Olympiad Maths Prep

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Problem 1183

National olympiad, first round
Number theory Difficulty 6.3 Prove it

Exercise 3. For any integer nNn \in \mathbb{N}^{*}, we denote v3(n)v_{3}(n) as the 3-adic valuation of nn, which is the greatest integer kk such that nn is divisible by 3k3^{k}. We set u1=2u_{1}=2 and un=4v3(n)+22un1u_{n}=4 v_{3}(n)+2-\frac{2}{u_{n-1}} for all n2n \geqslant 2 (provided that un1u_{n-1} is defined and non-zero).

Show that, for any positive rational number qq, there exists a unique integer n1n \geqslant 1 such that un=qu_{n}=q.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

## Solution to Exercise 3

First, note that u1=2,u2=1,u3=3,u4=32,u5=23u_{1}=2, u_{2}=1, u_{3}=3, u_{4}=\frac{3}{2}, u_{5}=\frac{2}{3} and u6=3u_{6}=3. We prove by induction that, for all integers n2n \geq 2, we have 0000, we set pq=p+q\left\|\frac{p}{q}\right\|=p+q. Then:

- φ(pq)=pqp=p<p+q=pq\left\|\varphi\left(\frac{p}{q}\right)\right\|=\left\|\frac{p}{q-p}\right\|=p<p+q=\left\|\frac{p}{q}\right\| if 0<p<q0<p<q;
- φ(1+pq)=2pqp2q<(p+q)+q=1+pq\left\|\varphi\left(1+\frac{p}{q}\right)\right\|=\left\|2 \frac{p}{q-p}\right\| \leq 2 q<(p+q)+q=\left\|1+\frac{p}{q}\right\| if 0<p<q0<p<q;
- φ(2+pq)=pq=p+q<(p+2q)+q=2+pq\left\|\varphi\left(2+\frac{p}{q}\right)\right\|=\left\|\frac{p}{q}\right\|=p+q<(p+2 q)+q=\left\|2+\frac{p}{q}\right\| if 0<p0<p and 0<q0<q.

In all cases, if xx is a strictly positive rational number such that x{1,2}x \notin\{1,2\}, we have φ(x)<x\|\varphi(x)\|<\|x\|. Now, for all strictly positive rational numbers xx and for all integers n1n \geq 1:

- if 0<x<10<x<1, then un=φ(x)u3n+2=xu_{n}=\varphi(x) \Leftrightarrow u_{3 n+2}=x;
- if 1<x<21<x<2, then un=φ(x)u3n+1=xu_{n}=\varphi(x) \Leftrightarrow u_{3 n+1}=x;
- if 2<x2<x, then un=φ(x)u3n=xu_{n}=\varphi(x) \Leftrightarrow u_{3 n}=x.

An induction on x\|x\| immediately shows that, for all strictly positive rational numbers xx, there exists a unique integer n1n \geq 1 such that un=xu_{n}=x, which concludes the exercise.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.