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 construct telescoping product out of positive integers: $\frac{a_2}{a_1}\cdot\frac{a_3}{a_2}\cdot\frac{a_4}{a_3} \cdot \frac{a_{k+1}}{a_k} = \frac{a_{k+1}}{a_1} = \frac{\left(a_1+2^{k}-1\right)}{a_1}$ where $a_1=n$ and 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 $m_i$ . All $\Delta_i$ will be different with $\Delta_i=2^j$ , $0\le j \le (k-1)$ . $\sum_{i=1}^{k}{\Delta_i}=\sum_{i=1}^{k}{2^{i-1}}=2^{k}-1$ .

Let's pull out factors of $2$ if there are any: $n=2^j(2z+1)$ , $z+1=2^q(2r+1)$ etc where $j\ge 0$ , $q\ge 0$ . Construct telescoping as $\frac{2^j(2z+2)}{2^j(2z+1)}\cdot \frac{2^{j+1+q}(2r+2)}{2^{j+1+q}(2r+1)} ...$ . Every $\Delta_i$ can be bigger then previous $\Delta_{i-1}$ by at least factor $2$ . The biggest needed $\Delta=2^{k-1}$ can be constructed in at most $k$ steps. After we constructed the fraction with the biggest needed $\Delta$ : $\frac{2^{k-1}(h+1)}{2^{k-1}h}$ we can construct any remaining needed $\Delta_i$ in the decresing order. That is if we need $\Delta=2^{d}$ where $d<k-1$ we can use as the next telescoping fraction $\frac{2^{d}(2^{k-1-d}(h+1)+1)}{2^{d}(2^{k-1-d}(h+1))}$