$ \sum_{n = 1}^{103} n^{p - 1}\equiv 0.[\frac {103}{p}] + 1(103 - [\frac {103}{p}](mod p)\Rightarrow 103\equiv[\frac {103}{p}](mod p)(1)$. Let $ 103 = p.q + r$ and $ 0\le r\le p - 1$ .Then $ (1)$ is equivalent to $ r \equiv q(mod p)$ 1 case: $ q < p$, then $ r = q$ and we have $ 103 = pr + r = (p + 1)r$ but since 103 is prime then $ r = 1,p + 1 = 103$ -contradiction because $ p$ is odd. 2 case: $ q\ge p$ , we have $ 103 = p.q\ge p^2$ and thus we obtain $ p\le \sqrt {103}\Rightarrow p\in { 3,5,7 }$ and using $ (1)$ we finally obtain that the only solution is $ p=3$

Could anyone explain the first step in above solutionbecause I don't understand it ? thanks a lot

matex (L)(L)(L) wrote: Could anyone explain the first step in above solutionbecause I don't understand it ? thanks a lot Think of Fermat's "little theorem" ...

An easy one If $(n,p)=1 \implies n^{p-1} \equiv 1$ mod $p$ $\implies \sum_{n=1}^{103}n^{p-1} \equiv 103 - [\frac{103}{p}]$ mod $p$ The rest is as @broniran