Show that the equation $x^2+xy-y^2=3$ does not have integer solutions. Show that the equation $$x^2+xy-y^2=3$$ does not have integer solutions. 
I solved the equation for $x$:
$x=\displaystyle \frac{-y\pm\sqrt{y^2+4(y^2+3)}}{2}$
$\displaystyle =\frac{-y\pm\sqrt{5y^2+12}}{2}$
I was then trying to show that $\sqrt{5y^2+12}$ can not be an integer using $r^2\equiv 12 \pmod{5y^2}$.
I got stuck here.

Generally, If $f(x) \equiv 0 \pmod{mn}$ has a solution, then that means $mn \mid f(x)$ for some $x$. But since $m \mid mn$ and $n \mid mn$ you can see $m,n \mid f(x)$, i.e. $f(x) \equiv 0 \pmod{m}$ and $f(x) \equiv 0 \pmod{n}$. So in your case you could conclude $r^2 \equiv 12 \pmod{5}$ and get rid of $y^2$ that is unknown. The rest is obvious as others said.
