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.



An equation involving perfect numbers
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 ]

```

