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Find $m,n\in \mathbb{Z}$ s.t.

$$ \sum\limits_{k = 0}^{mn - 1} {\left( { - 1} \right)^{\left\lfloor {\frac{k} {m}} \right\rfloor + \left\lfloor {\frac{k} {n}} \right\rfloor } } = 0 $$ See here for motivation, I said something there that I can not understand now

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  • $\begingroup$ It is easy to see that $m\,n-1$ must be even, and this implies that $m$ and $n$ must be odd. A little experimentation suggest that the solution is all pairs $(m,n)$ with $m,n$ odd $\ge3$ and $\gcd(m,n)=1$. $\endgroup$ – Julián Aguirre Jun 19 '17 at 10:55

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