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ELLIPTIC CURVES 111
that t t
3
is a square of a rational number. That is, we want to solve
for the solution set E(Q) where E is the elliptic curve given by
y
2
=−t
3
+ t.
5
By the way, the “extra” solution X =−1, Y = 0toX
2
+ Y
2
=
1 corresponds to the point O at inﬁnity on this elliptic curve.
6
However, it is not much of a triangle: One of the sides is 0. In
fact, we could have excluded the solutions where X or Y is0to
begin with. But we would not have seen how the point at inﬁnity on
the elliptic curve can correspond to a “degenerate solution” to our
original problem: a triangle with one side 0, the other side ,in
such a way that the area is 1.
But the real question here is whether there are any nondegener-
ate rational solutions in E(Q). In a course in elementary number
theory, you can prove that the answer is “no. So when D = 1,
the congruent number problem has a negative solution: There is
no right triangle with rational sides and area 1. For more on the
congruent number problem, see chapter 23.
Torsion and the Galois Group
It is very difﬁcult to prove that the ﬁve-step recipe that we have
given above actually deﬁnes a group law. It is not obvious that
it satisﬁes the associative law. We do have obvious inverses: the
inverse of (x
1
, y
1
)is(x
1
, y
1
). And we do have an identity element: O.
But you will have to take associativity on faith; the algebra needed
to verify it is rather involved.
7
(Commutativity of the group law is
not difﬁcult to check from the deﬁnition.)
The group law is a big help when studying E(Q). It can help you
derive new solutions from old ones, and sometimes can even help
5
You may have noticed that this does not conform to our template y
2
= x
3
+ Ax + B for
elliptic curves. To ﬁx this, deﬁne a new variable u =−t, and get the equivalent equation
y
2
= u
3
u.
6
Why? Well, we get this extra point if we set u =∞, as it were. That would correspond
to the point (, ) on the elliptic curve.
7
Associativity can also be proved in more elegant ways using algebraic geometry.
112 CHAPTER 9
to prove that there are no unobvious solutions—all depending on
which curve E you are studying.
There are many theorems about the set E(Z) and the group E(Q)
that can be found in the textbooks. For our purposes, in studying
representations of the absolute Galois group G of Q, we need to tell
you about a particular set of elements inside of E(C): the n-torsion.
DEFINITION: Let n be a positive integer. An element P of
E(C)ispartofthen-torsion if
n times

P + P +···+P = O.
The set of all n-torsion points is written E[n].
For example, an element P on the elliptic curve is part of the
5-torsion if P + P + P + P + P = O. This is usually written, rather
confusingly, as 5P = O. Do not confuse it with multiplying the x-
and y-coordinates of P by 5, because it means something much more
complicated. You can work out what it does mean: It will give some
Z-equation that the x-coordinate of P must satisfy, and another one
for the y-coordinate of P, for P to be 5-torsion. Therefore, the x- and
y-coordinates of P will be in Q
alg
.
In fact, the equation that the x-coordinate must solve will be a
rather messy degree-12 Z-equation, and so there will be no more
than 25 elements of the 5-torsion (for any particular elliptic curve,
of course; if you change the elliptic curve, then you will get a
different equation, and different solutions).
8
The general situation
is summarized as follows.
THEOREM 9.1: Let E be an elliptic curve, and let n be a
positive integer. All of the n-torsion points of E have
coordinates in Q
alg
, and the number of elements of E(Q
alg
) that
are n-torsion is n
2
.
8
How do we get 25 elements? The equation for the x-coordinate will be degree 12, whose
roots will give us at most 12 x-coordinates. Each of those corresponds to 2 y-coordinates,
which are ± of each other. That is 24 points. The remaining point is O.

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