Number Theory: An Introduction to Mathematics

6 Some Applications 579

The most celebrated application of the arithmetic of elliptic curves has been the
recent proof of Fermat’s last theorem. In his copy of the translation by Bachet of
Diophantus’ArithmeticaFermat also wrote “It is impossible to separate a cube into
two cubes, or a fourth power into two fourth powers or, in general, any power higher
than the second into two like powers. I have discovered a truly marvellous proof of
this, which this margin is too narrow to contain.”
In other words, Fermat asserted that, ifn>2, the equation

xn+yn=zn

has no solutions in nonzero integersx,y,z.In§2 of Chapter III we pointed out that it
was sufficient to prove his assertion whenn=4andwhenn=pis an odd prime, and
we gave a proof there forn=3.
A nice application to cubic curves of the casen=3 was made by Kronecker
(1859). If we make the change of variables

x= 2 a/( 3 b− 1 ), y=( 3 b+ 1 )/( 3 b− 1 ),

with inverse

a=x/(y− 1 ), b=(y+ 1 )/ 3 (y− 1 ),

then

x^3 +y^3 − 1 = 2 ( 4 a^3 + 27 b^2 + 1 )/( 3 b− 1 )^3.

Since the equationx^3 +y^3 =1 has no solution in nonzero rational numbers, the only
solutions in rational numbers of the equation

4 a^3 + 27 b^2 =− 1

area=−1,b=± 1 /3. Consequently the only cubic curvesCa,bwith rational coeffi-
cientsa,band discriminantd=−1areY^2 −X^3 +X± 1 /3.
We return now to Fermat’s assertion. In the present section we have already given
Fermat’s own proof forn=4. Suppose now thatp≥5 is prime and assume, contrary
to Fermat’s assertion, that the equation

ap+bp+cp= 0

does have a solution in nonzero integersa,b,c. By removing any common factor we
may assume that(a,b)=1, and then also(a,c)=(b,c)=1. Sincea,b,ccannot all
be odd, we may assume thatbis even. Thenaandcare odd, and we may assume that
a≡−1 mod 4.
We now consider the projective cubic curveEA,Bdefined by the polynomial

Y^2 −X(X−A)(X+B),

whereA=apandB=bp. By construction,(A,B)=1and

A≡−1mod4, B≡0 mod 32.