Prove that $x^4 \equiv a^2 \pmod p$ is solvable for all $a$

Question

Let $p$ be a prime such that $p \equiv 3 \pmod{4}$

Prove that $x^4 \equiv a^2 \pmod p$ is solvable for all $a$.

I'm asked to work out this problem and I'm wondering if my approach is correct?

My solution:

If $p|a$ then letting $x=0$ is a solution.

If $p \nmid a$, first factorize to get $(x^2 -a)(x^2-(-a)) \equiv 0 \pmod p$

and then $\left(\frac{a}{p}\right) = 1$ or $\left(\frac{a}{p}\right) = -1$.

If $\left(\frac{a}{p}\right) = 1$ then there is a solution.

If $\left(\frac{a}{p}\right) = -1$ then $\left(\frac{-a}{p}\right)= \left(\frac{-1}{p}\right)\left(\frac{a}{p}\right)= (-1)(-1) = 1$ because $\left(\frac{-1}{p}\right)=-1$ if $p\equiv 3 \pmod4$ and there is a solution.

Answer 1

Your solution is fine. Here's an another way to see it that doesn't rely on knowing how to manipulate quadratic residues.

Since $\Bbb Z_p^*$ is cyclic of order $p-1$, let $b$ be a generator. Then $a^2=b^{2k}$ for some $k \in \Bbb N$. If $k$ is even, we're done. If $k$ is odd, then $b^{2k}=b^{2k+p-1}$ and since $p \equiv 3 \pmod 4, p-1=2m$ for some $m$ odd, so that $k+m$ is even, $4 \mid 2(k+m)$ and again we're done.