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==================================== Topic =====================================
Link: https://artofproblemsolving.com/community/c6h1671291
Category: High School Olympiads
Title: IMO 2018 Problem 5
Source: IMO 2018
Post count: 57
View count: 18398
===================================== Post =====================================
Number: 1 (https://artofproblemsolving.com/community/c6h1671291p10632353)
Posted by: orthocentre (https://artofproblemsolving.com/community/user/343248)
Posted at: Jul 10, 2018, 6:19 AM (UTC-05)
Like count: 18
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=================================== Content ====================================
Let $a_1$, $a_2$, $\ldots$ be an infinite sequence of positive integers. Suppose
that there is an integer $N > 1$ such that, for each $n \geq N$, the number
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n} +
\frac{a_n}{a_1}$$
is an integer. Prove that there is a positive integer $M$ such that $a_m =
a_{m+1}$ for all $m \geq M$.

[i]Proposed by Bayarmagnai Gombodorj, Mongolia[/i]
===================================== Post =====================================
Number: 50 (https://artofproblemsolving.com/community/c6h1671291p24227866)
Posted by: mijail (https://artofproblemsolving.com/community/user/338783)
Posted at: Jan 26, 2022, 11:32 AM (UTC-05)
=================================== Content ====================================
Nice NT problem  :D

My solution is based on proving the following claim:
[color=#00f][b]Key claim:[/b][/color] For any $n \ge N$ we have that:
$\mathrm{lcm}(a_{n+1},a_1) \mid \mathrm{lcm}(a_n,a_1)$
[color=#00f]Proof:[/color] The difference $\frac{a_{n+1}}{a_1} - \frac{a_n}{a_1}
+ \frac{a_n}{a_{n+1}}-1=(a_{n+1}-a_n)(\frac{a_{n+1}-a_1}{a_{n+1}a_1})$ is a
integer so we can find positive integers $d,x,y,z,a,b,c$ such that:
$$a_{n+1}=dxya, a_n=dxzb, a_1=dyzc$$
With $(a_{n+1},a_n)=dx, (a_{n},a_1)=dz, (a_{n+1},a_1)=dy$ and also
$(a_{n+1},a_n,a_1)=d$ so we have that the next number is a integer:
$$\frac{(dxya-dxzb)(dxya-dyzc)}{d^2y^2xazc}=\frac{(ya-zb)(xa-zc)}{yazc}$$
$$\implies a\mid z^2bc  $$ $$\implies z\mid a^2xy  $$
But $(a,zbc)=1$ and $(z,axy)=1$ so $a=z=1$ but then
$\mathrm{lcm}(a_{n+1},a_1)=dyzxac=dyxc$ and
$\mathrm{lcm}(a_{n},a_1)=dzxybc=dxybc$ this implies the claim. $\square$
--------------------------------------------------------
[b][color=#f00]Corollary:[/color][/b] $\{a_n\}$ is bounded and
$\{\mathrm{lcm}(a_n,a_1)\}$ is eventually constant.
The collary implies that eventually in the claim we have $b=1$ for any $n \ge
N'$ so $a_{n+1}=dxy$ and $a_n=dx \implies a_{n+1} \ge a_n$ this implies that
$\{a_n\}$ is eventually non-decresing but also is bounded so its eventually
constant. $\blacksquare$
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