elementary number theory - $3$ never divides $n^2+1$

Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.

Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.

If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.

And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.

Answer

Instead of considering whether $n$ is even or odd, consider the remainder when $n$ is divided by $3$; as an example, if the remainder is $1$, we have $$n = 3k + 1 \implies n^2 + 1 = 9k^2 + 6k + 2$$

which is not divisible by $3$. There are two more cases.