There are no integers $x,y$ such that $x^2-6y^2=7$ How to show that there is no here are no integers $x,y$ such that $x^2-6y^2=7$?
Help me. I'm clueless.

We work modulo $8$. Clearly $x$ cannot be even. So $x^2\equiv 1\pmod{8}$.
If $y$ is even, then $6y^2\equiv 0\pmod{8}$, so $x^2-6y^2\equiv 1\pmod{8}$. But $7\not\equiv 1\pmod{8}$.
If $y$ is odd, then $y^2\equiv 1\pmod{8}$, so $6y^2\equiv 6\pmod{8}$. It follows that $x^2-6y^2\equiv -5\pmod{8}$. But $7\not\equiv -5\pmod{8}$. 
Remark: The $7$ in the problem, combined with the $6$, shout "work modulo $7$," so I would consider that the natural approach. This is particularly the case if the question is in the quadratic congruences chapter, since the key fact is that $-1$ is not a quadraic residue of $7$. 
