Question

2. Let p be an odd prime. We saw last week that the problem of counting solutions to the congruence (mod p) is only interesting when p has the form 4k1. For the rest of this problem let p 4k+1. (a) Show that (mod -1 5 (mod 8) (b) Show that pEl (mod 8 -1 p5 (mod 8) (c) Draw condlusions about the number of solutions to these congruences 14 (mod p) -1 (mod p) (mod p)

Answer 1

a) If
p1 mod 8
then $p=8k+1$, so that
= (-1)2k-1 (-1)
.

If
p 5 mod 8
then
p=8k + 5
, so that
2k+1-1
​​​​​​​.

b) We have

​​​​​​​

2 4

Therefore, if
p1 mod 8
then $p=8k+1$, so that

64k2-16k

and if
p 5 mod 8
then
p=8k + 5
, so that

4부 = (-1)수 = (-1)-부-= (-1)Ska +10k +4.1

c) Consider
ー-1 mod P
. There is a solution if and only if there is such that $x^8\equiv 1\mod p$ but $-1\equiv x^4\not\equiv 1\mod p$ , equivalently, iff
(Z/pZ)*
has an element of order . Since
(Z/pZ)*
is cyclic of order , there is an element of order iff
8
, which is equivalent to
p1 mod 8
. In that case, $x,x^3,x^5,x^7$ are distinct elements of order in
(Z/pZ)*
. Thus, the number of solutions to
ー-1 mod P
is if
p1 mod 8
, and if
p 5 mod 8
.

Consider $x^4\equiv 4\mod p$ . This is solvable if and only if $x^2\equiv \pm 2\mod p$ has a solution. By part c) we have

$\begin{pmatrix}\frac {\pm 2}p\end{pmatrix}={\pm 2}^{\frac{p-1}2}=4^{\frac{p-1}4}=-1$

if , and

p 5 mod 8

4 2

if
p1 mod 8
. Thus, the number of solutions is if
p1 mod 8
, and if
p 5 mod 8
.

Same for $x^4\equiv -4\mod p$ ​​​​​​​ by the previous two cases.