# 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 $n$, $k$.