fixed by $. Call this value p. Thus
To find a relation between X and ,u notice that
So we have the equation
We now show that K(X, p ) is the entire fixed field of $. To this end, we solve for
x , y in terms of X,p. We begin with
CHAPTER 2. ELLIPTIC CURVES
We also have
X=
x2
+ a x + b = x + - b+ a .
x
x
Combining these gives
x=
2
Thus K ( x , y) C K (A, p , a ) , so [K(x,y) : K(X, p)] 5 2. Clearly the field of invariants
contains K(X, p ) , so to show that K(X, p ) is the complete field of invariants, it remains
only to show that 11, is not the identity automorphism. This is clear though, since
$(x) = # x. Now we are in a position to define the isogeny. Notice that the map
11, has given us another curve El defined be equation (2.2), so we let
It remains to show that
4 is, in fact, an isogeny, i.e. 4 preserves the group law.
If P
and Q are points on E, then we have seen that there is a function f E Q(x, y) with
simple poles a t P, Q and simple zeros a t 0 , P
can simply multiply f by its conjugate
+ Q. To create a function in Q(X,p ) , we
11,( f ) . Thus f 11,( f ) is in Q(X, p ) , and f $( f )
+
has simple poles a t 4 ( P ) , #(Q) and simple zeros at O,4(P Q) since
4(0) = 0 .
Thus
f $(f ) corresponds to the equation
which gives us that
2.10
4 is a group homomorphism.
The Curve y 2 = x 3 - n 2x
We are now ready to begin examining t,he curve E defined by equation (1.2). If n
is a congruent number then we have shown how to construct a rational point on the
elliptic curve E , from the sides of the triangle with area n. If E has a rational point,
this is not enough to guarantee that n is a congruent number, and we will examine
the necessary and sufficient conditions below.
CHAPTER 2. ELLIPTIC CURVES
22
To see why every point on E does not correspond to a right hiangle with area
n, notice, for instance, the point,s that we can generate from a right triangle all have
a square x-coordinate. Points on E that have a non-square x-coordinate were not
generated from a right triangle. In fact, a point P on our curve comes froin a right
triangle if and only if P is twice a rational point, i.e. P = 2Q for some point Q E E.
This follows from basic calculations using the addition laws on E. We sketch the
correspondence here.
If Q = (xo,yo), then the addition law for points (equat,ion 2.1) on E gives
Using the fact, that y; = xi
-
n2xo,we have
z;
=
+ n,2 2
(yo)
Thus we find that n is a congruent number (this is just our second definition of
a congruent number).
To show the converse, we must find a point (x, y) such that
Writing out the addition law gives the two equations
and
Solving for x and y we find
CHAPTER 2. ELLIPTIC CURVES
Thus 2 ( x , y )
=
((1)2, (a2 - b 2 ) ; ) .
Clearly 2((.z,y)
+ P ) = 2 ( x ,y) for any two-
torsion point P. But we know that the two torsion of the curve E is (0,( 0 , O ) )( n ,0 ) , (-n, 0 ) )
(see Theorem (I))! so, in fact, we can find four points which are "half' of the point
coming from a right triangle. To effectively use this correspondence, we begin by analyzing which points on E can be twice another point. For our curve E , the sihation
is rather simple, and since by Theorem 1 we know Etors(Q)
= (0,(0,O),(f
n ,0 ) ) .
None of these is twice another point because there is no four-torsion. So a squarefree natural number n is a congruent number if and only if the curve E defined by
(1.2) has positive rank.
For the remainder of the paper we will concern ourselves with finding upper
bounds on the rank of E. For an overview of some of the methods and results in the
study of ranks of elliptic curves, see 1271.
Chapter 3
Two Descent
3.1
Overview
Our goal is to find an upper bound on the rank of the curve E. To that end we will find
a bound on the size of t,he group E ( Q ) / 2 E ( Q ) Since
.
we know E ( Q ) e E(Q)torsx Z T ,
we have
lE(Q)/2E(Q)l= lEtors(Q)/2Etors(Q)
I . 12'1 .
For the elliptic curve E
: y2 =
x3 - n2x,we know there are exactly four torsion
elements, and E,,,,(Q) = 2/22 x 2/22, so
In order t,o bound the size of E ( Q ) / 2 E ( Q )we
, construct a homomorpliism p from
the group E ( Q ) with ker(p) = 2E(Q). Thus we have an isomorphism
In particular ( E ( Q ) / 2 E ( Q )=J Ip(E(Q))I.It will be difficult to calculate the group
p(E ( 0 )directly,
)
so we will calculate p ( E ( Q ) )for various primes I . While Ip(E ( Q ) )
I#
Ip(E(Q))1, we can use information about p ( E ( Q l ) )in an attempt to bound the size
of P ( E ( Q ) ) .
CHAPTER 3. T W O DESCENT
3.2
Themapp
Consider the elliptic curve E : y2
Let K [Q] = K
[XI/(
f ( x ) ) .Set f
=
=
f ( x ) over a field K ; where f ( x ) = x3 + ax
n,"_,
be the factorization of f
fi
+ b.
into irreducibles
in K [ x ] . Since f has no repeated roots, f must square-free. In particular the irreducible components fi are distinct. Then by the Chinese Remainder Theorem,
n2,
K[Ql =
K[xl/(fi).
Let AK = K[B],A;( be the group of units in A K , and let A: be the multiplicative
subgroup of A> consisting of the squares of elements in A>.
Define the map p as follows:
We know that (xo - Q ) E A;( when ( x o ,yo)
6 E [ 2 ] ( K because
)
then
relatively prime t o f (x).
(xo- x ) is
-
When ( x o ,yo) E E [ 2 ] ( K we
) have f ( x o ) = 0, so f ( x ) = ( x - X O ) ~ ( X )and K[B]
K x K [ x ] / ( g ( x ) Here
) . we define
We now illustrate the important properties of p.
Lemma 4. The map p is a homomorphism.
Proof. We follow the proof in [19]Chapter 19. For an alternate proof see [5]Chapter
15 Lemma 1.
We begin by noting that if P = ( x ,y ) , then
because p(x, y ) is independent of y. We wish to show that P(P+Q) = p ( P ) p ( Q ) .Mul-
) , see that this is equivalent t o p(P+Q)p(P)p(Q)=
tiplying both sides by p ( p ) ~ ( Qwe
CHAPTER 3. T W O DESCENT
26
1. Then using the identity above, we see that it is enough t o prove p(P+Q)p(-P)p(--Q) =
1. Changing notation, we need to show that if A + B + C = 0 , then j ~ ( A ) p ( B ) p ( c =
) 1.
Recall that A
A
=
+ B + C = 0 is equivalent
( x l ,y l ) , B
=
distinct, for if p ( P
to stating that A, B, C are collinear. Let
(x2,y2) and C = (xg,y3). We can also assume that A, B , C are
+ Q ) = p ( P ) + p(Q) for distinct P,Q, then we have
Now, we divide the proof into cases
0
If xl = x2 then since A
#
B , we must have B = -A, which gives C = 0 , so we
have the equation
If xl
# x2 and none of the points have order two then, since A, B, C are collinear,
there is a line y = cx
+ d passing through A, B, C. Thus we have
Since both sides are monic polynomials of degree three with the same roots.
Reducing modulo f (x) and recalling that 8 denotes the residue of x, we have
Noticing the right hand side is just p ( A ) p ( B ) p ( C ) then gives the result.
If exactly one of the points has order two, then without loss of generality, we
may assume A = ( x l , 0). This means that xl is a root of f , and writing f (x) =
(x - xl)g(x) as above, we have K[8] = K[x]/(x - x l ) x K[x]/(g(x)), and we
check each conlponent separately. By the definition of p we have that
so the first component of p ( A ) p ( B ) p ( C ) is
CHAPTER 3. T W O DESCENT
Since y = cx
27
+ d goes through the point
(zl,O), we have czl
differentiating equation (3.1) and evaluating at x
= xl
+d
= 0, thus
gives
Thus the first con~ponentis just ( f l ( ~ ~ )The
) ~ .fact that the second component,
is a square follows as in the previous case by reducing equation (3.1) by g(x)
and noting that g (f .
0
If two of the points have order two, then the third point must as well, so that
leaves us in the final case that .4, B, C all have order two. Thus yl
=
y2 = y3 = 0.
So
K[8]
K[z]/(.x
K[z]/(n: - z2)x K[z]/(n: - x3).
- zl) x
We have
But differentiating Equation (3.1) and plugging in x = X I , x2 and x3 gives
Lemma 5. ker(p) = 2 E ( K ) .
Proof. We follow the proof in [19]. For an alternate proof see [5] Chapter 15 Lemma
2. The kernel of p clearly contains 2 E ( K ) because p ( 2 P ) = / I ( P ) ~= 1 E A;(lA4?,
so it only remains t o show the opposite inclusion. Let P E ker(p). If P
write P = (xo,y o ) Since p(P) = 1, if 2 P
#
# 0 , we can
0 , it must be that xo - 8 is a square in
A;(. On the other hand, if 2 P = 0 , then xo is a root of f , so one of the component of
xo - 19 is zero, and the the others must be squares since P E ker(p).
Thus we can write
xo - 8 = (uo
+ ulQ+ u2Q2)2
(3.3)
CHAPTER 3. T W O DESCENT
for some uo, u l , u2 E K Since f ( 0 ) = 0, we have O3 = -a0
- b,
so we have
So v , w E K . Squaring this equation gives
Then substituting equation (3.3) gives
We must have u, # 0, for otherwise equation (3.3) would not be satisfied. So dividing
by
U;
we have
(10 -
+
0)(v2- q2= (v18
~
1
+
~
where v2 = u 1 / u 2v1
, = v / u 2 ,w1 = w / u 2 . Thus (vlx
)
~
)
( x o~- x)(v2 - x ) is~ a
1- )
multiple of f ( x ) . Since they are both nlonic cubic polynon~ials,they must be equal,
thus
f ( x ) = (ulx
+
~
1 - ()2 0 ~
x)(u2 - x )2 .
Now, we can interpret this geometrically. If we consider the line L : y
=
vlx
+ wl,
we see that L intersects E a t x = xo and x = 212, with the latter intersection being of
multiplicity two. Since three points P, Q, R are collinear iff P
+ Q + R = 0 , this gives
for some t . Thus 2(v2,- t ) = ( x o ,y o ) , so P E 2 E ( K ) .
With these facts in hand we be describe the method of 2-descent.
3.3
The Two-descent
With the above definition of p we have the exact sequence
CHAPTER 3. TWO DESCENT
We will use the fact that
29
is an injection from E ( K ) / 2 E ( K )to A k / A $ to bound
the rank of E over K . To do this we will make heavy use of the commutative diagram
There is no known method for computing the image of p in A*/A*2,but it is
contained in a finite group which is con~putablein practice. We define the 2-Selmer
Group, denot,ed S ( 2 ) ( ~ / as
Q)
s ( ~ ) ( E / Q=) (6 E A*/A*2: $4(6) E p ( E ( Q I ) )for all 1 ) .
The 2-Selmer group contains the image of p, because if ( x ,y) E E ( Q ) ,then ( x ,y ) E
E ( Q ) for all 1. Since p ( E ( Q ) )c S ( 2 ) ( ~ / Qwe) ,will attempt to bound # p ( E ( Q ) )by
calculating the size of S ( 2 ) ( E / Q ) .
In other words, by intersecting the groups p ( E ( Q 1 ) with
)
the images Ab/A$ in
A & / A g for various 1 we hope to obtain a bound on the size of Ab/Ab2, and hence
the rank of E over Q.
For a further discussion of the method of 2-Descent see [29]Chapter X, [9]Chapter
3, [3],[19]Chapter 19 or [32].
Chapter 4
Two-descent applied to congruent
number curves
Let p be a prime number. We will be working with the elliptic curve:
Since f (z) = x ( x
(Q*/Q*2))".
So the map
/L
-
p)(x
+ p)
splits completely over Q , we have A ~ / A $ =
becomes
Lemma 6. If ( x , y ) E E ( Q ) then p ( x , y )
( 1 , -1; 2 , - 2 , p , -p, 2p, - 2 p ) , and b1b2b3 E
=
Q*2
( b 1 , b 2 ,b3) where each 6, is in the set
CHAPTER 4. TWO-DESCENT APPLIED TO CONGRUENT NUMBER CURVES31
Proof. We have that blb2b3 = y2. Consider the l-adic valuation of the hi's. We have
ordl(y2) = ordl(b162b3) = ordl (dl)
+ ordl(b2) + ordl(53) is even for any 1. If one of the
hi's has odd valuation, then two must. If bi has odd valuation, then 1 I hi, but the &'s
differ by p or 2p, so if I
#
2,p, none of the
working modulo squares, we see that if 1
#
&
can have odd valuation. Since we are
2 , p the hi's must have valuation 0 for all
1. So each bi E (1,-1,2,-2,p,-p12p,-2p).
0
Lemma 7.
where 1211 is the 1-adic valuation of 2.
Proof. This is [6] Lemma 5.1 and [7] equation 7.6.2. For an argument using Haar
measure see [12] page 451. First notice that since ker(p(E(Q1)))= 2E(Q1) we have
that #p(E(Qi)) = #E(Q)/2E(Q1). We will consider two cases, 1 < oo and 1 = oo.
If 1 < oo, then the group E(Ql) has a subgroup H of finite index such that I1
-. Z1.
See [29] Theorem VII.6.3. Consider the map
Since H is torsion free, the kernel of map [2IH is in one-to-one correspondence with
the kernel of the map [2],which is E[2](Q1),thus
This gives us that
The following equations are true of any abelian groups,
CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES32
combining these three equations we have
If I
# 2 then #Z1/2Z1 = 1, and if I
= 2 then #Z1/2Z1 = 2. This proves the lemma
for I < ca.
When I = ca, we distinguish two cases depending on how many roots f has in
R. If f splits completely over R, we call its roots, a l , a 2 , a3, and we can, without
loss of generality, assume a l < a 2 < as. In this case #E[2](R)
(1W*/1W*2)3
-. (iZ/2Z)3. If
=
4 and p : E ( R )
(x, y) is a point on the curve, t)hen a1 5 x 5
a 2
or a s 5 x,
which gives p(x, Y) = ( x - a l , x - a 2 , x - a s ) = (1, -1, -1) or ( I l l , 1). So #p(E(R))
2. If f has only one root, a l , over R then #E[2](R) = 2 and p(x, y)
=
--+
=
(1,1,1) for all
(x, y) E E ( R ) since al 5 x for all x , so #P(E(R)) = 1.
4.1
General Bounds
Here we try t o find a bound on the size of p ( E ( Q ) ) , and hence the rank of E.
By Lemma 6 we have that
Now Ak/Ak2 = {(f1,f 1,f 1)) since 1W*/IP2
and p((0,O)) = (-1, -1,l)
SO
=
Z/2Z. We know #p(E(R)) = 2
we conclude that p ( E ( R ) ) = {(1,1,1),(-1, -1,l)).
Then taking I = ca in diagram 3.5, we see that (-1,1, -1) $! p(E(Q)).
This shows that p ( E ( Q ) ) is generated by at most 5 elements, but two of these are
the images of two-torsion points, so at most three of these generators can be images
of points of infinite order, so we conclude that the rank of E ( Q ) is at most 3.
CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES33
4.2
Specific Bounds
In the previous section we found a bound on the size of p ( E ( Q ) ) that is independent
of p. Now we will impose congruence conditions on p t o further bound the size of
P(E(Q)).
For odd p, by (4.1): we have #E[2](Qp) = #p(E(Qp))
= 4, SO
we have that
p(E(QP)) = {(Il 1-11,(-1, -P,P)] (P, 212 ~ 1(-P,
,
-2~,2)}.
Note that this is just the image of the 2-torsion points of E. We only wish to
keep elements of ((-1, -1, I ) , (-1,1, -I), (p, p, I ) , (p, 1,p), (2,2, I ) , (2,1,2)) that are
equivalent to one of the above elements in
0,.
We distinguish 4 cases for p
If p
1 mod 8, then -1
If p
5 mod 8, then -1
I f p = 7 mod 8, then 2
= 2 .= -2 = 1 E Qg/Qg2
-
1 and 2 $ -1 E
= 1 and -1
SO
Qg/q2
we eliminate nothing.
SO
$ 1 EQg/Qg2
Finally we look at p(E(Q2)). By equation 4.1, #p(E(Q2))
=
8, so we need 3
generators for the group. The image of the 2-torsion provides 2 generators, the point
with x = 114 happens to map to an independent generator, (1/4,1/4
P(E(Q~)).
- p,
114
+ p) E
CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENT NUMBER CURVES34
(1/4,1/4
-
p, 114
+ p)
-
(1, 1 - 4p, 1
Thisisbecausepisoddandso1-4p=
+ 4p) = (1,5,5) E Q;/Qa2.
1+4p=5 mod8,so1-4p=1+4p=
5 E Q;/Qj12. This gives us our third generator, so we have that
Now we remove the elements that are not equivalent nlodulo squares (in Q2) to
one of the above elements.
We again distinguish 4 cases for p:
This is because (-1, -5,5) . (1,5,5) = (-1, - 1 , l ) E p(E(Q2)).
CHAPTER 4. TWO-DESCENT APPLIED T O CONGRUENTNUABER CURVES35
4.3
Results
Our results so far can be summarized in the following inequality
O i f p ~ 3m o d 8
rank(E) 5
1 if p
5 mod 8
From here we can already conclude that if p
=3
mod 8, p is not a congruent
number. This was already proven in the 19th century by Genocchi in [13] (see [lo]).
When p
=5
mod 8, or p
G
7 mod 8, p is in fact a congruent number. This was
stated in [31], and proven in [ll]for p
< lo6 (p not necessarily prime), and finally
proven for all prime p in [24]. Another derivation of the bounds we have obtained on
the size of the 2-Selmer group can be found in [17].
If U I ( E / K ) were known t o be finite, by [29]Chapter X Theorem 4.14, #UI(E/K)[2]
would have to be a perfect square, thus it would have an even number of generators.
The exact sequence
-
= )#UI(E/K)[2]. Because S ( 2 ) ( ~ / ~ )
gives us that # s ( ~ ) ( E / K ) / # E ( K ) / ~ E ( K
has a t most 3 generators for p
=5
mod 8, and p
7 mod 8, this implies that
III(E/K)[2] must be the trivial group. If ILLI(E/K)[2]1 = 1 then E ( K ) / 2 E ( K )
2.
S ( 2 ) ( ~ so
/ ~the
) ,rank of E ( Q ) would have to be exactly 1. For a slightly more in
depth discussion of this, see [29] Chapter X, Rernark 6.3. While it is not known that,
LU(E/K)[2] is finite, it is conjectured to be so, see [29] Chapter X, conjecture 4.13.
We will continue by further analyzing the case when p
=1
mod 8.
Chapter 5
The Case When p
5.1
-
1 mod 8
A Specific Two-Isogeny
We have done a 2-descent on the curve E : y2 = x(x
its rank. To further bound the rank in the case p
+ p)(x - p) in order to bound
fi
1 mod 8, we will perform a
2-descent on an isogenous curve.
The map
is an isogeny of degree two generated by the 2-torsion point (0, 0), from our curve
E : y2 = x3 - P
Since p
2 to
~
=1
such that p = a2
mod 4, we can write p as the sum of 2 squares in
Z.Let
a,b E
Z
+ b2. If we assume a , b > 0 and a is odd, then this representation is
unique. We will make frequent use of this decomposition of p.
Note that, unlike E , El has only t,wo rational 2-torsion points, (0:(0,O))
CHAPTER 5. THE CASE WHEN P
5.2
=1
MOD 8
Method
As before, we construct the exact sequence (3.4).
Since f ( x ) = x ( x 2
+ qp2)does not
split completely over Q , we now have two
distinct forms for the group Ak/A$ where p is defined as follows.
If i =
a
E K , i.e. f ( x ) splits completely over
K , then
When i @ K, we have
We proceed as before, making use of diagram (3.5),and equation (4.1).
5.3
General Bounds
We begin, as before, by identifying a finite set containing the image of p
+
Lemma 8. p ( E l ( Q ) )c ( ( 2 , l i), ( p , a
+ bi),( p ,a - bi),(1,i ) ) .
CHAPTER 5. THE CASE W H E N P
Proof. Suppose
=1
MOD 8
38
p ( z , y) = (61, 62). Then we have b1 . N(b2) = y2. Where N(J1) is
the norm of b2 as an element of the extension field Q(i), i.e. N(b2) = 6262 where
is the ordinary complex conjugate of 62. Consider the I-adic valuation of the 6,s.
f2
We have ordl(y2) = ordl(bl . N(62)) = ordl(bl)
+ ordl(N(b2))is even for any I.
If
bl or N(b2) has odd valuation, then they both must. If dl has odd valuation, then
I
1
dl, but since 62 = dl
we have that I
I
-
2pi, we have N(b2) = b1
+ 4p2.
So if I
1
dl and I
1
b2
4p2, so the only divisors bl are 1 , 2 , p . In Z[i], 4p2 factors into
+ i)2(1 - ~ ) ~+( bi)(a
a
- bi), since every factor of d2 is a factor
of 4p2 the only possible factors of 62 are 1 , i , 1 + i , 1 - i , a + bi, a - bi. The fact
that p(El(Q)) C ( ( 2 , l + i ) , (p, a + bi), (p, a - bi), (1, i)) follows from the fact that
irreducibles as (1
bl . N(d2) = 1 modulo squares.
0
It is also worthwhile t o notice that since we are working modulo squares, 22
(1
+ i)2 = 1 and so ( 2 , l + i ) ( l , i) = ( 2 , l - i)
5.4
=
mod squares.
p-adics
Since Ei[21(Qp) = (0, (0, O), (2pi, O), (-2pi, O)), (4.1) gives that #p(El(Qp)) = 4. The
map p looks like:
The last equivalences follow from the fact that 2 and i are squares in
Q,because
=1
CHAPTER 5. THE CASE WHEN P
p
=1
mod 8. Since Ip(E(Qp))l = 4, we must have that
Now since ( a + bi)(n - bi)
a
-
MOD 8
bi
=
p, we have that n
+ bi
-
p modulo squares and
-- 1 modulo squares in Qp (or vice-versa), and since (1 + i ) ( l- i) = 2, which
+ i = i - 1 n~odulosquares. Before we can eliminate
elements from p ( E ( Q ) ) , we must determine when 1 + i is a square in 0,.
is a square in Q,, we see that 1
-
5.4.1 The Quadratic Character of 1 + i
We have that p
=
a2 + b2,
SO
a
bi mod p. Using Quadratic Reciprocity and the
laws of the Jacobi symbol we have
(;) (E)
=
Then
=
(7)+
a2
(y)= (:) (y)= ( )
b2
=
(f) = 1.
So we just need to calculate
(y),which
we do using the Jacobi symbol.
By Quadratic Reciprocity
(5)
= ( -
quadratic residue mod p exactly when (a
by the quadratic charact,er of 1 + i.
1 ) ~ .This gives us that 1
+ b)2
E Qp (i.e. 1
p(E(Q))
2 1 i)
+
2 1 - 2)
(P, a
+ bi)
(PAP)
H
p(E(Qp)) C
( l , l , 1)
H
(1,1 , l )
-+
++
++
+ i is a
1 mod 16. We distinguish 2 cases,
+ i is a square mod p)
This occurs when (a + b)2 = 1 mod 16.
1. 1
+ i =-
-
(a+b12-1
(Q;/Q;2)3
(P, P, 1)
(P,LP).
In this case, we cannot further reduce the size of p(E(Q)).
CHAPTER 5 . THE CASE WHEN P
=1
MOD 8
2. 1 + i f 0 E Qp (i.e. 1 + 2 is not a square mod p)
This occurs when (a + b)2 = 9 mod 16.
Since El [2](Q2)= (0, (070)),we have, by (4.I ) , that #p(E1 ( 0 2 ) )
=4
and p is defined
as follows:
Since # p ( E 1(Q2)) = 4 we need to find 2 generators for the group. A little testing
yields t,he points x = 5 and x = 2p. To see t,hat these are in fact points on the curve
- -
2 / 2 2 x (2/82)*, where an element in Q2 is a square if
E1(Q2),recall that Q ; ~ / Q =
G~
and only if it has even valuation (the first tern? in the ~ r o d u c t )and
, the odd part is a
+
square mod 8 (i.e. 1 mod 8). Here we see that 5(52 4p2) = 125
+
mod 8, and 2 p ( ( 2 ~ 1 ) ~4p2) = gp3
mod 8.
+ 20p2
5
+4 = 1
+ 8p3 = 16p3 which is a square because p
1
C H A P T E R 5. T H E CASE W H E N P
=1
p ( x = 5 ) = (5,s- 2 p i )
p(z
=
2p)
=
-
MOD 8
( 1 , 5 - 2p.i)
mod squares
+i)
mod squares.
( 2 p , 2p - 2pi) = ( 2 , l
So we distinguish the same 2 cases as before.
-
1. 1 + i is a s q u a r e i n Qp
i.e. ( a
+ b)2
1 mod 16
p(E(Q))
(21+)
( p , a + bi)
(p,a-bi)
(1,~)
++
p(E(Q2))
(2,1+2)
++
(171)
-+
+
+
+
+
c Q2
x
Q2(i)
(L1)
(1.4 51 p ( E ( Q 2 ) ) .
i ) , which gives us that p ( E 1 ( Q ) ) has at most 3
generators, so we find that in this case the rank of El is at most 2.
We lose the generator ( 2 , l
-
-
2. 1 + i is not a square in Qp
i.e. ( a
+ b)2
-
9 mod 16
From this argument we have t'hat if p r 1 mod 8 , with a , b such that a2
and ( a + b)2
when ( a
+ b2 = p
9 mod 16, then p is not a congruent number. On the other hand,
+ b)2 = 1 mod 16, we cannot make any conclusions. This result was stated
CHAPTER 5. T H E CASE WHEN P
=1
MOD 8
42
as early as 1915 by L. Bastien [2]. This result can also be obtained as a consequence
of Tunnell's Theorem, as in [34] Proposition 6.
Chapter 6
Homogeneous Spaces
6.1
Method
In the previous sections, we have examined the image of the map
where, as before E : y2 = f ( x ) , A = Q [ x ] / ((fx ) )= Q[0]and 0 is the residue of
x. As f is a polynomial of degree three, we have that 1,0,O2 is a basis for A, so any
element in A can be written as uo
+ ule + u2Q2,where uo, ul, u2 E Q.
So for any 6 in the image of p we have the equation
+
+
x ( P ) - 0 = b(uo u18 ~ ~ 0 ~ ) ~ .
Expanding the square on the right hand side we have
for some quadratic forms Q6,iE Q[uo,ul , u2]
Equating powers of 0, we have that
CHAPTER 6. HOMOGENEOUS SPACES
So equat,ions (6.1) gives us a necessary condition for b to be in the image of p.
These equations are in fact sufficient as well, and if equations (6.1) are satisfied, then
Qs,o is the x-coordinate of a point on the elliptic curve E.
We would like to restrict our attention to integral solutions to equations (6.1),
and this is easily done by noting that Q6,1is homogeneous of degree two, and clearing
denominators, which gives us the two equations
We will begin by examining the equation Qa,2= 0. If we set
then C is a conic, and if we can find a rational point on C will allow us to
parametrize C .
Supposing that we were able to parametrize, C as (uo(A),ul (A), u2(A)), we can
move on to the second equation,
Qa,i
= -u$.
Using our parametrization for C , we
obtain the equation
Since Qa,i,QhI2are quadratic forms, the left hand side will be a quartic in A, call
it g(X). Then the curve g(A) = -u:
image of p.
has a rational point exactly when d is in the
CHAPTER 6. HOMOGENEOUS SPACES
6.2
Conics
We now examine the homogeneous spaces associated to the curve
E~: y2 = x3
+ 4p2x.
We have f (x) = x3 + 4p2x, which gives A = Q[x]/(x3+ 4p2x) = Q x Q(i).
In the preceding sections we showed that if p
=1
mod 8, then
P(E(Q)) c ( ( 2 , l + 4, (P, a + bi), (P, a
-
b4.
Remember, that when we calculated the 6 we used the natural basis (1,O), (0, I ) , (0, i)
of Q x Q(i). We have to convert these into the basis 1,9, O2 where 9 is the residue of
+
x in Q[x]/(x3 4p2x) Y Q x Q(i). TO do this, we make explicit the isomorphism
Now we are ready to try to solve the equation Qs,2 = 0 for the allowable values
of 6.
When 6
=
(1,p) = 1 -
so2,
i.e. 6 is the image of the two-torsion point, (0,O).
Then we have
1
Q6,2 = -(I
4
- P)U; -
This has a solution (O,2p, 1).
4p5u22
+ 2p uou2 + p ul.
3
3 2
(6.3)
CHAPTER 6. HOMOGENEOUS SPACES
0
When 6 = ( p ,a
+ bi) = p + $ 3 + yo2we have
When 6 = ( p ,a
- bi) = p
+
- LO
$$02
2~
When 6 = ( 2 , l + i ) . ( p , a + bi) = 2 p +
we have
$o+
we have
-Q2
+
+
Q6.2 = ( a b)puoul+ 2(a - b)p2uou2 ( a - b)p2u?
1
b-a
-pu;
4 ( b - a)p4u; - 4 ( a b ) p h l u 2 -uo.
2
4
+
0
+
When 6 = ( 2 , l + i ) . ( p , a
Qs,2 =
When 6
=
(a- ~
+
+
,
+ 2 0 + T O 2 have
+ 2(a + b ) p 2 u o u+~ ( a + b)p2u?
- bi) = 2p
(6.7)
2p-b-a
(6.8)
) P U ~ I L ~
( 2 , 1 + i ) . ( p , a - h i ) . ( p , a + bi) = ( 2 , p ( l + i ) ) = 2 +
;Q+3Q2
have
The first equation corresponds to a two-torsion point, so we analyze the only
other space with a rational point. From (6.4) for 6 = ( 2 , l + i ) , we have
We know that (-2p, 1 , 0 ) is a point on C , so letting L be the line uo = Xu2 - 2p, and
considering the intersections of L with Qs,2 as in [25],we arrive a t an equation
Setting this t o zero and solving for u2 we get
CHAPTER 6. HOMOGENEOUS SPACES
Plugging back in to L we have
So this gives a parametrization for C as
Thus
ul(X),u2(X)) = -ug becomes
Absorbing squares into u:, the equation Qs,l(uo(X),
So if we define
f (x) = p(x4 + 16p2x3 - 96p4x2 + 768p6x - 1792p8),
we have that f (z)is an irreducible quartic, and we would like t,o determine when the
equation
y2 = f ( 4
has solutions We can simplify this by making the change of variable x
y
++
++
p2x, and
p4y, and dividing both sides by p8, which gives us the equation
Making a further change of coordinates x
++
-42 and y
++
16y, and dividing both
sides by 28, we can reduce this to
Thus we conclude that (2:1
solution to equation (6.10).
+ i)
is in the image of
,LL
exactly when we have a
CHAPTER 6. HOMOGENEOUS SPACES
6.3
Follow up
The next step would be to find conditions on p such that equation (6.10) has a solution.
If we can find a rational solution to equation (6.10), then ( 2 , l + i ) is in the image of
p , which means that the #E1/2E1 > 2, so p is a congruent number. On the other
hand, if there is no rational solution to equation (6.10), then
We know also that
so in this case, we must have #LLI[2] > 1.
If, as it is conjectured, #LU < m then as we argued before, by [29] Chapter X,
Theorem 4.14, the order of IJ..I[2]must be a perfect square, so #LLI[2] 2 4. Since we
have shown in the previous sections that # s ( ~ ) ( E ~ / Q
5 )8, we must have #p(EIQ) =
2, which means that Rank(El/Q) = 0. Thus in this case we conclude that p is not a
congruent number.
This gives us a criterion to determine whether p is a congruent number: p is
congruent exactly when equation (6.10) has rational solutions.
We have already examined local conditions for the solubility of equation (6.10).
In sections 5.4 and 5.5 we examined when ( 2 , 1 + i ) was in the image of p ( E (Q)) and
p ( E ( Q 2 ) ) ,and we determined that ( 2 , l + i) was in both images when (a
mod 16. Thus equation (6.10) has solutions Q2 and Qp when (a
+ b)2 E 1
+ b)2 = 1 mod 16.
After testing for local solubility, the standard approach is to perform a second 2descent on (6.10) as described in [23] and [33] to determine when equation (6.10) has
rational solutions. We leave this for a future paper.
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