This is a 4 part question on my tutorial and if you prove part d) the rest follow, except there's no part d) in the question! I can prove a), b) and c) without using whatever d) is but just thought I'd see if you guys know what this proof is actually proving!

Q. Let p be an odd prime, and a, b be integers not divisible by p. Prove that

a) a \equiv b (mod p) implies that $\displaystyle \bigg{(}\frac{a}{p} \bigg{)} = \bigg{(}\frac{b}{p} \bigg{)}$

b) $\displaystyle \bigg{(}\frac{ab}{p} \bigg{)} = \bigg{(}\frac{a}{p} \bigg{)}\bigg{(}\frac{a}{p} \bigg{)}$.

c) $\displaystyle \bigg{(}\frac{a^2}{p} \bigg{)}=1$; $\displaystyle \bigg{(}\frac{a^2b}{p} \bigg{)} = \bigg{(}\frac{b}{p} \bigg{)}$.

(I though perhaps part d) was the second equality in part c) but I don't think the proof matches up.

PROOF.

a), b) and c) follow from d) so we should prove that first!

d) Take a primitive root g, with $\displaystyle a \equiv g^k$ mod $\displaystyle p$.

Then $\displaystyle a^{\tfrac{p-1}{2}} = g^{\tfrac{(p-1)k}{2}}$.

As $\displaystyle g^{\tfrac{(p-1)}{2}} \equiv -1$ mod $\displaystyle p$ we see that...

$\displaystyle g^{\tfrac{(p-1)k}{2}} \equiv 1$ mod $\displaystyle p$ if $\displaystyle k$ is even (i.e if a is a quadratic residue),

while $\displaystyle g^{\tfrac{(p-1)k}{2}} \equiv -1$ mod $\displaystyle p$ if $\displaystyle k$ is odd (i.e a non quadratic residue). Hence result.

I was thinking it was probably something like show that if g is a primitive root and k is even then,

$\displaystyle \bigg{(}\frac{g^k}{p} \bigg{)} = 1$. and if k is odd then ... = -1.