ELLIPTIC CURVES 111
that t − t
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
By the way, the “extra” solution X =−1, Y = 0toX
1 corresponds to the point O at inﬁnity on this elliptic curve.
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
). 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.
(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
You may have noticed that this does not conform to our template y
+ Ax + B for
elliptic curves. To ﬁx this, deﬁne a new variable u =−t, and get the equivalent equation
Why? Well, we get this extra point if we set u =∞, as it were. That would correspond
to the point (∞, ∞) on the elliptic curve.
Associativity can also be proved in more elegant ways using algebraic geometry.