Find all $n$ such that $\sqrt{5n+2}$ is an integer.

Question

Here is my solution. There is no such $n$.

If $n$ is odd, then, then $5n+2 \equiv 7 \pmod {10}$. Else, $5n+2 \equiv 2\pmod {10}$.

But, the quadratic residues of $10$ are only $0,1,4,9,6,5$. Therefore, the expression is never a perfect square.

Is it correct?

Answer 1

Note that if $k$ is a perfect square then $k\equiv0,1,4\pmod5$:

• $m\equiv0\pmod5 \implies m^2\equiv0^2\equiv0\pmod5$
• $m\equiv1\pmod5 \implies m^2\equiv1^2\equiv1\pmod5$
• $m\equiv2\pmod5 \implies m^2\equiv2^2\equiv4\pmod5$
• $m\equiv3\pmod5 \implies m^2\equiv3^2\equiv4\pmod5$
• $m\equiv4\pmod5 \implies m^2\equiv4^2\equiv1\pmod5$

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$\forall{n\in\mathbb{N}}:5n+2\equiv2\pmod5\implies5n+2\text{ is never a perfect square}$

Answer 2

For it to be an integer, $5n+2$ must be a perfect square, Now the last digit of $5n+2$ can only be either a $2$ or a $7$, None of which can be the last digits of perfect squares, hence $5n+2$ can never be a perfect square.