Prove that there exist infinitely many integers \(n\) which satisfy \(2017^2 | 1^n + 2^n + ... + 2017^n\).
Problem
Source: China Northern Mathematical Olympiad 2017
Tags: Divisibility, number theory
29.07.2017 16:31
we have $1^n+2^n+...+2017^n=\sum_{k=0}^{1008} (2017-k)^n+k^n$ take $n=2017^k.m$ where $m,k \geq 1$ by LTE : $v_{2017}(k^n+(2017-k)^n)=v_{2017}(2017)+v_{2017}(n)=1+v_{2017}(m)+k \geq 2 $
29.07.2017 21:35
Faster: Note that n=3 is a solution, and that the value of $1^n + 2^n + ... + 2017^n$ must eventually become cyclic modulo $2017^2$, hence we have an infinite arithmetic progression of solutions.
21.06.2018 13:00
21.06.2018 14:28
@Devastator It is called Euler's theorem, Fermat Euler theorem, or the Euler Totient theorem.
21.06.2018 14:37
Muradjl wrote: we have $1^n+2^n+...+2017^n=\sum_{k=0}^{1008} (2017-k)^n+k^n$ take $n=2017^k.m$ where $m,k \geq 1$ by LTE : $v_{2017}(k^n+(2017-k)^n)=v_{2017}(2017)+v_{2017}(n)=1+v_{2017}(m)+k \geq 2 $ it's awesome!!!