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Prove that number $1^{2005}+2^{2005}+\cdots+n^{2005}$ is not divisible by number $n+2$ for every $n\in \mathbb N$

I have solution $2(1^{2005}+2^{2005}+ \cdots +n^{2005})=2+(2^{2005}+n^{2005})+(3^{2005}+(n-1)^{2005})+\cdots+(n^{2005}+2^{2005})=2+(n+2)M$, but I do not know how he get that result and i try to do on another way. I know that $1+2+\cdots+n=\frac{n(n+1)}{2}$and that $1+2+\cdots+n| 1^{2005}+2^{2005}+ \cdots +n^{2005}$.

So $\frac{n(n+1)}{2}|1^{2005}+2^{2005}+ \cdots +n^{2005}.$ that mean $n(n+1)|2(1^{2005}+2^{2005}+ \cdots +n^{2005})$ so $2(1^{2005}+2^{2005}+ \cdots +n^{2005})=M_1n(n+1)$ now if $n+2|1^{2005}+2^{2005}+ \cdots +n^{2005}$ that mean that $1^{2005}+2^{2005}+ \cdots +n^{2005}=M_2(n+2)+r$, where $r=0$ but I need to show that is true or not so I do this.

$2(M_2(n+2)+r)=M_1n(n+1)$ , but I stuck here so maybe this is not good option to show that $r \not =0,$ do you have some idea?

Marko Škorić
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    If you are familiar with congruences then the first solution can be verified simply by using the fact that for all relevant $a$ we have ($2005$ is odd) $$(n+2-a)^{2005}\equiv(-a)^{2005}=-a^{2005}\pmod{n+2}.$$ Implying that for all $a$, the sum $a^{2005}+(n+2-a)^{2005}$ is divisible by $n+2$. This leaves you with only $1^{2005}$ and, when $n$ is even, $((n+2)/2)^{2005}$ to work with. – Jyrki Lahtonen Nov 11 '18 at 09:10
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    See this similar problem and links there on cancelling inverses in sums and products, Wilson's theorem, Gauss's trick and related (reflection) symmetry exploits. – Bill Dubuque Nov 12 '18 at 03:16

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The solution in your statement is using the following fact:

Let $k$ be an odd natural number. Then $x+y$ divides $x^k+y^k$.

The simplest way to get this is perhaps to view this as a polynomial in $x$ and then if we substitute $x=-y$, we get $x^k+y^k=x^k+(-x)^k=0$.

In your question each of the term: $(m^{2005}+(n-m+2)^{2005})$ will be divisible by $m+(n-m+2)=n+2$. This shows that \begin{align*} 2(1^{2005}+2^{2005}+ \cdots +n^{2005}) &=2+(2^{2005}+n^{2005})+(3^{2005}+(n-1)^{2005})+\cdots+(n^{2005}+2^{2005})\\ &=2+A_2(n+2)+A_3(n+2)+\dotsb+A_n(n+2)\\ &=2+(n+2)[A_2+\dotsb+A_n]\\ &=2+(n+2)M \end{align*}

Anurag A
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