Quadratic Residues with $p \equiv {3 \bmod 4}$

Question

When $p$ is a prime of the form $4m+3$ for integer $m$, I've noticed that it is never the case that $a$ and $p-a$ are both quadratic residues. Why is this?

Answer 1

If they were both quadratic residues (or indeed both quadratic non-residues) we would have that $-a^2$ is a quadratic residue, and therefore $-1$ is a quadratic residue. But it is a standard fact that $-1$ is not a quadratic residue of $p$ if $p\equiv 3\pmod{4}$.

Answer 2

Use the Legendre symbol and the law of quadratic reciprocity: $$\biggl(\frac{p-a}p\biggr)=\biggl(\frac{-a}p\biggr)=\biggl(\frac{-1}p\biggr)\biggl(\frac ap\biggr)=-\biggl(\frac ap\biggr). $$ Thus, if one of them is equal to $1$, the other is $-1$.