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Find all integer $n$ such that $n^2$ is of the form $$3k+2.$$

In observing I find that it has no integer solution, but I can't solve it mathematically. Any hint regarding this is appreciated. Thanks in advance.

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Hint: Write $n$ as $3m+r$ with $m\in\Bbb Z$ and $r\in\{0,1,2\}$

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  • $\begingroup$ m(3m+2)+$r^2+r$=$n^2$.how I say it is not of form 3k+2? $\endgroup$ – Sathasivam K Sep 18 '16 at 16:29
  • $\begingroup$ Please explain it sir? $\endgroup$ – Sathasivam K Sep 18 '16 at 16:38
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    $\begingroup$ $(3m+r)^2=9m^2+6mr+r^2=3(3m^2+2mr)+r^2$, but none of the possible values of $r$ give $r^2=2$ $\endgroup$ – John Doe Sep 18 '16 at 20:31
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$$n^2 = 3k+2$$

Solving for k:

$$k = \frac{n^2 - 2}{3}$$

For this to be an integer,

$$n^2 - 2 \equiv 0 \mod 3$$ $$n * n \equiv 2 \mod 3$$

The fundamental property of multiplication in modular arithmetic states:

$$(a\ \%\ m) * (b\ \%\ m) \equiv (ab\ \%\ m) \mod m$$

where % is the modulo operator. Therefore:

$$n * n \equiv 2 \equiv (n\ \% \ 3)^2 \mod 3$$

$n\ \%\ 3$ can be either $0$, $1$, or $2$, so $(n\ \%\ 3)^2$ can be either $0$, $1$, or $4$ respectively. None of these are $2$ (mod $3$), so $k$ can never be an integer, regardless of what you choose n to be.

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Any perfect square , say $n^2$, when divided by $3$ leaves either $0$ or $1$ as remainder. ( Check yourself by taking $n$ as $3m$, $3m+1$ or $3m+2$ ). Here $(3k+2)$ when divided by $3$, leaves $2$ as remainder. Hence $(3k+2)$ cannot be a perfect square for any integer $k$.

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    $\begingroup$ Welcome to MSE. What is the point of posting this as an answer to a three year old question with an accepted answer? $\endgroup$ – José Carlos Santos Dec 31 '19 at 12:02

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