Let $p$ be an odd prime and $q$ be a positive integer that is not divisible by $p$. Show that
\[\sum_{k=1}^{p-1} \left\lfloor (-1)^k\cdot\frac{k^2q}{p}\right\rfloor=\frac{(p-1)(q-1)}{2}\]
Again Sum with floor
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Unread post by mutasimmim » Sat Oct 18, 2014 10:25 am
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