Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 225

Our job is to eliminate the remaining 56 possibilities.
Observe that

x − 5= bv 2 <x = au 2 <x +5= cw 2 .

If a< 0, then b< 0. If a> 0then c> 0, hence b> 0since abc is a square.
Therefore, a and b have the same sign. This leaves 32 possible pairs a, b .
We now consider, and eliminate, three special pairs a, b . The fact that
φ is a homomorphism will then su ce to eliminate all but the eight pairs
corresponding to known points.
(a,b)=(2,1). We have

x =2 u 2
x − 5= v 2
x +5=2 w 2 .

Therefore,

2 u 2 − v 2 =5 , 2 w 2 − 2 u 2 =5 .

If one of u or v has an even denominator, then so does the other. However,
2 u 2 has an odd power of 2 in its denominator, while v 2 has an even power

of 2 in its denominator. Therefore, 2 u 2

− v 2 is not an integer, contradiction.
It follows that u, v have odd denominators, so we may work with them mod
powers of 2. Since v 2 ≡− 5(mod2),wemusthave v odd. Therefore, v 2 ≡ 1

(mod 8), so

2 u 2 ≡ 6(mod .

This implies that u 2 ≡ 3 (mod 4), which is impossible. Therefore, the pair

( a, b )=(2 , 1) is eliminated.
(a,b)=(5,1). We have

x =5 u 2

5= v 2
x +5=5 w 2 .

x −

Therefore,

5 u 2 v 2 =5 , 5 w 2 5 u 2 =5 . − −

If the denominator of one of u or v is divisible by 5, then so is the other.
But 5 u 2 then has an odd power of 5 in its denominator, while v 2 has an even
power of 5 in its denominator. This is impossible, so the denominators of
both u and v are not divisible by 5. Since w 2

− u 2 = 1, the same holds for w .
Therefore, we can work with u, v, w mod 5. We have v

≡ 0(mod5),sowe

can write v =5 v 1 .Then

u 2 5 v 1 =1 ,

−

so u 2 ≡ 1 (mod 5). Therefore, w 2 =1+ u 2 ≡ 2 (mod 5). This is impossible.

Therefore, the pair ( a, b )=(5 , 1) is eliminated.