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.
 A: Hint: Show that, if there are any solutions $x$ and $y$, then it cannot be the case that both $x$ and $y$ are multiples of $7$. Then work modulo $7$. Which of $0,1,2,3,4,5,6$ is a square modulo $7$? What is $-6$ modulo $7$?
A: 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 $7$. 
A: 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.
A: I will generalize  $7$ with $4n-1$ where $n$ is any integer
$\implies 6=4n-2$
$$\text{so   }x^2-(4n-2)y^2=4n-1\implies x^2+y^2=(4n-1)(1+y^2)\ \ \ \ (1)$$
From this and Prove that n is a sum of two squares,  each of $x^2+y^2$ and $1+y^2$ has an even number (may be $0$) of factors $\equiv-1\pmod4$ (Proof 1 , 2)
But as $(4a+1)(4b+1)\equiv1$ and $(4a-1)(4b-1)\equiv1\pmod 4, 4n-1$ must have an odd number of  factors $\equiv-1\pmod4$
