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

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.

Answer 1

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.