# 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?
you are not to delete your problems. simply edit your thread saying you have solved it. or mark the thread as "Solved" using Thread Tools.

### x^2-10y^2=2

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