Let $s,x_1,x_2,\cdots, x_s$ be natural numbers not neccesarily distinct.
I am interested in solving the equation $$(x_1+x_2+\cdots +x_s)^s=2^s(x_1\cdot x_2\cdots x_s)^2$$
Some Notes:

I have found two solutions $(x_1,\cdots ,x_s)$
1) We can see that equality is satisfied if $x_s= 2^{p-1}(2^p-1)$ is an even perfect number and all the other $x_i$ are the proper divisors of $x_s$. (That is why i started to investigate the equation)
Of course we can disregard the restriction that $x_s$ is perfect and show easily that we also get a solution if: $$x_1=1,x_2=2^1,...,x_n=2^{n-1},x_{n+1}=2^n-1,x_{n+2}=2^1(2^n-1),...$$

and $x_s=2^{n-1}(2^n-1)$.(The number $n$ is not necessarily prime )

2) A trivial one: $$x_1=x_2=\cdots =x_s=\frac{s}{2}$$ .

Is it possible to find other solutions or to prove that there are only 2 solutions, those mentioned above?
Thanks in advance.

  • 1
    $\begingroup$ If I understand well your identity can be written in terms of the arithmetic and geometric means as $sA(x_1,x_2,\ldots,x_s)=2G(x_1,x_2,\ldots,x_s)^2$. And it is possible to deduce also a different expression using the expression from the Wikipedia Harmonic divisor number. I did it, thus your solutions also satisfy the equation that I evoke. I think (it is just my opinion from my ignorance) that maybe these facts are known. On the other hand your equation seems interesting and I wondered if you can to edit the post to fix punctuation or add tags (diophantine-equation) and (divisors-multiples) $\endgroup$
    – user142929
    Oct 6, 2019 at 18:03

2 Answers 2


I think the computer found other solutions for $s=4$:

[ 2 , 27 , 150 , 1 ]
[ 2 , 3 , 6 , 121 ]
[ 27 , 150 , 2 , 1 ]
[ 50 , 6 , 3 , 1 ]
  • $\begingroup$ Thank you a lot for these results.(By the way,the third solution is the same with the first one).Any ideas about what family of solutions these number may represent in general? $\endgroup$ Jan 26, 2014 at 19:44
  • $\begingroup$ @KonstantinosGaitanas I don't know. There might be sporadic solutions, not sure. $\endgroup$
    – joro
    Jan 27, 2014 at 12:24

I will prove that there are no other solutions for $s\le2$.

For $s=1$, this equation becomes $$x_1=2x_1,$$ giving $x_1=0$, a contradiction.

For $s=2$, this equation becomes:



We can take the square roots of both sides since they are positive integers. We then get:






Since $x_1$ and $x_2$ are integers, then so are $2x_1-1$ and $2x_2-1$. Hence


which gives


the only solution in positive integers being $$(x_1,x_2)=(1,1).$$

Hence $x_1=x_2=1=\frac{2}{2}=\frac{s}{2}$, a contradiction. Hence $s\ge3$ if there are other solutions.


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