Prove x^2 ? a (mod p), x^2 ? b (mod p), and x^2 ? ab (mod p) – Solvability

Question

Let $p$ be an odd prime and $a, b \in \Bbb Z$ with $p$ doesn't divide $a$ and $a$ doesn't divide $b$. Prove that among the congruence's $x^2 \equiv a \mod p$, $\ x^2 \equiv b \mod p$, and $x^2 \equiv ab \mod p$, either all three are solvable or exactly one.

Please help I'm trying to study for final in number theory and I can't figure out this proof.

Answer 1

What you should prove directly is the following:

• The product of two quadratic residues is a quadratic residue.
• The product of a quadratic residue and a quadratic non-residue is a quadratic non-residue.

A counting argument then yields additionally:

• The product of two quadratic non-residues is a quadratic residue.

These facts combined yield the desired answer, by breaking into cases as to whether or not $a,b$ are quadratic residues.