IMO 2011 #5

by Wolstenholme, Oct 30, 2014, 3:38 PM

Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proof:

In the following proof, $ (x, y) $ will be used to denote the positive greatest common divisor of integers $ x $ and $ y $.

Lemma 1: $ f(n) \vert f(kn) $ for all $ k, n \in \mathbb{N} $

Proof: We proceed with induction on $ k $. Note that the base case of $ k = 1 $ is trivial. Since $ f(n) = f(kn - (k - 1)n) \vert f(kn) - f((k - 1)n) $ we have that $ f(n) \vert f((k - 1)n) \Longrightarrow f(n) \vert f(kn) $ which completes the induction.

Lemma 2: $ f(n) = f(-n) $ for all $ n \in \mathbb{N} $

Proof: By an easy extension Lemma 1 we have that $ f(n) \vert f(-n) $ and similarly $ f(-n) \vert f(n) $ which implies the desired result.

Lemma 3: For all $ m, n \in \mathbb{Z} $, there exists $ r, s \in \mathbb{Z} $ such that $ mr + ns = (m, n). $

Proof: This is Euclid's Lemma

Now consider any two positive integers $ m, n $ and the corresponding integers $ r, s $ such that $ mr + ns = (m, n) $. Then $ f(ns) \vert f(mr) - f((m, n)) $ and $ f(mr) \vert f(ns) - f((m, n)). $ By Lemma 1 we have that $ f((m, n)) $ divides both $ f(mr) $ and $ f(ns) $ so one of $ f(ns) $ and $ f(mr) $ is equal to $ f((m, n)). $ Assume WLOG that $ f(mr) = f((m, n)). $ Then $ f(m) \vert f(mr) = f((m, n)) \vert f(n) $ by repeated applications of Lemma 1 as desired.

Note I kind of ignored cases and negatives because I'm lazy but hopefully this general idea works.

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