# 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) Oct 6, 2019 at 18:03

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

[x1,x2,x3,x4]
[ 2 , 27 , 150 , 1 ]
[ 2 , 3 , 6 , 121 ]
[ 27 , 150 , 2 , 1 ]
[ 50 , 6 , 3 , 1 ]

• 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? Jan 26, 2014 at 19:44
• @KonstantinosGaitanas I don't know. There might be sporadic solutions, not sure.
– 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:

$$(x_1+x_2)^2=4(x_1x_2)^2$$

$$(x_1+x_2)^2=(2x_1x_2)^2.$$

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

$$x_1+x_2=2x_1x_2$$

$$x_1+x_2-2x_1x_2=0$$

$$-2x_1x_2+x_1+x_2-\frac{1}{2}=-\frac{1}{2}$$

$$-\frac{1}{2}(4x_1x_2-2x_1-2x_2+1)=-\frac{1}{2}$$

$$(2x_1-1)(2x_2-1)=1$$

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

$$(2x_1-1,2x_2-1)\in\{(1,1),(-1,-1)\},$$

which gives

$$(x_1,x_2)\in\{(1,1),(0,0)\},$$

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.