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I am currently working on the following problem:

Find all positive integers $r$ such that, for any natural number $k \geq 1$, $$(1+2^r+3^r+\cdots+k^r) = (1+2+3+\cdots+k)^{r-1}$$

I've tried to look at this with a sequence and series lens but it didn't get me far.

Any help is highly appreciated.

bru1987
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    We know it is true for $r=3$. Also we know that left side is a polynomial of degree $r+1$, and the right side is a polynomial of degree $2(r-1)$. Thus the only possible $r$ is $r=3$. – Dyako Soltanian Apr 24 '23 at 16:10
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    Even if you restrict this to $k=2$, you are looking for solutions to $1+2^r=3^{r-1}$ and it is not difficult to see there is only one solution for that, namely $r=3$ which you can then check for other $k$ – Henry Apr 24 '23 at 16:20
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    Does this answer your question? Proof $\sum\limits_{a=1}^{n} a^3 = {\left(\sum\limits_{a=1}^{n} a \right)}^2$? (Note that this question is actually asking about the general case, not the $r=3$ case that the title suggests.) – MJD Apr 24 '23 at 16:23
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    @Henry, restriction is a good idea! If we restrict it to $k=2$ we get $1+2^r=3^{r-1}$, and if we restrict it to $k=3$, we get $1+2^r+3^r=6^{r-1}$, which together with the first one we get $1+3=2^{r-1}$, and again $r=3$. – Dyako Soltanian Apr 24 '23 at 16:38
  • @DyakoSoltanian why is the left side a polynomial of degree $r+1$ and not $r$? – bru1987 Apr 24 '23 at 18:26
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    @bru1987 See https://en.wikipedia.org/wiki/Faulhaber%27s_formula#Examples - you get more than $\int_0^k x^r dr = \frac1r x^{r+1}$ – Henry Apr 24 '23 at 19:12
  • @bru1987, letting $p_r(n)=\sum\limits_{a=0}^{n} a^r$ and using $(1+a)^{r+1}-a^{r+1}=\sum\limits_{k=0}^r{\binom{r+1}{k}a^k}$, one can show that $$(r+1)p_r(n)=(n+1)^{r+1}-\sum_{k=0}^{r-1}{\binom{r+1}{k}p_k(n)}.$$ Now using induction on $r$ we can see that $p_r(n)$ is a polynomial of degree $r+1$ and even the coefficient of $n^{r+1}$ is $\displaystyle\frac{1}{r+1}$. – Dyako Soltanian Apr 24 '23 at 19:43

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