# Thread: Solve x^2 - 10y^2 = 2

1. ## Solve x^2 - 10y^2 = 2

I want to show that $x^2-10y^2 = \pm 2$ is insoluble in integers. I've tried factorisation in $\mathbb{Z}[\sqrt{10}]$ but this isn't a UFD so I can't get things to work. I've also tried considering norms. Is there a more elementary idea that can show this? I assume we should look for a contradiction. By simple "odd/even" arguments it's clear that $x$ must be even and $y$ must be odd, but I can't get much further. If anyone's able to give a hint then that would be most appreciated!

EDIT: OK I've spotted it - just look at the equation modulo 10 ! Incidentally, is there a way of deleting a question of yours so that other people won't waste time trying to answer something you've already done?

2. Try modulo 4. For n any even integer, $n^2$ is congruent to 0 mod 4. For n any odd integer, $n^2$ is congruent to 1 mod 4.

3. Originally Posted by Boysilver
...Incidentally, is there a way of deleting a question of yours so that other people won't waste time trying to answer something you've already done?