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.

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.

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.

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.