Show that $a^{2n}$ ($n$ is a natural number) is a quadratic residue of a prime $p$, provided $p \not \mid a$.
We consider
$$ x^2 \equiv a^{2n} \pmod p $$
By proposition (3.14b) we have
$$ x \equiv \pm a^n \pmod p $$
Here $a^n$ is an integer is not divisible by $p$, because $p \not \mid a \implies p \not \mid a^n $.
And so $a^{2n}$ is a quadratic residue of prime $p$.