Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 239

By Proposition 8.26, there exist integers m, n with gcd( m, n ) = 1 such that

s = m 2 − n 2 , t = m 2 + n 2 .

2 e =2 mn,

Therefore,

pr 2 = s 2 +2 e 2 =( m 2 n 2 ) 2 +2( mn ) 2 = m 4 + n 4 . −

Let q be a prime dividing r . Proposition 8.26 says that m ≡ n (mod 2), which
implies that pr 2 must be odd. Therefore, q =2. Sincegcd( m, n ) = 1, at least
one of m, n is not divisible by q . It follows that both m, n are not divisible by
q ,since m 4 + n 4 ≡ 0(mod q ). Therefore,

( m/n ) 4 ≡− 1(mod q ) .

It follows that m/n has order 8 in F q ,so q ≡ 1 (mod 8). Since r is a positive
integer and all prime factors of r are 1 mod 8, we obtain

r ≡ 1(mod .

Therefore, r 2

≡ 1 (mod 16), so

m 4 + n 4 = pr 2 ≡ 9 (mod 16) .

But, for an arbitrary integer j ,wehave j 4 ≡ 0 , 1 (mod 16). Therefore,

m 4 + n 4 ≡ 0 , 1 , 2 (mod 16) ,

so pr 2 = m 4 + n 4 . This contradiction proves that C 1 ,p,p has no rational points.

We now need to show that C 1 ,p,p has q -adic points for all primes q ≤∞ .
The proof breaks into four cases: q = ∞ , q =2, q = p ,andallother q .
The case of the reals is easy. Let u be large enough that u 2 > 2 p .Then

choose v, w satisfying (8.14).
For q =2,write

u =1 / 2 , v = v 1 / 2 , w = w 1 / 2 .

The equations for C 1 ,p,p become

1 − pv 1 =8 p, 1 − pw 1 = − 8 p.

We need to solve

v 1 =(1 w 1 =(1+8 p ) /p − 8 p ) /p,

in the 2-adics. Since

(1 ± 8 p ) /p ≡ 1(mod ,

and since any number congruent to 1 mod 8 has a 2-adic square root (see
Appendix A), v 1 ,w 1 exist. Therefore, C 1 ,p,p has a 2-adic point.