Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 220

follows that there is no prime dividing c 1 c 2 and d ,sothegcdinthelemmais
1.

We can apply the lemmas to a 3 ,a 4 ,b 3 ,b 4 .Sin e cd a 3 ,b 3 )=1and
gcd( a 4 ,b 4 )=1,wehave

gcd( a 3 a 4 ,a 3 b 4 + a 4 b 3 ,b 3 b 4 )=1 .

Therefore, there exist integers x, y, z such that

a 3 a 4 x +( a 3 b 4 + a 4 b 3 ) y + b 3 b 4 z =1 .

Since

g 3 ( a 3 b 4 + a 4 b 3 )= g 1 ( b 3 b 4 )and g 3 ( a 3 a 4 )= g 2 ( b 3 b 4 ) , (8.8)

we have

g 3 = g 3 ( a 3 a 4 ) x + g 3 ( a 3 b 4 + a 4 b 3 ) y + g 3 ( b 3 b 4 ) z
= g 2 ( b 3 b 4 ) x + g 1 ( b 3 b 4 ) y + g 3 ( b 3 b 4 ) z.

Therefore, b 3 b 4 |g 3 ,so

|b 3 b 4 |≤|g 3 |.

Similarly,

| a 3 a 4 |≤| g 2 | .

Equation 8.8 and the fact that |b 3 b 4 |≤|g 3 | imply that

| a 3 b 4 + a 4 b 3 |≤| g 1 | .

In terms of the nonlogarithmic height H , these inequalities say that

H ( P + Q ) · H ( P − Q )=Max( |a 3 |, |b 3 | ) · Max( |a 4 |, |b 4 | )
≤ 2Max( | a 3 a 4 | , | a 3 b 4 + a 4 b 3 | , | b 3 b 4 | )

≤ 2Max( | g 2 | , | g 1 | , | g 3 | ) .

Let H 1 =Max( |a 1 |, |b 1 | )and H 2 =Max( |a 2 |, |b 2 | ). Then

|g 1 | = | 2( a 1 b 2 + a 2 b 1 )( Ab 1 b 2 + a 1 a 2 )+4 Bb 1 b 2 |
≤

H 1 H 2

2( H 1 H 2 + H 2 H 1 )( | A | H 1 H 2 + H 1 H 2 )+4 | B |

) H 1 H 2 . ≤ 4( | A | +1+ | B |

Similarly,

|g 2 |≤ ((1 + |A| ) 2 +8 |B| ) H 1 H 2 , |g 3 |≤ 4 H 1 H 2 .

Therefore,

H ( P + Q ) · H ( P − Q ) ≤ CH 1 H 2 = CH ( P ) 2 H ( Q ) 2