Is there positive integer $n$, such that $n+2$ divides $S=1^{2019}+2^{2019}+...+n^{2019}$
Problem
Source: Saudi Arabia JBMO training test 2 2019, P3
Tags: number theory
twinbrian
17.04.2020 21:14
I believe the answer is no. If $n+2$ is even, we notice that $n^{2019} \equiv -2^{2019}(\text{mod} n+2)$ and so this cancels out with $2^{2019}$. Similarly, we apply the similar process for $n-3$ and $3$ and so on. Thus, since both sides are symmetric(excluding the $1^{2019}$), we obtain that $1^{2019}+2^{2019}...+n^{2019} \equiv 1^{2019}+(\frac{n}{2})^{2019} \equiv 0(\text{mod} n+2)$. This means that $\frac{n}{2} \equiv -1(\text{mod} n+2)$, a contradiction when $n$ is even. If $n+2$ is odd, we obtain that $1^{2019}+2^{2019}...+n^{2019} \equiv 1^{2019}(\text{mod} n+2)$, a contradiction as well.
@below, may I ask why? I mean yeah you can use order and primitive roots, or even Bernoulli's or LTE pairing to deduce that $n$ must be odd and $n|2019$, but in the end, that will lead to my case $2$.
Al3jandro0000
18.04.2020 04:14
twinbrian wrote:
I believe the answer is no. If $n+2$ is even, we notice that $n^{2019} \equiv -2^{2019}(\text{mod} n+2)$ and so this cancels out with $2^{2019}$. Similarly, we apply the similar process for $n-3$ and $3$ and so on. Thus, since both sides are symmetric(excluding the $1^{2019}$), we obtain that $1^{2019}+2^{2019}...+n^{2019} \equiv 1^{2019}+(\frac{n}{2})^{2019} \equiv 0(\text{mod} n+2)$. This means that $\frac{n}{2} \equiv -1(\text{mod} n+2)$, a contradiction when $n$ is even. If $n+2$ is odd, we obtain that $1^{2019}+2^{2019}...+n^{2019} \equiv 1^{2019}(\text{mod} n+2)$, a contradiction as well.
I think your solution is wrong. The idea in this kind of problem is ever consider the primitive root modulo $p $ where $p $ is a prime divisor of $n+2$. Using this we get if $n $ is odd so $n\mid S $. Sorry my bad, your solution is right!