Let p = 2017.
Find the remainder when
$\lfloor 1^p/p \rfloor + \lfloor 2^p/p \rfloor + \lfloor 3^p/p \rfloor + .... + \lfloor 2015^p/p \rfloor$
is divided by p.
How do I solve this number theory question using Fermat's Little Theorem? Any help would be really appreciated.
Since $n^p\equiv n\pmod p$ for all $n=1,2,\dots,p-1$, we have $$n^p=\lfloor n^p/p\rfloor p+n,$$ from which $$\sum_{n=1}^{p-2}n^p=p\sum_{n=1}^{p-2}\lfloor n^p/p\rfloor+\sum_{n=1}^{p-2}n,$$ whence \begin{align} S &=\sum_{n=1}^{p-2}\lfloor n^p/p\rfloor\\ &=\sum_{n=2}^{p-2}\frac{n^p-n}p\\ &=\sum_{n=2}^{(p-1)/2}\left(\frac{n^p+(p-n)^p}p-\frac{n+(p-n)}p\right)\\ &=\sum_{n=2}^{(p-1)/2}\left(\frac{n^p+(p-n)^p}p-1\right)\\ \end{align} Since $(p-n)^p\equiv -n^p\pmod{p^2}$, we get \begin{align} S &\equiv -\sum_{n=2}^{(p-1)/2}1\\ &=-\left(\frac{p-1}2-1\right)\\ &=-\frac{p-3}2\\ &\equiv\frac{p+3}2\pmod p \end{align} For $p=2017$ we obtain $S\equiv 1010\pmod{2017}$.