Elliptic Curves over Finite Fields - Elliptic Curves: Number Theory and Cryptography - page 118

This yields the first equality of the proposition.
To prove the second equality, multiply the Jacobi sum by its complex con-
jugate to obtain

2 =
a

a )
b

| J ( χ 2 ,χ 4 ) | χ 2 ( a ) χ 4 (1 − χ 2 ( b ) χ 4 (1 − b )

=0 , 1 =0 , 1

b χ 4 1

.

χ 2 a

=
a

− a

1 − b

=0 , 1 b =0 , 1

We have used the fact that χ 4 ( x )= χ 4 ( x ) − 1 .Wenowneedthefollowing.

LEMMA 4.28
Let S = { ( x, y ) | x, y ∈ F p ; x, y =1; x = y} .Themap

x
y ,

1 − x
1

σ :( x, y ) →

− y

is a perm utation of S .

PROOF Let c = x/y and d =(1 − x ) / (1 − y ). Then x = 0 yields c =0
and x = 1 yields d = 0. The assumption that x = y yields c, d =1and c = d .
Therefore, ( c, d ) S .

To show that σ is surjective, let c, d

∈

∈ S .Let

x = c d

− 1

y = d

− 1

c .
It is easily verified that ( c, d ) ∈ S implies ( x, y ) ∈ S and that σ ( x, y )=( c, d ).

c ,

d − d −

Returning to the proof of the proposition, we find that

b χ 4 1

+
( a,b )

b χ 4 1

χ 2 a χ 2 a

2 =
a = b

− a − a

|J ( χ 2 ,χ 4 ) |

1 − b 1 − b

∈ S

=( p − 2) +
( c,d )

χ 2 ( c ) χ 4 ( d )

∈ S

⎛
⎝

⎞

=( p − 2) +
d

⎠

χ 4 ( d ) χ 2 ( c ) − χ 2 (1) − χ 2 ( d )

=0 , 1

∈ F p c

=( p − 2) +
d

χ 4 ( d )(0 − 1 − χ 4 ( d ) 2 )

=0 , 1

χ 4 ( d ) 3 =( p − 2) − χ 4 ( d ) −

d =0 , 1 d =0 , 1

=( p − 2) + χ 4 (1) + χ 4 (1) 3 = p.