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.

${\rm mod}\ 17\!:\ 5(n^2\!+3)\equiv0\,\Rightarrow\,n^2\equiv-3\,\Rightarrow\,n^4\equiv 9\,\Rightarrow\, n^8\equiv -4\,\Rightarrow\,n^{16}\equiv -1$ contra little Fermat
