It is a lemma used in `On the Reducibility of Cyclotomic Polynomials over Finite Fields' by Brett Harrison.
TP: For $a\in \mathbb{Z}$, $x^2 = a$ has a solution $x \in \mathbb{Z}$ IFF $x^2 \equiv a \pmod n$ for all $n \in \mathbb{Z}$.
P: "$\Rightarrow$". Assume $x^2 = a$, if we reduce this equality $\pmod n$, we obtain $x^2 \equiv a \pmod n$.
"$\Leftarrow$". Assume $x^2 \equiv a \pmod n$ for all $n \in \mathbb{Z}$. Let $b \in \mathbb{Z}$, s.t. $x^2 = a + b$, and reduce this expression $\pmod a$, using the assumption we find $b \equiv 0 \pmod a$, i.e. $b$ is a multiple of $a$. So we can write $x^2 = n a$ for some $n \in \mathbb{Z}$.
Now I'm stuck, probably the square needs to be used now.