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    $\begingroup$ And for $p\equiv3\pmod4$ with $N=p^n$ it appears that $$2S_N=p^n(p^n-p^{\lfloor n/2\rfloor}-2((p^{\lceil n/2\rceil}-1)/(p-1))h(-p)$$ where $h(-p)$ is the class number of $Q(\sqrt{-p})$. $\endgroup$ Commented Mar 11, 2023 at 13:49
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    $\begingroup$ If we put $f(n)=\sum_{i=1}^{n-1}\lfloor i^2/n\rfloor-(n-1)(n-2)/3$ then direct computer calculation shows that $f(n)\geq 0$ for $n\leq 10^5$. Also it seems experimentally that if $f(n)=0$ then $n-1$ is a square mod $24$. However, that congruence is satisfied 25% of the time, whereas $f(n)=0$ about 14% of the time for $n<10^5$. $\endgroup$ Commented Mar 12, 2023 at 13:49
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    $\begingroup$ les-mathematiques.net/vanilla/index.php?p=/discussion/2333199/… $\endgroup$ Commented Mar 12, 2023 at 15:05
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    $\begingroup$ It seems that $S_n = n(n-1)/2$ if and only if $n$ is a square-free sum of two squares. Is there an explanation for this? $\endgroup$ Commented Mar 14, 2023 at 22:44
  • $\begingroup$ Apart from these values of $n$, it seems that $S_n$ slowly falls away from $n(n-1)/2$. For example, for $n\geqslant 7$, they seem to satisfy $S_n \leqslant (n-1)(n-2)/2$, with equality only for $n=7$, $11$, $14$, $19$, $22$, $38$, $43$, $67$, $86$, $134$, $163$ and $326$. $\endgroup$ Commented Mar 15, 2023 at 9:01