Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 232

This is the same equation we started with. Since

0 <t≤ t 4 = m 2 <c, (8.13)

we have proved that for every triple ( a, b, c )with a 4 + b 4 = c 2 , there is another
solution ( r, s, t )with0 <t<c . We therefore have an infinitely descending
sequence c>t>... of positive integers. This is impossible. Therefore, there
is no solution ( a, b, c ).
Observe that m 2 >n 2 ,so c< 2 m 2 =2 t 4 . Combining this with (8.13) yields

t 4 <c< 2 t 4 .

This implies that the logarithmic height of t is approximately one fourth the
logarithmic height of c . Recall that the canonical height of 2 P is four times
the height of P . Therefore, we suspect that Fermat's procedure amounts to
halving a point on an elliptic curve. We'll show that this is the case.
We showed in Section 2.5.3 that the transformation

x = 2( z +1)
w 2

y = 4( z +1)
w 3

,

maps the curve

C : w 2 = z 4 +1

to the curve

E : y 2 = x 3 − 4 x.

If we start with

a 4 + b 4 = c 2 ,

then the point

( z, w )=( a

b , c b 2 )

lies on C . It maps to a point ( x, y )on E ,with

2( c

b 2 +1)
( a/b ) 2

= 2( c + b 2 )
a 2

x =

2( t 4 +4 r 4 s 4 +( r 4

− s 4 ) 2 )

=

(2 rst ) 2

= t
rs

2

.

This implies that

= r 2

2

x − 2= t 2

− 2 r 2 s 2
( rs ) 2

− s 2
rs

= r 2 + s 2
rs

2

x +2= t 2 +2 r 2 s 2
( rs ) 2

.