How to show that $x^2 - 37y^2 =2$ does not have integer solutions We need to prove that $x^2 - 37y^2 =2$ does not have integer solutions.
I have two angles I thought about approaching it from:


*

*Since 37 is prime, I can show that for $x$ not divisible by $37$, we have $x^{36} ≡ 1mod(37)$ but I don't see how that's useful





*I could manipulate the equation and make it to: $x^2 - 4 = 37y^2 - 2$
$\implies (x-2)(x+2) = 37y^2 - 2$ 


Then if the RHS is even, then $y^2$ is even $\implies$ $y^2$ ends with $0, 4,$ or $6$ $\implies$ $37y^2$ ends with $0, 8,$ or $2$ 
$\implies 37y^2 -2$ ends with $0, 6,$ or $8$
But then I reach a dead end here too

Any suggestions or ideas?

$x^2 - 37y^2 = 2$ is even.  So as $odd \pm even = odd$, $even + even = even$, $odd + odd = even$ we can see that either $x^2$ and $37y^2$ are either both even or both odd and we can pursue that and get a contradiction.
But now would be a nice time to point out that for all integer $m^2 \not \equiv 2 \mod 4$ and $m^2 \not \equiv 3 \mod 4$ and that either $m^2 \equiv 0 \mod 4$ or $m^2 \equiv 1 \mod 4$.
Prook:  Let $m = 2k + i$ where $i= 0,1$.  Then $m^2 \equiv 4k^2 + 4ki +i^2 \equiv i^2 \mod 4$.  And $i^2$ is either $0$ or $1$.
So $x^2 \equiv \{0,1\} \mod 4$ and $37y^2 \equiv y^2 \equiv \{0,1\} \mod 4$.
So $x^2 - 37y^2 \equiv \{0,1\} - \{0,1\} \equiv \{0-0,0-1,1-0,1-1\} \equiv \{0,3,1,0\} \mod 4$.  And $x^2 - 37y^2 \equiv 2 \mod 4$ is just about the only possibility  that can never happen.
