Let E be an elliptic curve defined over Q. The following
is a deep theorem
about the group E (Q).
Theorem 6.5.1 (Mordell). The group E(Q) is finitely generated. That
is, there are points 
such that every element of E(Q) is
of the form 
for integers
.
Mordell's theorem implies that it makes sense to ask whether or not
we can compute E(Q), where by "compute" we mean find a finite set
of points on E that generate E(Q) as an abelian group. There
is a systematic approach to computing E(Q) called "descent" (see e.g.,
[Cre97, Cre, Sil86]). It is widely believed that descent will always succeeds,
but nobody has yet proved that it does. Proving that descent works for
all curves is one of the central open problem in number theory , and is
closely related to the Birch and Swinnerton-Dyer conjecture (one of the
Clay Math Institute 's million dollar prize problems). The crucial difficulty
amounts to deciding whether or not certain explicitly given curves have any
rational points on them or not (these are curves that have points over R
and modulo n for all n).
The details of using descent to computing E(Q) are beyond the scope
of this book. In several places below we will simply assert that E(Q) has
a certain structure or is generated by certain elements. In each case, we
computed E(Q) using a computer implementation of this method .
6.5.1 The Torsion Subgroup of E(Q) and the Rank
For any abelian group G, let
be the subgroup of elements of finite
order. If E is an elliptic curve over Q, then 
is a subgroup of
E(Q), which must be finite because of Theorem 6.5.1 (see Exercise 6.6).
One can also prove that is finite by showing that
there is a prime
p and an injective reduction homomorphism
, then
noting that E(Z/pZ) is finite. For example, if E is y2 = x3 − 5x + 4, then
The possibilities for are known.
Theorem 6.5.2 (Mazur, 1976). Let E be an elliptic curve over Q. Then
is isomorphic to one of the following 15 groups:
Z/nZ for n ≤ 10 or n = 12,
Z/2 × Z/2n for n ≤ 4.
The quotient 
is a finitely generated free abelian group,
so it is isomorphism to Zr for some integer r, called the rank of E(Q).
For example, using descent one finds that if E is y2 = x3 − 5x + 4, then
is generated by the point (0, 2). Thus
The following is a folklore conjecture, not associated to any particular
mathematician:
Conjecture 6.5.3. There are elliptic curves over Q of arbitrarily large
rank.
The "world record" is the following curve, whose rank is at least 24:
It was discovered in January 2000 by Roland Martin and William McMillen
of the National Security Agency.
6.5.2 The Congruent Number Problem
Definition 6.5.4 (Congruent Number). We call a non zero rational
number n a congruent number if ±n is the area of a right triangle with
rational side lengths. Equivalently , n is a congruent number if the system
of two equations
has a solution with a, b, c ∈ Q.
For example, 6 is the area of the right triangle with side lengths 3, 4,
and 5, so 6 is a congruent number. Less obvious is that 5 is also a congruent
number; it is the area of the right triangle with side lengths 3/2, 20/3, and
41/6. It is nontrivial to prove that 1, 2, 3, and 4 are not congruent numbers.
Here is a list of the integer congruent numbers up to 50:
5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37, 38, 39, 41, 45,
46, 47.
Every congruence class modulo 8 except 3 is re presented in
this list ,
which incorrectly suggests that if n ≡ 3 (mod 8) then n is not a congruent
number. Though no n ≤ 218 with n ≡ 3 (mod 8) is a congruent number,
n = 219 is a congruent number congruent and 219 ≡ 3 (mod 8).
Deciding whether an integer n is a congruent number can be subtle since
the simplest triangle with area n can be very complicated. For example,
as Zagier pointed out, the number 157 is a congruent number, and the
"simplest" rational right triangle with area 157 has side lengths
This solution would be difficult to find by a brute force search.
We call congruent numbers "congruent" because of the following proposi-
tion, which asserts that any congruent number is the common "congruence"
between three perfect squares .
Proposition 6.5.5. Suppose n is the area of a right triangle with rational
side lengths a, b, c, with a ≤ b < c. Let A = (c/2)2. Then
A - n, A, and A + n
are all perfect squares of rational numbers.
Proof. We have
Add or subtract 4 times the second equation to the first to get
The main motivating open problem related to congruent numbers, is to
give a systematic way to recognize them.
Open Problem 6.5.6. Give an algorithm which, given n, outputs whether
or not n is a congruent number.
Fortunately, the vast theory developed about elliptic
curves has some-
thing to say about the above problem. In order to understand this connec-
tion, we begin with an elementary algebraic proposition that establishes a
link between elliptic curves and the congruent number problem.
Proposition 6.5.7 (Congruent numbers and elliptic curves). Let n
be a rational number. There is a bijection between
and
given explicitly by the maps
and
The proof of this proposition is not deep, but involves substantial (ele-
mentary) algebra and we will not prove it in this book.
For n ≠ 0, let 
be the elliptic curve y2 = x3 − n2x.
Proposition 6.5.8 (Congruent number criterion). The rational num-
ber n is a congruent number if and only if there is a point P = (x, y) ∈
with y ≠ 0.
Proof. The number n is a congruent number if and only if the set A from
Proposition 6.5.7 is nonempty. By the proposition A is nonempty if and
only if B is nonempty.
Example 6.5.9. Let n = 5. Then En is y2 = x3 − 25x, and we notice that
(−4,−6) ∈ . We next use the bijection of Proposition 6.5.7 to find
the corresponding right traingle:
Multiplying through by −1 yields the side lengths of a rational right triangle
with area 5. Are there any others?
Observe that we can apply g to any point in with y
≠ 0. Using
the group law we find that 2(−4,−6) = (1681/144, 62279/1728), and
Example 6.5.10. Let n = 1, so
is defined by y2 = x3 −
x. Since 1 is not
a congruent number, the elliptic curve has no point with y ≠ 0. See
Exercise 6.10.
Example 6.5.9 foreshadows the following theorem.
Theorem 6.5.11 (Infinitely Many Triangles). If n is a congruent
number, then there are infinitely many distinct right triangles with rational
side lengths and area n.
We will not prove this theorem, except to note that one proves it by
showing that 
, so the elements of the
set B in Proposition 6.5.7 all have infinite order, hence B is infinite so A
is infinite.
Tunnell has proved that the Birch and Swinnerton-Dyer (alluded to
above), implies the existence of an elementary way to decide whether or
not an integer n is a congruent number. We state Tunnell's elementary way
in the form of a conjecture.
Conjecture 6.5.12. Let a, b, c denote integers. If n is an even square-free
integer then n is a congruent number if and only if
If n is odd and square free then n is a congruent number if and only if
Enough of the Birch and Swinnerton-Dyer conjecture is known to prove
one direction of Conjecture 6.5.12. In particular, it is a very deep theorem
that if we do not have equality of the displayed cardinalities, then n is not
a congruent number. For example, when n = 1,
The even more difficult (and still open!) part of Conjecture 6.5.12 is the
converse: If one has equality of the displayed cardinalities, prove that n
is a congruent number. The difficulty in this direction, which appears to
be very deep, is that we must somehow construct (or prove the existence
of) elements of . This has been accomplished in some cases do to
groundbreaking work of Gross and Zagier ([GZ86]) but much work remains
to be done.
The excellent book [Kob84] is about congruent numbers and Conjec-
ture 6.5.12, and we encourage the reader to consult it. The Birch and
Swinnerton-Dyer conjecture is a Clay Math Institute million dollar millen-
nium prize problem (see [Cla, Wil00]).
