<s>[INST] Solving for an inverse diophantine equation $\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$ 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.
 [/INST] 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*}
</s>