Prove $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$

Question

Prove that for any odd natural number $n$, the number $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$.

Answer 1

We know, $$r^{2m+1}+(n+2-r)^{2m+1}\equiv0\pmod{n+2}$$ for any integer $m\ge0$

Let $r=1,2\cdots,n,n+1 $ and add to get $$T=2\sum_{1\le r\le n+1}r^{2m+1}\equiv0\pmod{n+2}$$

$$\text{If the given expression } S=\sum_{1\le r\le n}r^{2m+1}\equiv0\pmod{n+2},$$

$n+2$ will definitely divide $T-2S=2(n+1)^{2m+1}$

Observe that $(n+1,n+2)=1$

Answer 2

Hint $ $ If $\,f(x)\,$ is an integer coefficient polynomial that is odd $\,f(-n) = - f(n),\,$ then we can employ this reflection symmetry to group the sum into pairs that cancel (sum $\equiv 0)$, i.e.

$\!\begin{eqnarray} \rm mod\ n\!+\!2\!:\,\ \color{#c00}{n\equiv\color{#0a0}{ -2}}\ \Rightarrow &&\ f(1)+f(2)+f(3)+\cdots+f(\color{#c00}n-1)&+&\ f(\color{#c00}n)\\[3px] &\equiv&\ f(1)+\color{#0a0}{f(2)}+\color{blue}{f(3)}+\cdots+\ \ \color{blue}{f(-3)}& +& \color{#0a0}{f(-2)}\end{eqnarray}$

But by $\,f\,$ odd, $\ \color{#0a0}{f(-2) = -f(2)},\ \color{blue}{f(-3) = -f(3)},\ldots$ all cancel out, leaving only $\,f(1)\equiv 1$.

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Remark $\ $ Here is a similar example from this now deleted question

How to prove that the number $$1^{2015}+2^{2015}+3^{2015}+\cdots+2015^{2015}$$ is divisible by $2015$?

Hint $ $ If $\,f(x)\,$ is a polynomial with integer coefficients that is odd, i.e. $\,f(-x) = -f(x),\,$ e.g. $\,f(x) = x^{2015},\,$ then $\,f(n)+f(-n) = 0,\,$ so applying Gauss / Wilson reflection

$$\begin{align}{\rm mod}\,\ 2k\!+\!1\!:\ \ \color{#c00}{2k\equiv -1}&,\ \color{#0a0}{2k\!-\!1\equiv -2},\ldots,\ \ \ \color{#a0f}{k\!+\!1\equiv -k}\\[4px] \Longrightarrow\qquad\ \ \ f(1)&\ \ \,+\,\ \ f(2)\ + \cdots + f(k)\\ +\ f(\color{#c00}{2k})&\!+\!f(\color{#0a0}{2k\!-\!1})+\cdots+f(\color{#a0f}{k\!+\!1})\\[4px] \equiv\quad f(1)&\ \ \,+\,\ \ f(2)\ + \cdots + f(k)\\ f(\color{#c00}{-1})&\ +\ f(\color{#0a0}{-2})\ +\cdots+f(\color{#a0f}{-k})\\[4px] \equiv\quad\ \color{#c00}0\quad & \ +\quad \color{#0a0}0\quad\ \ + \cdots +\ \ \ \color{#a0f}0 \end{align}\qquad\qquad\qquad$$

Note $ $ Inserting into the sum the term $\,f(0) = 0\,$ then the sum is over the complete system of residues $\, 0,1,2\ldots {2k},\,$ and the reflection method amounts to replacing this by the symmetric system $\, -k,\ldots,-1,0,1,\ldots, k,\,$ where the reflection about the midpoint $\,0\,$ is given simply by negation $\, n \mapsto -n \pmod{2k\!+\!1}.\,$ Such reflection (involution) symmetry is ubiquitous, e.g. at the heart of Wilson's theorem in its various forms.

Guass's grade school trick for summing an arithmetic progression is the special case $\,f(x) = x$.

Above we used standard congruence arithmetic rules, esp. the Polynomial Congruence Rule $\, A\equiv a\,\Rightarrow\,f(A)\equiv f(a)\,$ for any polynomial $\,f(x)\,$ with integer coefficients. In your case we have that $\,f(x) = x^{2015},\,$ which is odd, so the Hint applies.

Answer 3

By taking modulo $n + 2$, and partition the terms as groups of two each,

$$\begin{align*} 1^{2007}+2^{2007}+3^{2007}+\cdots n^{2007} \\ \equiv1+\big\lfloor 2^{2007}+(n)^{2007}\big\rfloor+\cdots +\cdots \big\lfloor \ \Big(\frac{n+1}{2}\Big)^{2007}+\Big(\frac{n+3}{2}\Big)^{2007}\big\rfloor\\ \equiv1+\big\lfloor 2^{2007}+(-2)^{2007}\big\rfloor+\cdots +\cdots \big\lfloor \ \Big(\frac{n+1}{2}\Big)^{2007}+\Big(-\frac{n+1}{2}\Big)^{2007}\big\rfloor\\ \equiv 1\pmod{n+2} \end{align*} $$ Thus, the conclusion is proven