**Question** : Is the following conjecture true?

**Conjecture** : Let $a,b(\ge 2),c,n(\ge 2)$ be natural numbers. If $$\left(\sum_{k=1}^nk^a\right)^b=\sum_{k=1}^nk^c\ \ \ \ \ \cdots(\star)$$
for **some** $n$, then $(a,b,c)=(1,2,3).$

**Remark** : This question has been asked previously on math.SE without receiving any answers.

**Motivation** : This question comes from

$$\left(\sum_{k=1}^nk\right)^2=\sum_{k=1}^nk^3.$$

This got me interested in $(\star)$. I've got the followings :

*1.* If $(\star)$ for **any** $n\in\mathbb N$, then $(a,b,c)=(1,2,3).$

We can easily prove this by considering the limitation $n\to\infty$ of the both sides of $$\frac{n^{(a+1)b}}{n^{c+1}}\left\{\sum_{k=1}^n\frac 1n\left(\frac kn\right)^a\right\}^b=\sum_{k=1}^n\frac 1n \left(\frac kn\right)^c.$$

*2.* If $(\star)$ for $n=2$, then $(a,b,c)=(1,2,3).$

*3.* If $(\star)$ for $n=3$, then $(a,b,c)=(1,2,3).$

Since both *1* and *2* are easy to prove, I'm going to prove *3*.

**Proof** : Supposing $c\le ab$, since $b\ge 2$, we get
$$(1+2^a+3^a)^b=1+2^{ab}+3^{ab}+\cdots\gt 1+2^c+3^c.$$ This is a contradiction. Hence, $c\gt ab$. Supposing $b\ge 3$, we get $c\gt ab\ge 3$.

Here, since $3^c+1\equiv 4,2$ (mod $8$) for any $c\in\mathbb N$, $3^c+1$ is not a multiple of $8$. By the way, since $1+2^a+3^a$ is even, $(1+2^a+3^a)^b$ is a multiple of $8$. Since $2^c$ is a multiple of $8$, this leads that $3^c+1$ is a multiple of $8$, which is a contradiction. Hence, $b=2, c\gt 2a$.

If $a\ge 3$, since $$\left(\frac 23\right)^a+\left(\frac 13\right)^a\le\left(\frac 23\right)^3+\left(\frac 13\right)^3=\frac13,$$ $2^a+1\le \frac{3^a}{3}.$ Hence, $$3^c\lt 1+2^c+3^c=(1+2^a+3^a)^2\le \left(\frac{3^a}{3}+3^a\right)^2=3^{2a}\left(\frac 43\right)^2=3^{2a}\cdot\frac {16}{9}\lt 3^{2a+1}.$$ $3^c\lt 3^{2a+1}$ leads $0\lt c-2a\lt 1$, which means that $c-2a$ is not an integer. This is a contradiction. Hence, we know $a=1$ or $a=2$.

The $(a,b)=(1,2)$ case leads $c=3$.

The $(a,b)=(2,2)$ case leads $c\ge5\Rightarrow 1+2^c+3^c\gt 196$, which is a contradiction. Now the proof is completed.

After getting these results, I reached the above conjecture. Can anyone help?

mayhelp you in reducing thesearch space/number of casesyou need to consider. $\endgroup$ – Jose Arnaldo Bebita-Dris Oct 11 '13 at 19:26