Exercise 1.2.11 ($4$ doesn't divide $n^2 + 2$)

Proof. Assume that $4\mid (n^2+2)$.

• If $n$ is even, $n=2k$ for some integer $k$. Then $4\mid 4k^2+2$, thus $4\mid 2$. Since $4\nmid 2$, this is a contradiction.
• If $n$ is odd, $n=2k+1$ for some integer $k$. Then $4\mid {(2k+1)}^2+2=4k^2+4k+3$, thus $4\mid 3$. This is a contradiction.

Therefore the hypothesis $4\mid (n^2+2)$ leads to a contradiction in both cases. We can conclude that $4\nmid (n^2+2)$ for any integer $n$. □