Elliptic Curves over Q - Elliptic Curves: Number Theory and Cryptography - page 234

On y 2 = x 3 4 x , the points of order 2 played a role in the descent procedure

in Section 8.2. We showed that the image of the map φ was equal to the image
of E [2] under φ . If we start with a possible point P

−

∈ E ( Q ), then φ ( P )= φ ( T )

for some T E ( Q ). In Fermat's

method, the points of order 2 appear more subtly. If ( x, y )on E corresponds
to the solution a, b, c of a 4 + b 4 = c 2 , then a calculation shows that

∈ E [2]. Therefore, P − T =2 Q for some Q ∈

( x, y )+(0 , 0) ←→ − a, b, −c
( x, y )+(2 , 0)

←→ − b, a, c

( x, y )+( − 2 , 0) ←→ b, a, − c.

Since we assumed that a was even and b was odd, we removed the solutions
±b, a, ∓c from consideration. The solution −a, b, −c was implicitly removed
by the equation c = m 2 + n 2 , which required c to be positive. Therefore,
the choices that were made, which seemed fairly natural and innocent, were
exactly those that caused φ ( P ) to be trivial and thus allowed us to halve the
point.
Finally, we note that in the descent procedure for E in Section 8.2, we elim-
inated many possibilities by congruences mod powers of 2. The considerations
also appear in Fermat's method, for example, in the argument that n is even.
In Fermat's descent, the equation

b 2 = t 4 4 u 4 −

appears in an intermediate stage. This means we are working with the point
( w, z )=( u/t, b/t 2 )onthecurve

C : w 2 = − 4 z 4 +1 .

The transformation (see Theorem 2.17)

x = 2( z +1)
w 2

y = 4( z +1)
w 3

,

maps C to the elliptic curve

E : y 2 = x 3 +16 x .

Thereisamap ψ : E → E given by
( x ,y )= ψ ( x, y )= y 2

.

y ( x 2 +4)
x 2 x 2 ,

Thereisalsoamap ψ : E → E given by
( x, y )= ψ ( x ,y )= y 2

.

y ( x 2 − 16)

8 x 2

4 x 2 ,