Suppose that $p>3$ is a prime, and suppose also that $3|(p−1)$. Let $m= \frac{(p−1)}{3}$. Find and prove a simple condition or set of conditions for $p$ mod 12 that determine when $m$ is a quadratic residue mod $p$.
I know the obvious truth that $3|(p-1)$ means that $(p-1)=3k$ where $k$ is some integer. Is this just dictated by the fact that $m=\frac{(p-1)}{3}$? Is this just saying that $k$ and $m$ are the same thing?
If so, how does any of this help me? Quadratic residues have always been tough for me.
Assuming $p$ is a prime number such that $p \not= 2,3$ then using the idea that (Lord Shark pointed out) if $-3$ is quadratic residue $\mod p$ then also $m$ so we will work out when $-3$ is quadratic residue and the laws of Legendre symbol (found in this Wikipedia page Legendre symbol).
which are :
1 ) $\left( \frac{-1}{p} \right) = (-1)^{\frac{p-1}{2}} $ which is equal $1$ if $p= 1 \mod 4$ and equals $-1$ if $p=3 \mod 4$.
2 ) $\left( \frac{q}{p} \right) \left( \frac{p}{q} \right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}$
and substitute in the second law $q=3$, we get $\left( \frac{3}{p} \right) \left( \frac{p}{3} \right) = (-1)^{\frac{p-1}{2} \frac{3-1}{2}}$ which means that :
$\left( \frac{3}{p} \right) = \left( \frac{p}{3} \right) (-1)^{\frac{p-1}{2}}$
Now we know that $(-1)^{\frac{p-1}{2}}$ which is the first law, and $\left( \frac{p}{3} \right)$ equals $1$ when $p=1 \mod 3$ and equals $-1$ when $p=2 \mod 3$.
Combining all the above we can say that $m$ is quadratic residue $\mod p$ when : $p=1,7 \mod 12$ and quadratic non-residue $\mod p$ when $p =5,11 \mod 12$.
(Hint) in your case because $5,11-1 \mod 3 \not=0 $ then if we restrict $m$ to be integer for all primes which fulfill the condition that $m$ is integer,$m$ will be quadratic residue $\mod p$