# prove that $5v^2+10$ cannot be a perfect square

I was able to prove that $$5v^2+10$$ cannot be a perfect square by working on the integers modulo 4 where the value of perfect squares are either 0 or 1. Initially, I tried to work in the integers modulo 5 but I couldn't arrive at a contradiction there. I would like to understand what kind of intuition a mathematician uses to pick the right modulo space where the contradiction can be shown. Is it simply trial and error or is there some kind of deeper reasoning / cleverness involved? And if so, how should I go about building this kind of intuition? Number theory books are full of clever solutions but there appears to be little in the way of methodology unlike other (undergraduate-level) math areas.

For the record my proof was that, in the integers modulo 4, perfect squares are either 0 or 1. Therefore $$5*a+2$$ must be congruent to 0 or 1 (mod 4), with $$a$$ being either 0 or 1, which is impossible.

• Hint: can $25$ divide any of those numbers? – lulu Jun 14 '20 at 13:00
• $v^2 \neq 3 \pmod 5$ – Mohammad Zuhair Khan Jun 14 '20 at 13:03
• to address your question (picking the right modulus), note that there aren't very many different squares modulo $3$ or $4$ (their totient is $2$), and there aren't very many different cubes modulo $7$ and $9$ (their totient is $6$) – J. W. Tanner Jun 14 '20 at 13:23

Let $$k^2=5v^2+10,\qquad k\in\Bbb Z$$ Now, $$5\bigg|k^2\implies5\bigg|k$$, since $$5$$ is prime. Therefore, let $$k=5l, \qquad l\in\Bbb Z\\ \implies 5l^2=v^2+2$$ Now note that $$LHS$$ is divisible by $$5$$, whereas $$RHS$$ is not. We get a contradiction.
• did you mean $5q+\color{red}4$ where you wrote $5q+2$? – J. W. Tanner Jun 14 '20 at 13:25