# Solving for an inverse diophantine equation!

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.

• 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. – JKnecht Feb 15 '16 at 8:05

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