Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 208

with rational numbers a, b, c, u, v, w .Then y 2 = abc ( uvw ) 2 ,so

abc is a square.

By adjusting u, v, w , we may assume that a, b, c are squarefree integers. In
fact, we claim that

a, b, c ∈{± 1 , ± 2 } .

Suppose that p is an odd prime dividing a .Since a is squarefree, p 2 a ,so

the exact power p k dividing x = au 2 has k odd. If k< 0, then p k is the

2, so p 3 k is the power of p in the
denominator of y 2 = x ( x − 2)( x + 2). Since 3 k is odd and y 2 is a square, this
is impossible. If k> 0then x ≡ 0(mod p ), so x ± 2 ≡ 0(mod p ). Therefore,
p k is the power of p dividing y 2 .Since k is odd, this is impossible. Therefore,
p a . Similarly, no odd prime divides b or c . Therefore, each of a, b, c is, up
to sign, a power of 2. Since they are squarefree, this proves the claim.
The procedure we are following is called descent , or, more precisely, a
2-descent . Suppose x is a rational number with at most N digits in its
numerator and denominator. Then u, v, w should have at most N/ 2 digits
(approximately) in their numerators and denominators. Therefore, if we are
searching for points ( x, y ), we can instead search for smaller numbers u, v, w .
This method was developed by Fermat. See Section 8.6.
We have four choices for a and four choices for b .Since a and b together
determine c (because abc is a square), there are 16 possible combinations for
a, b, c . We can eliminate some of them quickly. Since x ( x− 1)( x +2) = y 2 > 0,
we have cw 2 = x +2 > 0, so c> 0. Since abc > 0, it follows that a and b
must have the same sign. We are now down to 8 possible combinations.
Let's consider ( a, b, c )=(1 , 2 , 2). We have

exact power of p in the denominator of x ±

x = u 2 , 2=2 v 2 , x +2=2 w 2 x −

with rational numbers u, v, w . Therefore,

u 2 2 v 2 =2 ,

2

2 w 2 =

− − − 2 .

If v has 2 in its denominator, then 2 v 2 has an odd power of 2 in its denomi-
nator. But u 2 has an even power of 2 in its denominator, so u 2 2 v 2 cannot

be an integer. This contradiction shows that v and u have odd denominators.
Therefore, we may consider u, v modpowersof2. Since2

−

u 2 ,wehave2 | | u ,

u 2 . Therefore, 2 v 2 hence 4 | − ≡ 2 (mod 4), which implies that 2 v . Similarly,

− 2 w 2 ≡− 2(mod4),so2 w . It follows that v 2 ≡ w 2 ≡ 1(mod8),so

u 2 2 v 2 u 2 u 2 2 w 2 2 ≡ − ≡ − 2 ≡ − ≡− 2(mod ,

which is a contradiction. It follows that ( a, b, c )=(1 , 2 , 2) is impossible.
Similar considerations eliminate the combinations ( − 1 , − 1 , 1), (2 , 1 , 2), and