Number Theory: An Introduction to Mathematics

6 Some Applications 575

In the same paper in which he proved his theorem, Mordell (1922) conjectured that
if a non-singular irreducible projective curve, defined by a homogeneous polynomial
F(x,y,z)with rational coefficients, has infinitely many rational points, then it is bira-
tionally equivalent to a line, a conic or a cubic. Mordell’s conjecture was first proved by
Faltings (1983). Actually Falting’s result was not restricted toplanealgebraic curves,
and on the way he proved two other important conjectures of Tate and Shafarevich.
Falting’s result implies that the Fermat equationxn+yn = zn has at most
finitely many solutions in integers ifn>3. In the next section we will see that Wiles’
result that semi-stable elliptic curves are modular implies that there arenosolutions in
nonzero integers.

6 SomeApplications

The arithmetic of elliptic curves has an interesting application to the ancient problem
of congruent numbers. A positive integernis (confusingly) said to becongruentif it is
the area of a right-angled triangle whose sides all have rational length, i.e. if there exist
positive rational numbersu,v,wsuch thatu^2 +v^2 =w^2 ,uv= 2 n. For example, 6 is
congruent, since it is the area of the right-angled triangle with sides of length 3, 4 ,5.
Similarly, 5 is congruent, since it is the area of the right-angled triangle with sides of
length 3/ 2 , 20 / 3 , 41 /6.
In the margin of his copy of Diophantus’ArithmeticaFermat (c. 1640) gave a
complete proof that 1 is not congruent. The following is a paraphrase of his argument.
Assume that 1 is congruent. Then there exist positive rational numbersu,v,wsuch
that

u^2 +v^2 =w^2 , uv= 2.

Since an integer is a rational square only if it is an integral square, on clearing denom-
inators it follows that there exist positive integersa,b,c,dsuch that

a^2 +b^2 =c^2 , 2 ab=d^2.

Choose such a quadruplea,b,c,dfor whichcis minimal. Then(a,b)=1. Sinced
is even, exactly one ofa,bis even and we may suppose it to bea.Then

a= 2 g^2 , b=h^2

for some positive integersg,h.Sincebandcare both odd and(b,c)=1,

(c−b,c+b)= 2.

Since

(c−b)(c+b)=a^2 = 4 g^4 ,

it follows that

c+b= 2 c^41 , c−b= 2 d 14 ,