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For $N$ an integer, let $$\Omega_N:=\left\{\frac{\alpha}{\|\alpha\|}|\alpha \in\textbf{Z}^n~\text{and}~\|\alpha\|^2=N\right\}.$$ For $n=3$, Linnik asked if the set $\Omega_N$ was uniformly distributed with respect to the normalised Lebesgue's mesure on the sphere as $N$ goes to infinity. In fact this is true if $N$ goes to infinity and affords values for which $\Omega_N$ is empty.

For $n=4$ the result is also true when the limit is taken such as the odd part of $N$ goes to infinity.

For $n=5$ this result is unconditionaly true.

Unfortunatly I can't find any references for the case $n=2$, and it doesn't seem trivial to me. I am asking whether the set $$\Omega_N:=\left\{\frac{\alpha}{\|\alpha\|}|\alpha \in\textbf{Z}^2~\text{and}~\|\alpha\|^2=x_1^2+x_2^2=N\right\}$$ is uniformly distributed with respect to the Lebesgue's mesure on $\textbf{S}^1$. I can guess that it will not be true for all $N$ thanks to the formula $$|\Omega_N|=4\sum_{\substack{d|N \\ d~\text{odd}}}{(-1)^{\frac{d-1}{2}}}. $$

Many thanks for your help!

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    $\begingroup$ One problem is that if $N=1 \pmod 4$ is prime, then $|\Omega_N|$ is bounded for all $N$, so obviously it is impossible for points to equidistribute. $\endgroup$
    – Matt Young
    Jun 24, 2016 at 17:25
  • $\begingroup$ Right, as @MattYoung notes, the nature of the issue is different for $n=2$, and we cannot expect equidistribution. An even more extreme situation is when $N=1 \mod 4$ is prime, so there are exactly $4$ points on the circle of radius $\sqrt N$, and certainly cannot be equidistributed. $\endgroup$ Jun 24, 2016 at 21:38
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    $\begingroup$ Have you looked at mathoverflow.net/questions/159863/… ? $\endgroup$ Jun 25, 2016 at 5:17
  • $\begingroup$ Thank you my friend Don! Nice link... Ok so this isn't a trivial question $\endgroup$
    – Stabilo
    Jun 25, 2016 at 17:26

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