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I thought of this problem after thinking of how $n=3$ seems to have no solution to the below problem.

What numbers $n$ exist such that there are $n$ integers $a_1,a_2,a_3,a_4,...,a_n$ such that their standard deviation is a natural number? (Take the numbers as a population)

Overall progress: $n=1$ and $n=2k$ for $k$ is an integer both work.

Progress for the current case I'm working on: $n=3$ is very difficult to find, but thinking a lot, the current problem is to find $x^2+xy+y^2=6p^2$ for an integer $p.$ From there, I could use complex factoring, but the sum of squares is just as unwieldy.

I think that this can be proved/disproved using NT, but I'm stuck on how to go from there.

What do I do now?

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    $\begingroup$ If your $n$ integers are all equal, then the standard deviation is zero. Do you not allow your $n$ integers to be equal? Do you not consider zero to be a natural number? $\endgroup$ – JMoravitz Sep 19 '18 at 21:40
  • $\begingroup$ Also for the case of $n=3$, if you have any three consecutive integers, their standard deviation will be $1$ $\endgroup$ – WaveX Sep 19 '18 at 21:44
  • $\begingroup$ Hm... I got it as $\sqrt{\frac23}...$ am I using a different function? $\endgroup$ – Jason Kim Sep 19 '18 at 21:47
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    $\begingroup$ Biaised or unbiaised ? $\endgroup$ – Yves Daoust Sep 19 '18 at 22:00
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    $\begingroup$ It depends on what kind of standard deviation we are taking. If these sets of $n$ integers are treated as a sample, then divide by $n-1$. But if you treat them as a population, then your formula is correct dividing by $n$ so it depends on the context of the problem $\endgroup$ – WaveX Sep 19 '18 at 22:04

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