Proving true or false integral solution to equation $X^2-13Y^2 = 7$ The equation $X^2 −13Y^2 = 7$ has a integer solution, such that there is a pair $(x,y) \in\Bbb Z^2$
such that $x^2 −13y^2 = 7$.
I'm struggling to understand how this is supposed to be proved without brute forcing? I am relatively a beginner to mathematical proofs, so I would like to know how you would solve a proof like this as well.
 A: Easier than $\bmod 7$ is $\bmod 13$, whereby $x^2-13y^2=7 \Rightarrow x^2\equiv 7 \bmod 13$. The squares $\bmod 13$ are $\{0,1,3,4,9,10,12\}$. Hence there can be no integer solution to the original equation.
A: We can prove that $x^2\equiv\{0,1,2,4\}\mod7$, for all $x\in\mathbb Z$. We can also prove that $13y^2\equiv\{0,3,5,6\}\mod7$ for all $y\in\mathbb Z$. So the only solutions happen when $x^2\equiv13y^2\equiv0\mod7$. This happens if and only if $x\equiv y\equiv0\mod7$.
So the left side of $x^2-13y^2$ must be a multiple of $7^2$, which is not possible as it must be equal to $7$.
There is no solution.
A: This is a generalized Pell's Equation and their theory is very heavy.
For this equation, let's work modulo 7.
Then the equation is $x^2+y^2=0$ modulo 7.
$x^2$ can be 0,1,2,4 modulo seven but then $y^2$ must be 0,6,5, or 3 modulo 7 respectively. So $x$ and $y$ must be both divisible by 7.
But then from the integral equation we get $u^2-13v^2=\frac{1}{7}$ for some integers $u,v$. Absurd.
So there is no solution.
