Integer solutions to $\prod\limits_{i=1}^{n}x_i=\sum\limits_{i=1}^{n}x_i^2$ Given integers $x_1,\dots,x_n>1$.
Let's assume WLOG that ${x_1}\leq\ldots\leq{x_n}$.
I want to prove that the only integer solutions to any equation of this type are:


*

*$x_{1,2,3  }=3\implies\prod\limits_{i=1}^{3}x_i=\sum\limits_{i=1}^{3}x_i^2=27$

*$x_{1,2,3,4}=2\implies\prod\limits_{i=1}^{4}x_i=\sum\limits_{i=1}^{4}x_i^2=16$


I have a partial proof below, but there are a few holes in it that I would be happy to get help with.

For $n=1$:


*

*$x_1<x_1^2$

*$\color{green}{\text{Done.}}$



For $n=2$:


*

*${x_1x_2}\leq{x_2x_2}={x_2^2}<{x_1^2+x_2^2}$

*$\color{green}{\text{Done.}}$



For $n=3$ and $x_1=2$:


*

*$\color{red}{\text{This seems to be the most difficult case.}}$

*$\color{red}{\text{I suspect that there might even be a solution here.}}$



For $n=3$ and $x_{1,2,3}=3,3,3$:


*

*$\prod\limits_{i=1}^{3}x_i=27=\sum\limits_{i=1}^{3}x_i^2$

*$\color{blue}{\text{This is a solution.}}$



For $n=3$ and $x_{1,2,3}=3,3,4$:


*

*$\prod\limits_{i=1}^{3}x_i=36>34=\sum\limits_{i=1}^{3}x_i^2$

*$\color{red}{\text{How do we prove that it holds for every }x_i\text{ being replaced with a larger value?}}$



For $n=4$ and $x_{1,2,3,4}=2,2,2,2$:


*

*$\prod\limits_{i=1}^{4}x_i=16=\sum\limits_{i=1}^{4}x_i^2$

*$\color{blue}{\text{This is a solution.}}$



For $n=4$ and $x_{1,2,3,4}=2,2,2,3$:


*

*$\prod\limits_{i=1}^{4}x_i=24>21=\sum\limits_{i=1}^{4}x_i^2$

*$\color{red}{\text{How do we prove that it holds for every }x_i\text{ being replaced with a larger value?}}$



For $n>4$ and $x_{1,\dots,n}=2,\dots,2$:


*

*$\prod\limits_{i=1}^{n}x_i=2^n>4n=\sum\limits_{i=1}^{n}x_i^2$ can prove by induction

*$\color{red}{\text{How do we prove that it holds for every }x_i\text{ being replaced with a larger value?}}$



UPDATE:
Based on @sciona's comments:


*

*For $n=3$ and $x_1=2$, we can show that there are no solutions:
$\sum\limits_{i=1}^{n}x_i^2-\prod\limits_{i=1}^{n}x_i=4+x_2^2+x_3^2-2x_2x_3=4+(x_2-x_3)^2>0$
$\sum\limits_{i=1}^{n}x_i^2-\prod\limits_{i=1}^{n}x_i>0\implies\sum\limits_{i=1}^{n}x_i^2>\prod\limits_{i=1}^{n}x_i\implies\sum\limits_{i=1}^{n}x_i^2\neq\prod\limits_{i=1}^{n}x_i$

*We can use the solution $(3,3,3)$ in order to generate infinitely many more:
If $(a,b,c)$ is a solution, then so is $(b,c,bc-a)$
For example, $(3,3,3)\rightarrow(3,3,6)\rightarrow(3,6,15)\rightarrow\dots$

*We can use the solution $(2,2,2,2)$ in order to generate infinitely many more:
If $(a,b,c,d)$ is a solution, then so is $(b,c,d,bcd-a)$
For example, $(2,2,2,2)\rightarrow(2,2,2,6)\rightarrow(2,2,6,22)\rightarrow\dots$
So my question narrows-down to proving that there are no other types of solutions.
 A: In this answer we will present a method to solve the equations below over the set of integers for $n>2$  :
$$(E_n)\,\,\, \ \ \ \ \  x_1^2+x_2^2+\cdots+x_n^2=x_1x_2\cdots x_n \\ x_1\leq x_2\leq \cdots \leq x_n$$
First, as it was mentioned above, given $(x_1,\cdots,x_n)$ of $E_n$ we can construct another solution by replacing  an element $x_i$ by $ \prod_{j=1,j\neq i}^{n} x_j-x_i$. In order to use this observation properly we define :
$$ P(X)=X^2-x_1x_2\cdots x_{n-1}X+x_1^2+x_2^2+\cdots+x_{n-1}^2$$
we have clearly $P(x_n)=0$, if the other root of this $P$ is $t$, then $(x_1,x_2,\cdots,x_{n-1},t)$ is a solution of $E_n$. 
The idea of this proof is to look for minimal solutions: the solutions for which $\sum_{i=1}^n x_i$ is minimal. notice that :
$$P(x_{n-1})<0 \Rightarrow  t<x_n$$
So if $(x_1,\cdots,x_n)$ is a minimal solution , then $0\leq P(x_{n-1})$ and this inequality will reduce dramatically the number of solutions in fact we will prove that almost all $x_i$ are equal to $1$.


*

*Case $n=3:$
$$P(x_2) \geq 0 \Rightarrow x_2^2(3-x_1)+x_1^2-x_2^2 \geq 0  $$
and because $x_1\leq x_2$ and $x_2^2+x_3^2\geq 2x_2x_3$ we conclude that $x_1=3$ , and $$ (2x_2-3x_3)^2-5x_3^2=-36 \ \ \ \ \ (P_1)$$
This is a Pell_Fermat equation which have an infinity of solutions (see here).
The set of solution of $E_3$ is constituted by the values taken by one of the sequences defined by :


*

*$\{a_0,b_0,c_0\}=\{3,x,y\}$ such that $(x,y)$ is a solution to $P_1$

*$(a_{n+1},b_{n+1},c_{n+1})\in \{ (a_n,b_n,a_nb_n-c_n),(a_n,c_na_n-b_n,c_n),(b_nc_n-a_n,b_n,c_n)\}$.


*Case $n>=4$ assume that $ 2^k \leq n \leq 2^{k+1}$ ($k=[log_2(n)]$) :



$$P(x_{n-1}) \geq 0 \Rightarrow x_{n-1}^2(n-\prod_{i=1}^{n-2}x_i)+\sum_{i=1}^{n-2}x_i^2-(n-2)x_{n-1}^2 \geq 0  $$
and because $x_i$ is ordered $\sum_{i=1}^{n-2}x_i^2-(n-2)x_{n-2}^2 \leq 0 $ so obviously 
$$\prod_{i=1}^{n-2}x_i \leq n $$ 
This result is very powerful, it proves that for any minimal solution, the first $n-2$ integers are bounded, and we can notice that at least the $n-k$ first elements of the sequence $(x_i)$ are equal to $1$ (we can prove more than that :only (n+1)/5 elements are $\geq 2$ ). so to construct solutions to $E_n$ we take all tuples $(x_1,x_2,\cdots, x_{n-1})$ such that :
    $$\prod_{i=1}^{n-2}x_i \leq n $$ 
 and we verify if $P(x_{n-1}) \geq 0 $ and finally we solve the associated Pell Fermat equation: 
 $$\sum_{i=1}^{n-2}x_i^2+ A^2+B^2=(\prod_{i=1}^{n-2}x_i)AB$$ 
where $A=x_n$ and $B=x_{n-1}$
we generate all solutions using the transformations $x_i\rightarrow \prod_{j=1,j\neq i}^{n} x_j-x_i$ as showed in the case $n=3$.
Finally, some remarks:


*

*for $n=4$ the only associated Pell Fermat equation here.

*for $n=5$ the only associated Pell Fermat equation here.

*for $n\geq 21$ we will have more than one equation.
This method can be used to prove that the equation 
$$x_1^2+x_2^2+\cdots+x_n^2=\alpha x_1x_2\cdots x_n $$
has no integer solutions (Edit : when $\alpha>n$).
