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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.

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    $\begingroup$ Hint: can $25$ divide any of those numbers? $\endgroup$ – lulu Jun 14 '20 at 13:00
  • $\begingroup$ $v^2 \neq 3 \pmod 5$ $\endgroup$ – Mohammad Zuhair Khan Jun 14 '20 at 13:03
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    $\begingroup$ 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 $6$) $\endgroup$ – J. W. Tanner Jun 14 '20 at 13:23
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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.

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S=5(v²+2) , as S is a perfect square therefore it must be divisible by 25 implying v²+3 is divisible by 5 .

So v² should be of form 5q+3 .

But a perfect square is only of the form 5q, 5q+1 or 5q+2

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    $\begingroup$ did you mean $5q+\color{red}4$ where you wrote $5q+2$? $\endgroup$ – J. W. Tanner Jun 14 '20 at 13:25

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