15 Squares mod p

n(p) is the number of integers s such that
Explicitly,
p − 1
4
< s < p − 1
2 .
2 1, p = ±1 (mod8)
=
p −1, if p = ±5 (mod8)
Proof. Apply Gauss’ lemma to S = {1, 2,..., p−1
} with a =2. Then
2
es(2) =
1, if 2s ≤ p−1
2
−1, otherwise
Since ( 2
= p s∈S es(a) (modp), ( 2
p )=(−1)n(p) . The rest follows.
Definition: Ifx∈ R, its integral part [x] is the largest integer ≤ x.
Proposition (Formulation III of Gauss’ Lemma) Let p odd prime, and a ∈ Z
with p |a. Then
a
=(−1)
p
t (p−1)/2
, where t = [ ja
p ].
j=1
Proof: For every j ∈{1, 2,..., p−1
} it is easy to see that
2
Easy exercise:
aj = qjp +āj, with 0 < āj