Solving for an inverse diophantine equation $\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$

Question

How can I prove that the Diophantine equation $$\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$$ has at most one solution? All $x_i$ and $n$ are natural numbers.

My attempt was:
For example consider equation for $n=3$:

$$\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac1{x_1x_2x_3} =1$$ then $$x_1x_2 + x_1x_3 + x_2x_3 +1 = x_1x_2x_3 \tag{1} $$ and \begin{cases} x_2x_3+1\equiv 0\pmod {x_1} \\ x_1x_3+1\equiv 0\pmod {x_2} \\ x_1x_2+1\equiv 0\pmod {x_3} \\ \end{cases} so \begin{cases} x_2x_3=k_1x_1-1 \\ x_1x_3=k_2x_2-1 \\ x_1x_2=k_3x_3-1 \\ \end{cases} substitute in (1) gives this $$k_1x_1 + k_2x_2 + k_3x_3 = x_1x_2x_3 + (3-1) $$ general form will be $$k_1x_1 + k_2x_2 + k_3x_3 + ... + k_nx_n = x_1x_2x_3...x_n + (n-1) $$ I know to solve this but $x_1x_2x_3...x_n$ term is the problem.

Answer 1

Suppose that $x_1,x_2….,x_r$ is a solution then $$\frac1 x_1 + \frac1 x_2 + … + \frac 1 x_{r+1} + \frac1{x_1x_2… x_{r+1}} =\\ 1- \frac 1 {x_1 x_2 … x_r} + \frac 1 x_{r+1} + \frac1{x_1x_2… x_{r+1}} =\\ 1+\frac 1 x_{r+1}-\frac {x_{r+1}-1}{x_1x_2… x_{r+1}} = 1$$

so $$\frac 1 x_{r+1}-\frac {x_{r+1}-1}{x_1x_2… x_{r+1}} = 0$$ $$x_{r+1} = 1+x_1x_2… x_r $$ the solution is $$x_1 = 2, x_{r+1} = 1+x_1x_2… x_r, 1\le r \le n $$ for $n =1,2$ solution is unique(regardless of permutations). for every $n$, $n>2$, number of solutions as @Anurag said is not unique.

Answer 2

Assuming that you want the number of solutions (you have referred to as root so it is a bit confusing).

This is NOT true, for example if $n=5$, then there are $3$ solutions.

\begin{align*} 1 & = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\frac{1}{2 \cdot 3 \cdot 7 \cdot 43 \cdot 1807}\\ & = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{47}+\frac{1}{395}+\frac{1}{2 \cdot 3 \cdot 7 \cdot 47 \cdot 395}\\ & = \frac{1}{2}+\frac{1}{3}+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}+\frac{1}{2 \cdot 3 \cdot 11 \cdot 23 \cdot 31}\\ \end{align*}