<s>[INST] 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.
 [/INST] First show that since $a^2 \equiv 0,1,4$
$$7 \vert x^2 + y^2 \implies 7 \vert x \text{ and } 7 \vert y$$
We have
$$x^2 - 6y^2 = 7 \implies (x^2+y^2) - 7y^2 = 7 \implies 7 \vert (x^2+y^2) \implies 7 \vert x \text{ and } 7 \vert y$$
Hence, we have $x = 7x_1$ and $y=7y_1$, which gives us
$$7(x_1^2 - 6y_1^2) = 1$$
which is clearly impossible.
</s>