Integer solutions for $x^2-2y^2=10$ Prove there is no integer solutions $(x,y)$ for the equation $x^2-2y^2=10$.
This seems like a basic question and I had a few techniques in my mind. I tried to look at the equation modulo 2 to 13 and show there is no solution but none of them worked. I also tried to derive a contradiction by looking at the parity of $x$ and $y$ but I didnt get that to work either. Is there some theory I don't know about that would help (this looks a lot like pell's equation) or maybe I made a mistake in my efforts? Thanks for the help!
 A: Note that $x = 2k$ (otherwise LHS will be odd). Hence you have:
$4k^2 - 2y^2 = 10$, now this is equal to $2k^2 - y^2 = 5$
Consider this equation modulo $5$. $2k^2 $gives you reminders equal to $\{0,2,3\}$, and $y^2$ gives you $\{0,1,4\}$.
So it should be $5|k,y$. Now it's almost done.
A: If you consider the equation modulo $5$, you get $x^2 = 2 y ^2$, from which you can deduce $5 | x, 5 | y$ (if you haven't done that already, you may want to check your calculations). This is very pretty since it doesn't involve multiple cases. So it is probable that we are on the right track. Now you can set $x' = x/5, y'=y/5$, and you get $5(x')^2 - 10(y')^2 = 2$. Consider the equation modulo 5 again.
Or, alternatively, consider the equation modulo 25, this solves the problem instantly.
A: The argument mod $5$, as given by others, is probably the easiest way to go, but here's a parity-only proof that there are no solutions.
If $x^2-2y^2=10$, it's clear that $x$ must be even, and easy to see that $y$ must be odd. Writing $x=2u$ and $y=2v+1$, the equation becomes $4u^2-2(2v+1)^2=10$, which simplifies to
$$u^2-2v(v+1)=3$$
We now see that $u$ must be odd. Writing $u=2w+1$, the equation now simplifies to
$$2w(w+1)-v(v+1)=1$$
But $v(v+1)$ is necessarily even, so this last equation has no solutions.
Remark: Knowing that $v(v+1)$ is even, one might also go from $u^2-2v(v+1)=3$ to $u^2\equiv3$ mod $4$ and cite the fact that $3$ is not a square mod $4$; the parity-only approach taken here is essentially a proof of that fact.
