Introduction - Elliptic Curves: Number Theory and Cryptography - page 7

(all possible combinations of plus and minus signs are allowed). There are
32 solutions in all. There are 16 solutions with z even and 16 with z odd.
Therefore, we expect a triangle with area 41. The same method as above,
using the tangent line at the point ( 9 , 120) to the curve y 2 = x 3 41 2 x , − −

yields the triangle with sides (40 / 3 , 123 / 20 , 881 / 60) and area 41.
For much more on the congruent number problem, see [64].
Finally, let's consider the quartic Fermat equation. We want to show that

a 4 + b 4 = c 4 (1.1)

has no solutions in nonzero integers a, b, c . This equation represents the easiest
case of Fermat's Last Theorem, which asserts that the sum of two nonzero
n th powers of integers cannot be a nonzero n th power when n 3. This

general result was proved by Wiles (using work of Frey, Ribet, Serre, Mazur,
Taylor, ...) in 1994 using properties of elliptic curves. We'll discuss some of
these ideas in Chapter 15, but, for the moment, we restrict our attention to
the much easier case of n = 4. The first proof in this case was due to Fermat.
Suppose a 4 + b 4 = c 4 with a =0. Let

≥

x =2 b 2 + c 2
a 2

y =4 b ( b 2 + c 2 )
a 3

,

(see Example 2.2). A straightforward calculation shows that

y 2 = x 3 − 4 x.

In Chapter 8 we'll show that the only rational solutions to this equation are

( x, y )=(0 , 0) , (2 , 0) , ( − 2 , 0) .

These all correspond to b = 0, so there are no nontrivial integer solutions of
(1.1).
The cubic Fermat equation also can be changed to an elliptic curve. Suppose
that a 3 + b 3 = c 3 and abc =0. Since a 3 + b 3 =( a + b )( a 2 ab + b 2 ), we must −

have a + b =0. Let

c
a + b ,

y =36 a

b
a + b .

−

x =12

Then

y 2 = x 3 − 432 .

(Where did this change of variables come from? See Section 2.5.2.) It can be
shown (but this is not easy) that the only rational solutions to this equation
are ( x, y )=(12 , ± 36). The case y = 36 yields a−b = a + b ,so b = 0. Similarly,
y = − 36 yields a = 0. Therefore, there are no solutions to a 3 + b 3 = c 3 when
abc =0.