Prove the following results for an odd prime $p$:
(a) The product of two quadratic non-residues of $p$ is a quadratic residue of $p$.
(b) The product of a quadratic residue and quadratic non-residue of $p$ is a quadratic non-residue.
(c) The square of a quadratic residue of $p$ is a quadratic residue.
(a) Consider two quadratic non-residues, $a$ and $b$, of odd prime $p$.
By Euler's Criterion (7.5) and Proposition (7.6) we have
$$ a^{\frac{p-1}{2}} \equiv b^{\frac{p-1}{2}} \equiv -1 \pmod p $$
And so,
$$ (ab)^{\frac{p-1}{2}} \equiv 1 \pmod p $$
By Euler's Criterion (7.5), this tells us the product of two quadratic non-residues is a quadratic residue.
(b) Consider a quadratic residue $a$, and a quadratic non-residue $b$, of odd prime $p$.
By Euler's Criterion (7.5) and Proposition (7.6) we have
$$ \begin{align} a^{\frac{p-1}{2}} & \equiv 1 \pmod p \\ \\ b^{\frac{p-1}{2}} & \equiv -1 \pmod p \end{align}$$
And so,
$$ (ab)^{\frac{p-1}{2}} \equiv -1 \pmod p $$
By Euler's Criterion (7.5), this tells us the product of a quadratic residue and quadratic non-residue of $p$ is a quadratic non-residue.
(c) Consider a quadratic residue $a$ of odd prime $p$.
By Euler's Criterion (7.5) and Proposition (7.6) we have
$$ a^{\frac{p-1}{2}} \equiv 1 \pmod p $$
And so,
$$ (a^2)^{\frac{p-1}{2}} \equiv 1 \pmod p $$
By Euler's Criterion (7.5), this tells us the square of a quadratic residue of $p$ is a quadratic residue.
Note: I think the author's solution to part (c) is wrong.