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Another modest suggestion: Set $f(n)=\sum_{i=1}^{n-1}\lfloor i^2/n\rfloor-(n-1)(n-2)/3$. Then it seems to be the case that $f(ab)\ge f(a)+f(b)$ for all positive integers $a$ and $b$. (Checked for all $a,b$ where $ab\le30000$.) Since $f(n)\ge0$ for all primes $n$, one is reduced to prove this inequality. Maybe some clever manipulation of \begin{equation} \sum_{i=1}^{ab-1}\left\lfloor\frac{i^2}{ab}\right\rfloor=\sum_{0\le i<b, 0\le j<a}\left\lfloor\frac{(ai+j)^2}{ab}\right\rfloor \end{equation} yields some progress/insight?