The question states: If $a \in \mathbb{Z}$, prove that $a^2$ is not congruent to 2 modulo 4 or 3 modulo 4.
Our work: It's an if then proof, so I am thinking we can do a contradiction proof. Suppose $a^2 \equiv 2 \pmod{4}$ and $a^2 \equiv 3 \pmod{4}$. Then you can write by definition \begin{align*} a^2 &= 4k + 2 \\\ a^2 &= 4k + 3 \end{align*}
but hold on, another way to write 2 mod 4 is $[2]_{4}$ which means the infinite set of all the integers in the form of $2 + 4k$ and $[3]_{4}$ which means the infinite set of all the integers in the form of $3 + 4k$.
so we one set is a set of even integers and one set is of odd integers. but our a^2 can't be in both right?