Divisible by what?

Question

$1^n+2^n+\ldots+2006^n$ is divisible by :

$1.~2006$

$2.~2007$

$3.~2008$

$4.~\text{None}$

Please Explain.

Answer 1

For odd $n$, this answer says that the sum is divisible by $\frac{2006\cdot2007}{2}$.

For even $n$, none need be true. For example, $$ \sum_{k=1}^{2006}k^2=2692751091 $$ which is divisible by none of $2006$, $2007$, or $2008$.

Answer 2

Let $f(n)=1^n+2^n+\ldots+2006^n$

For $n=0$,

$f(n) = 1^n+2^n+\ldots+2006^n=\underbrace{1+1+\ldots+1}_\text{2006 times} = 2006$

So $2006 \mid f(n)$ and $2007,2008 \not\mid f(n)$

For $n > 0$

$\begin{align*} 1^n+2^n+\ldots+2006^n &= (1^n+2^n+3^n+4^n)+\ldots+(2001^n+2002^n+2003^n+2004^n)+2005^n+2006^n\\ &\equiv (1+0+1+0)+\ldots+(1+0+1+0)+1+0 \mod 2\\ &\equiv 1 \mod 2 \end{align*}$

So $2006,2008 \not\mid f(n)$

For $n > 0$, odd

$\begin{align*} 1^n+2^n+\ldots+2006^n &= (1^n+2006^n)+(2^n+2005^n)+\ldots+(1003^n+1004^n)\\ &\equiv (1^n+(-1)^n)+(2^n+(-2)^n)+\ldots+(1003^n+(-1003)^n) \mod 2007\\ &\equiv (1^n-1^n)+(2^n-2^n)+\ldots+(1003^n-1003^n) \mod 2007\\ &\equiv 0 \mod 2007 \end{align*}$

So $2007 \mid f(n)$

For $n > 0, n$ even

From Faulhaber's formula

$f(2) = \dfrac{2006 * (2006+1) * (2*2006 + 1)}{6} \equiv 6 \mod 9$

$f(4) = f(2)*(\dfrac{3*2006^2 +3*2006 - 1}{5}) \equiv 6 \mod 9$

$f(6) = f(2)*(\dfrac{3*2006^4 +6*2006^3- 3*2006 +1}{7}) \equiv 6 \mod 9$

So $9 \not\mid f(2),f(4),f(6)$

Hence $2007 \not\mid f(2),f(4),f(6)$

Now note, since

$a^n \equiv \left.\begin{align*} & 0 & a \equiv 0,3,6 \mod 9 \end{align*}\right.$

and

$a^6 \equiv \left\{\begin{align*} & 0 & a \equiv 0,3,6 \mod 9\\ & 1 & a \not\equiv 0,3,6 \mod 9 \end{align*}\right.$

It follows $n,m > 0, n,m$ even,

$\begin{align*} n \equiv m \mod 6 & \Rightarrow & a^n \equiv a^m \mod 9\\ & \Rightarrow & f(n) \equiv f(m) \mod 9 \end{align*}$

Hence $2007 \not\mid f(n)$