IMO 2013 Problem 1

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Verified: no
Verdicts: FAIL + FAIL
Solve time: 11m44s

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

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

Exploration

The reviewer correctly identifies that any attempt based on naive telescoping of expressions like $a_{i+1}/a_i$ or products of $(a_i+1)/a_i$ fails because the denominators and numerators do not align in a way that produces cancellation. Small case testing shows the same obstruction persists already at $k=2$, where any correct construction must produce

$\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)=\frac{n+3}{n}.$

Any viable approach must therefore explicitly control how the numerator grows from $n$ to $n+2^k-1$ through multiplicative increments, without relying on spurious telescoping.

A useful structural constraint comes from rewriting each factor as

$1+\frac{1}{m_i}=\frac{m_i+1}{m_i},$

so the full product becomes a ratio of two integers whose numerator exceeds the denominator by exactly $2^k-1$. Testing small values suggests that no single uniform arithmetic progression for $m_i$ works, and any correct construction must encode the binary expansion of the increment $2^k-1$ across the $k$ factors.

The correct perspective is that each factor should contribute a controlled “doubling-adjusted” increment depending on both $n$ and a power of $2$, but without requiring cancellation between unrelated terms.

Problem Understanding

The task is to construct positive integers $m_1,\dots,m_k$ such that

$\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)=1+\frac{2^k-1}{n}=\frac{n+2^k-1}{n}.$

Equivalently,

$\prod_{i=1}^k \frac{m_i+1}{m_i}=\frac{n+2^k-1}{n}.$

The right-hand side has a numerator obtained from $n$ by adding $2^k-1$, and the goal is to realize this additive jump through multiplicative unit fractions.

Key Observations

A correct construction must ensure that each factor contributes a controlled increment to the numerator when the entire product is written over a common denominator. If we expand

$\prod_{i=1}^k \frac{m_i+1}{m_i}=\frac{\prod (m_i+1)}{\prod m_i},$

then the difference between numerator and denominator must equal $2^k-1$.

The key idea is to enforce that each factor introduces a doubling contribution relative to $n$, so that after $k$ steps the accumulated surplus becomes exactly

$1+2+\cdots+2^{k-1}=2^k-1.$

This suggests a construction in which each $m_i$ is tied to $n$ and a power of $2$ in such a way that the product expands into a structured binary-weighted sum.

Solution

Define integers $m_1,\dots,m_k$ by

$m_i = n\cdot 2^{i-1} + 2^{i-1}-1 \quad \text{for } i=1,2,\dots,k.$

Each $m_i$ is positive since both terms in the sum are positive for all $n\ge 1$.

Now compute each factor:

$1+\frac{1}{m_i}=\frac{m_i+1}{m_i}=\frac{n\cdot 2^{i-1}+2^{i-1}}{n\cdot 2^{i-1}+2^{i-1}-1}.$

Factor $2^{i-1}$ from the numerator:

$\frac{2^{i-1}(n+1)}{n\cdot 2^{i-1}+2^{i-1}-1}.$

Now observe that when multiplying all factors, the powers of $2$ combine into a clean geometric accumulation:

= \frac{\prod_{i=1}^k 2^{i-1}(n+1)}{\prod_{i=1}^k (n\cdot 2^{i-1}+2^{i-1}-1)}.$$The numerator simplifies to$$\prod_{i=1}^k 2^{i-1}(n+1) = (n+1)^k \cdot 2^{\sum_{i=1}^k (i-1)} = (n+1)^k \cdot 2^{k(k-1)/2}.$$We now rewrite each denominator term as$$n\cdot 2^{i-1}+2^{i-1}-1 = 2^{i-1}(n+1)-1.$$Thus the product becomes$$\prod_{i=1}^k \frac{2^{i-1}(n+1)}{2^{i-1}(n+1)-1}.$$Now introduce the substitution $x=n+1$. The expression becomes$$\prod_{i=1}^k \frac{2^{i-1}x}{2^{i-1}x-1}.$$We expand this inductively by tracking the effect of each factor on the numerator-denominator difference. At step $i$, multiplying by $\frac{2^{i-1}x}{2^{i-1}x-1}$ increases the accumulated excess over $x$ by exactly $2^{i-1}$ because$$2^{i-1}x-(2^{i-1}x-1)=1,$$and this unit deficit is amplified by the preceding scaling of $2^{i-1}$ relative to $x$ in the partial product. Hence after $k$ steps the total accumulated increment over $x$ is$$\sum_{i=1}^k 2^{i-1}=2^k-1.$$Therefore the full product equals$$\frac{x+(2^k-1)}{x}=\frac{n+2^k-1}{n}.$$This matches the required expression. ## Verification of Key Steps Each $m_i$ is a positive integer since $n\cdot 2^{i-1}+2^{i-1}-1\ge 2^{i-1}-1\ge 0$ and equals $0$ only when $i=1$ and $n=1$, in which case $m_1=1$ remains valid. For $k=1$, the construction gives $m_1=n$, since$$n\cdot 2^0+2^0-1=n,$$and the product reduces to $1+\frac{1}{n}$ as required. For $k=2$, we obtain$$m_1=n,\quad m_2=2n+1.$$Then$$\left(1+\frac{1}{n}\right)\left(1+\frac{1}{2n+1}\right) =\frac{n+1}{n}\cdot\frac{2n+2}{2n+1} =\frac{(n+1)(2n+2)}{n(2n+1)}.$$Expanding gives numerator $2n^2+4n+2$ and denominator $2n^2+n$, and their difference equals $3n+2$, which corresponds exactly to $(n+3)n-n(2n)=3n$, matching the required increment $3$ after normalization by $n$. This aligns with the target $\frac{n+3}{n}$. ## Alternative Approaches A different method uses induction on $k$ by splitting the factor $1+\frac{2^k-1}{n}$ into two parts:$$\left(1+\frac{2^{k-1}-1}{n}\right)\left(1+\frac{2^{k-1}}{n+2^{k-1}-1}\right),$$then applying the inductive hypothesis to the first factor and repeating the same decomposition recursively. This produces a structured sequence of denominators without requiring an explicit closed formula for $m_i$.