The Basic Theory - Elliptic Curves: Number Theory and Cryptography - page 36

The parameter λ for E is not unique. In fact, each of

{λ, 1

1 λ

1 , λ

− 1

λ , 1 − λ, λ ,

}

1 − λ − λ

yields a Legendre equation for E . They correspond to the six permutations
of the roots e 1 ,e 2 ,e 3 . It can be shown that these are the only values of
λ corresponding to E ,sothemap λ → E is six-to-one, except where λ =

1 , 1 / 2 , 2, or λ 2 − − λ + 1 = 0 (in these situations, the above set collapses; see

Exercise 2.13).

2.5.2 Cubic Equations

It is possible to start with a cubic equation C ( x, y ) = 0, over a field K of
characteristic not 2 or 3, that has a point with x, y ∈ K and find an invertible
change of variables that transforms the equation to Weierstrass form (although
possibly 4 A 3 +27 B 2 = 0). The procedure is fairly complicated (see [25], [28],
or [84]), so we restrict our attention to a specific example.
ConsiderthecubicFermatequation

x 3 + y 3 + z 3 =0 .

The fact that this equation has no rational solutions with xyz = 0 was conjec-

tured by the Arabs in the 900s and represents a special case of Fermat's Last
Theorem, which asserts that the sum of two nonzero n th powers of integers
cannot be a nonzero n th power when n ≥ 3. The first proof in the case n =3
was probably due to Fermat. We'll discuss some of the ideas for the proof in
the general case in Chapter 15.
Suppose that x 3 + y 3 + z 3 =0and xyz =0. Since x 3 + y 3 =( x + y )( x 2 −

xy + y 2 ), we must have x + y =0. Write
x
z

y
z

= u + v, = u − v.

Then ( u + v ) 3 +( u − v ) 3 +1=0, so 2 u 3 +6 uv 2 + 1 = 0. Divide by u 3 (since
x + y =0,wehave u = 0) and rearrange to obtain

6( v/u ) 2 = (1 /u ) 3 − − 2 .

Let

x 1 = −

6
u

z
x + y ,

y 1 = 36 v =36 x

y
x + y .

−

= − 12

u

Then

y 1 = x 1 −

432 .

It can be shown (this is somewhat nontrivial) that the only rational solutions
to this equation are ( x 1 ,y 1 )=(12 , ± 36), and ∞ . The case y 1 = 36 yields