2013 IMO Problems/Problem 1

Problem

Prove that for any pair of positive integers $k$ and $n$ , there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that

$1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})$ .

Solution

We prove the claim by induction on $k$ .

Base case: If $k = 1$ then $1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}$ , so the claim is true for all positive integers $n$ .

Inductive hypothesis: Suppose that for some $m \in \mathbb{Z}^{+}$ the claim is true for $k = m$ , for all $n \in \mathbb{Z}^{+}$ .

Inductive step: Let $n$ be arbitrary and fixed. Case on the parity of $n$ :

[Case 1: $n$ is even] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m} - 1}{\frac{n}{2}} \right) \cdot \left( 1 + \frac{1}{n + 2^{m+1} - 2} \right)$

[Case 2: $n$ is odd] $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^{m}-1}{\frac{n+1}{2}} \right) \cdot \left( 1 + \frac{1}{n} \right)$

In either case, $1 + \frac{2^{m+1} - 1}{n} = \left( 1 + \frac{2^m - 1}{c} \right) \cdot \left( 1 + \frac{1}{a_{m+1}} \right)$ for some $c, a_{m+1} \in \mathbb{Z}^+$ .

By the induction hypothesis we can choose $a_1, ..., a_m$ such that $\left( 1 + \frac{2^m - 1}{c} \right) = \prod_{i=1}^{m} (1 + \frac{1}{a_i})$ .

Therefore, since $n$ was arbitrary, the claim is true for $k = m+1$ , for all $n$ . Our induction is complete and the claim is true for all positive integers $n$ , $k$ .

Alternative Solution

We will prove by constructing telescoping product: $\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3} \cdot \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(a_1+2^{k}-1\right)}{a_1}$ where each fraction $\frac{a_{i+1}}{a_i}=\frac{a_{i}+\Delta_i}{a_i}$ can also be written as $\frac{m_i+1}{m_i}$ for some positive integer $m_i$ . Telescoping property implies $\sum{\Delta_i}=2^{k}-1$ . We will show that the set of all $\Delta_i$ can be taken to be a collection of $2^j$ for $1\le j \le (k-1)$