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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.

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    $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – JKnecht Feb 15 '16 at 8:05
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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.

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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*}

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  • $\begingroup$ And that's not even considering permutations of the inputs. $\endgroup$ – Morgan Rodgers Feb 15 '16 at 15:01

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