Prime Numbers

7.7 Exercises 379
This reduction ignites a chain of exact results for the Frobenius relation,
as we shall see.
(3) Show that x p can now be given the closed form
x p ≡ (−4b) ⌊p/3⌋ x p mod 3 (mod Ψ3),
where our usual mod notation is in force, so p mod 3 = 1 or 2.
(4) Show that xp2 canalsobewrittendownexactlyas
x p2
≡ (−4b) (p2 −1)/3 x (mod Ψ3),
and argue that for p ≡ 2 (mod 3) the congruence here boils down to
x p2
≡ x, independent of b.
(5) By way of binomial series and the reduction relation from (2) above,
establish the following general identity for positive integer d and γ ≡ 0
(mod p):
(x 3 + γ) d ≡ γ d
1 − x3 d
(1 − 4b/γ) − 1
4b
(mod Ψ3).
(6) Starting with the notion that y p ≡ y(x 3 + b) (p−1)/2 , resolve the power y p
as
y p ≡ yb (p−1)/2 q(x) (modΨ3),
where q(x) =1or(1+x 3 /(2b)) as p ≡ 1, 2 (mod 3), respectively.
(7) Show that we always have, then,
y p2
≡ y (mod Ψ3).
Now, given the above preparation, argue from Theorem 7.5.5 that for p ≡ 2
(mod 3) we have, independent of b,
#E ≡ p +1≡ 0(mod3).
Finally, for p ≡ 1 (mod 3) argue, on the basis of the remaining possibilities
for the Frobenius
(c1x, y) + [1](x, y) =t(c2x, yc3)
for b-dependent parameters ci, that the curve order (mod 3) depends on the
quadratic character of b (mod p) in the following way:
#E ≡ p +1+
b
≡ 2+
p
b
(mod 3).
p
An interesting research question is: How far can this “symbolic Schoof”
algorithm be pushed (see Exercise 7.30)?