<s>[INST] 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?
 [/INST] $$x^2-37y^2=2\Rightarrow x^2-y^2\equiv2\mod(4)\Rightarrow x^2\equiv 2+y^2\mod(4)$$
but $2+y^2\equiv 2\mod(4)$ or $2+y^2\equiv 3\mod(4)$ since any square has remainder $0$ or $1$ when divided by $4$. Either case is impossible since $x^2$ is a square itself. 
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