An equation involving perfect numbers

Question

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.

Answer 1

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 ]

Answer 2

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.