# How to prove that $\frac{1}{x_1}+\frac{1}{x_2}+…+\frac{1}{x_n}-\frac{1}{x_1x_2…x_n}\in \mathbb{N}\cup \{0\}$

Question:

Show that for every natural number $n$ there exist $n$ natural numbers $x_1 < x_2 < ... < x_n ,$ such that $$\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}-\frac{1}{x_1x_2...x_n} \in \mathbb{N}\cup \{0\}.$$

My idea: For $n=1$ we can take any natural $x_1\geq 1$, since $\dfrac {1}{x_1} - \dfrac {1}{x_1} = 0$.

For $n=2$ we can take $x_1=1$ and any natural $x_2>1$, since $\dfrac {1}{1} + \dfrac {1}{x_2} - \dfrac {1}{1\cdot x_2} = 1$.

For $n=3$, we can take $x_1=2,\,x_2=3$ and $x_3=5$.

But for $n>3$, I can't find any example, Now I have read this links,But I can't understand

can you someone have easy example?

Thank you for you help.

• I hope that it can help you but for $n=3$ there is a solution $2, 3, 5$ – PenasRaul Oct 4 '14 at 9:51
• This is a recent Iran national math olympiad(3rd round) problem 1 on number theory section. There is a solution posted here. – karvens Oct 4 '14 at 12:24
• $2,3,5$; $2,3,7,41$; $2,3,7,43,x$; I'll leave it to you to find $x$ and to see the developing pattern and to prove it continues. – Gerry Myerson Oct 4 '14 at 12:51
• You always have interesting problems. Thank you for sharing! – Petite Etincelle Oct 4 '14 at 12:55
• Have you made any progress with my comment, china math? – Gerry Myerson Oct 6 '14 at 0:18

The sequence shown by Gerry Myerson, ie. $a_1=2$, $a_{k+1}=a_1\cdots a_k+1$ satisfies $$\frac1{a_1}+\dots+\frac1{a_n} = 1 - \frac1{a_1\cdots a_n}.$$ (Induction works due to $\frac1{a_1\cdots a_n} - \frac1{a_1\cdots a_{n+1}} = \frac{a_{n+1}-1}{a_1\cdots a_{n+1}}=\frac1{a_{n+1}}$.)
Based on this sequence, for $n\ge3$ the numbers $$(x_1,x_2,\ldots,x_n) = (a_1,\ldots,a_{n-1},a_n-2)$$ solves the problem: $$\frac1{x_1}+\dots+\frac1{x_{n-1}}+\frac1{x_n} = \left(\frac1{a_1}+\dots+\frac1{a_{n-1}}\right)+\frac1{a_n-2} = \\ = \left(1-\frac1{a_1\cdots a_{n-1}}\right)+\frac1{a_1\cdots a_{n-1}-1} = 1+\frac1{a_1\cdots a_{n-1}(a_1\cdots a_{n-1}-1)} = \\ = 1+\frac1{a_1\cdots a_{n-1}\cdot (a_n-2)}. = 1+\frac1{x_1\cdots x_{n-1}\cdot x_n}.$$