Show $17$ does not divide $5n^2 + 15$ for any integer $n$

Question

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.

Answer 1

You can transform the equation a bit.

$5n^2 + 15 \equiv 0 \pmod{17}$ is equivalent to $5n^2 \equiv 2 \pmod{17}$ and further to $n^2 \equiv 14 \pmod{17} $ and finally $n^2 \equiv -3 \pmod{17} $.

This is already more convenient to check. And, if you know about quadratic reciprocity you can finish quite directly by invoking that $-3$ is a quadratic residue only for primes congruent $1 \mod 3$ while $17$ is $2 \mod 3$.

Answer 2

$5(n^2 + 3) \equiv 0 \bmod \ 17$ if and only if $n^2 \equiv 14 \bmod \ 17$ but $\forall\ n \ n^2 \equiv 1, 4, 9, 16, 8, 2, 15, 13\ \bmod \ 17$

Answer 3

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 $n^2 +3$.

Answer 4

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