$3$ 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.

Since $n-1,n,n+1$ are three successive integers so one of them must be divisible by 3 hence there product must be divisible by $3$ i.e. $3|(n-1)n(n+1) \implies 3|n^3-n$ , now 
$3|n^2+1 \implies 3|n(n^2+1) \implies 3|n^3+n \implies 3|n^3+n-(n^3-n) \implies 3|2n$ , since 3 does 
not divide $2$ so $3|n \implies 3|n^2 \implies 3|n^2+1-n^2 \implies 3|1$ , contradiction ! 
