I'm trying to prove that $17$ does not divide $c^2$, given $c^2=a^2-7b^2$, where $a$, $b$, and $c$ are three relatively prime positive integers.
I initially tried to look at the four cases given by the two possible forms of $a^2$ and $b^2$, namely, $4k+1$ and $4k$ (all squares are of one of these two forms). So we have:
(1) $a^2\equiv 0$ and $b^2\equiv 0$ (mod 4);
(2) $a^2\equiv 1$ and $b^2\equiv 0$ (mod 4);
(3) $a^2\equiv 0$ and $b^2\equiv 1$ (mod 4); and
(4) $a^2\equiv 1$ and $b^2\equiv 1$ (mod 4).
For cases (1), (3), and (4), I was able to find a contradiction, if we assume 17 does indeed divide $c^2$. For example, for case (1), we know $a^2-7b^2 \equiv 0$, thus, if $c=17x$ for some nonzero integer $x$, then $289x^2 \equiv 0$ i.e. $x^2\equiv 0$. $x$ was defined not to be zero, so we have a contradiction.
But for case (2), this method doesn't work. We have $a^2-7b^2 \equiv 1$ and thus $x^2\equiv 1$, which doesn't seem to be a contradiction. Or is it? I'd love some direction here; or re-direction, if this is just totally the wrong way of trying to approach this problem.
Suppose $a^2\equiv7b^2; a,b\not\equiv0\bmod17$. Then $b$ has a $\bmod17$ multiplicative inverse $d$ so $a^2d^2\equiv(ad)^2\equiv7\bmod 17$, making $ad$ a nonzero $\bmod 17$ residue whose square is $\equiv7$. Now try to find what this nonzero residue could be. Uh-oh.