Let $p=3 \ (mod \ 4)$ and assume $a$ is a square in $\mathbb Z/p \mathbb Z $. Compute an explicit square root of $a$.
I am confused when it gives us that $p$ is modulo 4. How is this any different from saying that $p=3$? I know we are supposed to explain methods that we have already attempted, but I am pretty lost on how to even begin this problem.
For $p\equiv 3 \bmod 4$ you can compute the root explicit as $a^{\frac{p+1}{4}} \bmod p$ (see e.g. here; with fast modular exponentiation it is effective even for large $p$).
This assumes the $a$ is quadratic residue, for general $a$ the formula does not necessarily give a square root, e.g. $6$ has no square root $\bmod 11$, and $(6^3)^2 \equiv 5 \not \equiv 6 \pmod {11},$ where as for $a=3$ you have $a^3 \equiv 5 \bmod {11}$ and $5^2 \equiv 3 \pmod {11}$
