<s>[INST] Does $(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$ imply $(a,b,c)=(1,2,3)$? 
Question : Is the following proposition true?
Proposition : For positive integers $a,b,c$ where $b\ge 2$, if
$$(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$$then $(a,b,c)=(1,2,3)$.

This is an unsolved case of this question where it has been proven that each of the following propositions is true for positive integers $a,b,c$ where $b\ge 2$ :

*

*$\text{If $(1^a+2^a)^b=1^c+2^c$, then $(a,b,c)=(1,2,3)$.}$


*$\text{If $(1^a+2^a+3^a)^b=1^c+2^c+3^c$, then $(a,b,c)=(1,2,3)$.}$


*$\text{If $(1^a+2^a+3^a+4^a)^b=1^c+2^c+3^c+4^c$, then $(a,b,c)=(1,2,3)$.}$


*$\text{If $(1^a+2^a+\cdots +{11}^a)^b=1^c+2^c+\cdots+11^c$, then $(a,b,c)=(1,2,3)$.}$


*$\text{If $(1^a+2^a+\cdots +{12}^a)^b=1^c+2^c+\cdots+12^c$, then $(a,b,c)=(1,2,3)$.}$
$$\vdots$$
I've been trying to use mod and some inequalities, but every attempt has failed. Can anyone help?
Added : I'm going to add the background of this question.
We know that
$$\left(1^1+2^1+\cdots+n^1\right)^2=1^3+2^3+\cdots+n^3$$
holds for every $n\in\mathbb N$. Also, it is known that, for positive integers $a,b,c$ where $b\ge 2$, if
$$\left(1^a+2^a+\cdots+n^a\right)^b=1^c+2^c+\cdots+n^c$$
holds for every $n\in\mathbb N$, then $(a,b,c)=(1,2,3)$.
Then, I've been interested in the following similar, but completely different question :

For positive integers $a,b,c$ where $b\ge 2$, if
$$\left(1^a+2^a+\cdots+n^a\right)^b=1^c+2^c+\cdots+n^c$$
holds for a specific $n\color{red}{\ge 2}\in\mathbb N$, then can we say that $(a,b,c)=(1,2,3)$?

This question has been asked here. It has been proven that the answer is yes for $n=2,8k-5,8k-4$ where $k\in\mathbb N$.
However, the question has not received any complete answers. For example, it is not known if the answer is yes for $n=5$, which is the smallest unsolved case I'm asking here.
 [/INST] In general we have

$$
\sum_{k=1}^n k^a = \frac{1}{a+1} \prod_{\jmath=1}^{a+1} \big( n + n_\jmath\big)
\tag 1
$$

Examples

$$
\begin{eqnarray}
\sum_{k=1}^n k &=& \frac{1}{2} n \big( n + 1 \big)\\\
\sum_{k=1}^n k^2 &=& \frac{1}{3} n \big( n + 1 \big) \big( n + 1/2 \big)\\\
\sum_{k=1}^n k^3 &=& \frac{1}{4} n^2 \big( n + 1 \big)^2\\\
\sum_{k=1}^n k^4 &=& \frac{1}{5} n \big( n + 1 \big)
   \big( n + 1/2 \big)
   \Big( n + 1/2 - \sqrt{7/12} \Big)
   \Big( n + 1/2 + \sqrt{7/12} \Big)
\end{eqnarray}
$$

So we can only get

$$
\left( \sum_{k=1}^n k^a \right)^b = \sum_{k=1}^n k^c,\\
\quad \textrm{if at least} \quad
\left( \frac{1}{a+1} \right)^b = \frac{1}{1+c},
\quad \textrm{and} \quad
\big( a + 1 \big) b = c + 1. \tag 2
$$

So the question is, when do we have
$$
\big( a + 1 \big)^{b-1} = b ?
$$
So only $a=1$, $b=2$ and $c=3$ are the solutions.
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