Show $17$ does not divide $5n^2 + 15$ for any integer $n$ Claim: $17$ does not divide $5n^2 + 15$ for any integer $n$.
Is there a way to do this aside from exhaustively considering $n \equiv 0$, $n \equiv 1 , \ldots, n \equiv 16 \pmod{17}$ and showing $5n^2 + 15$ leaves a remainder of anything but $0$. It's easy but tedious. Since if $n \equiv 0$ then $5n^2 + 15 \equiv 15$ so that 17 does not divide $5n^2 + 15$. I did four more cases successfully, got bored, and skipped to the last case which also worked. Thanks in advance for any responses.

Let $ n=17k + p $ where $ p, k \in \mathbb{N}$ and $0\le p \le 16 $.
Therefore $ 5n^2+15=5 \cdot 17^2 k^2+170pk +5p^2 +15 \equiv 5p^2 -2 \; (mod \; 17)$
PS. You can do it for $n^2 +3$.
