# An equation involving perfect numbers

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?

• 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) – user142929 Oct 6 '19 at 18:03

I think the computer found other solutions for $s=4$:
[x1,x2,x3,x4]