$3$ 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.

If $3$ divides $n^2+1$ then it must have solution modulo $3$. But clearly
$0^2+1\equiv 1 \pmod 3$
$1^2+1 \equiv 2 \pmod 3$
$2^2+1 \equiv 5 \equiv 2 \pmod 3$ 
Otherwise put $n=3k,3k+1,3k+2$ and see that $3$ never divides it
