Given are a positive integer $n$ and an infinite sequence of proper fractions $x_0 = \frac{a_0}{n}$, $\ldots$, $x_i=\frac{a_i}{n+i}$, with $a_i < n+i$. Prove that there exist a positive integer $k$ and integers $c_1$, $\ldots$, $c_k$ such that \[ c_1 x_1 + \ldots + c_k x_k = 1. \] Proposed by M. Dubashinsky
Problem
Source: Tuymaada 2005, Day 2, Problem 6
Tags: algebra unsolved, algebra
29.08.2005 14:54
But what do we know about $a_i$ do they have to be integer, rational or it doesn't matters :
29.08.2005 15:13
I'm not really sure... here http://www.mathlinks.ro/Forum/viewtopic.php?t=44430 isn't said nothing but i think that the meaning of "proper fractions" is "integer/integer", so i guess the $a_{i}$s are integers...
27.11.2005 12:43
Please someone post a solution to this interesting problem. Does anyone has the official solution ?It is unsolved for 4 months.
27.11.2005 13:48
Let's try the following: We can assume that $a_i$ are positive integers where $a_i=\epsilon _i (n+i)x_i$ where $\epsilon _i \in \{-1,1\}$. As each of $x_i$ is proper fraction so $(a_i,n+i)=1$. Now our statement is equivalent to: $\exists _k (a_0,...,a_k)=1$ As the sequence is infinite we can take large enough $k$ such that $n+k=p!$ where $p>a_0$. What more $(a_k,p!)=1$ so $a_k>p$ and $a_k$ has no prime divisors smaller than $p$. This lead us to $(a_k,a_0)=1$ and hence $(a_0,...,a_k)=1$