Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Unsymmetric FE
Lahmacuncu   0
20 minutes ago
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfies $f(x^2+xy+y)+f(x^2y)+f(xy^2)=2f(xy)+f(x)+f(y)$ for all real $(x,y)$
0 replies
Lahmacuncu
20 minutes ago
0 replies
Divisibility on 101 integers
BR1F1SZ   5
N 27 minutes ago by Grasshopper-
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
5 replies
BR1F1SZ
Aug 9, 2024
Grasshopper-
27 minutes ago
Nice number theory problem
ItsBesi   9
N 29 minutes ago by Jupiterballs
Source:  Kosovo Math Olympiad 2025, Grade 8, Problem 3
Let $m$ and $n$ be natural numbers such that $m^3-n^3$ is a prime number. What is the remainder of the number $m^3-n^3$ when divided by $6$?
9 replies
ItsBesi
Nov 17, 2024
Jupiterballs
29 minutes ago
My Unsolved Problem
ZeltaQN2008   1
N 41 minutes ago by wh0nix
Let \(f:[0,+\infty)\to\mathbb{R}\) be a function which is differentiable on \([0,+\infty)\) and satisfies
\[
\lim_{x\to+\infty}\bigl(f'(x)/e^x\bigr)=0.
\]Prove that
\[
\lim_{x\to+\infty}\bigl(f(x)/e^x\bigr)0.
\]
1 reply
ZeltaQN2008
2 hours ago
wh0nix
41 minutes ago
Find the value
sqing   8
N 44 minutes ago by xytunghoanh
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
8 replies
+1 w
sqing
Mar 17, 2025
xytunghoanh
44 minutes ago
Simple inequality
sqing   22
N an hour ago by ND_
Source: JBMO 2011 Shortlist A3
$\boxed{\text{A3}}$If $a,b$ be positive real numbers, show that:$$ \displaystyle{\sqrt{\dfrac{a^2+ab+b^2}{3}}+\sqrt{ab}\leq a+b}$$
22 replies
sqing
May 15, 2016
ND_
an hour ago
greatest volume
hzbrl   0
an hour ago
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
0 replies
hzbrl
an hour ago
0 replies
Serbia national Olympiad Day 2 Problem 2
IgorM   19
N 2 hours ago by IndexLibrorumProhibitorum
Source: Serbia national Olympiad Day 2 Problem 2
Let $x,y,z$ be nonnegative positive integers.
Prove $\frac{x-y}{xy+2y+1}+\frac{y-z}{zy+2z+1}+\frac{z-x}{xz+2x+1}\ge 0$
19 replies
IgorM
Mar 28, 2015
IndexLibrorumProhibitorum
2 hours ago
find angle
TBazar   2
N 2 hours ago by sunken rock
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
2 replies
TBazar
4 hours ago
sunken rock
2 hours ago
Geometry marathon
HoRI_DA_GRe8   845
N 2 hours ago by leon.tyumen
Ok so there's been no geo marathon here for more than 2 years,so lets start one,rules remain same.
1st problem.
Let $PQRS$ be a cyclic quadrilateral with $\angle PSR=90°$ and let $H$ and $K$ be the feet of altitudes from $Q$ to the lines $PR$ and $PS$,.Prove $HK$ bisects $QS$.
P.s._eeezy ,try without ss line.
845 replies
HoRI_DA_GRe8
Sep 5, 2021
leon.tyumen
2 hours ago
polonomials
Ducksohappi   0
2 hours ago
$P\in \mathbb{R}[x] $ with even-degree
Prove that there is a non-negative integer k such that
$Q_k(x)=P(x)+P(x+1)+...+P(x+k)$
has no real root
0 replies
Ducksohappi
2 hours ago
0 replies
Four tangent lines concur on the circumcircle
v_Enhance   35
N 3 hours ago by bin_sherlo
Source: USA TSTST 2018 Problem 3
Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$.

Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$.

Evan Chen and Yannick Yao
35 replies
v_Enhance
Jun 26, 2018
bin_sherlo
3 hours ago
tangent and tangent again
ItzsleepyXD   0
4 hours ago
Source: holder send me this
Given an acute non-isosceles triangle $ABC$ with circumcircle $\Gamma$. $M$ is the midpoint of segment $BC$ and $N$ is the midpoint of $\overarc{BC}$ of $\Gamma$ (the one that doesn't contain $A$). $X$ and $Y$ are points on $\Gamma$ such that $BX \parallel CY \parallel AM$. Assume there exists point $Z$ on segment $BC$ such that circumcircle of triangle $XYZ$ is tangent to $BC$. Let $\omega$ be the circumcircle of triangle $ZMN$. Line $AM$ meets $\omega$ for the second time at $P$. Let $K$ be a point on $\omega$ such that $KN \parallel AM, \omega_b$ be a circle thaat passes through $B$, $X$ and tangents to $BC$ and $\omega_C$ be a circle that passes through $C,Y$ and tangents to $BC$. Prove that circle with center $K$ and radias $KP$ is tangent to circles $\omega_B,\omega_C$ and $\Gamma$.
0 replies
ItzsleepyXD
4 hours ago
0 replies
Geometry
Lukariman   10
N 4 hours ago by Captainscrubz
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
10 replies
Lukariman
Tuesday at 12:43 PM
Captainscrubz
4 hours ago
IMO 2018 Problem 5
orthocentre   80
N Yesterday at 10:10 PM by OronSH
Source: IMO 2018
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 $m \geq M$.

Proposed by Bayarmagnai Gombodorj, Mongolia
80 replies
orthocentre
Jul 10, 2018
OronSH
Yesterday at 10:10 PM
IMO 2018 Problem 5
G H J
Source: IMO 2018
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Markas
105 posts
#77
Y by
From the condition we get that $S(n) = \frac{a_{n+1}}{a_1} - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}}$ is an integer for all $n > N$. Now if $p \not \mid  a_1$, then the first two terms of $S(n)$ have $\nu_p \geq 0$ $\Rightarrow$ $\nu_p(\frac{a_n}{a_{n+1}}) \geq 0$ $\Rightarrow$ $\nu_p(a_n) \geq \nu_p(a_{n+1}) $ for $n \geq N$. Now we will prove that if $p \mid a_1$, then $\nu_p(a_n)$ will be constant at some point. We have that $\nu_p(a_1) > 0$. Let $\nu_p(a_k) \ge \nu_p(a_1)$ for $k > N$. We will show that for all $n \ge k$ we have $\nu_p(a_1) \leq \nu_p(a_{n+1}) \leq \nu_p(a_n)$, which can be done by induction on n. Now from $\nu(\frac{a_n}{a_1}) \geq 0$, we have that $\nu_p(\frac{a_{n+1}}{a_1} + \frac{a_n}{a_{n+1}}) \geq 0$ which gives us the inequality we wanted, if thats not true we will have exactly one term of $S(n)$ with $\nu_p \geq 0$ $\Rightarrow$ $\nu_p(a_n)$ is monotonic and at the same time is lower bounded by $\nu_p(a_1)$ $\Rightarrow$ it will eventually be constant.
Now let $\nu_p(a_k) < \nu_p(a_1)$ for every $k > N$. Also get any $n > N$. We have $\nu_p(\frac{a_{n+1}}{a_1}) < 0$, and also $\nu_p(\frac{a_n}{a_1}) < 0$, so from the three terms of $S(n)$, two must have equal $\nu_p$'s. We will check all 3 cases. 1) $\nu_p(\frac{a_{n+1}}{a_1}) = \nu_p(\frac{a_n}{a_1})$ $\Rightarrow$ ${\nu_p(a_{n+1}) = \nu_p(a_{n})}$, which is enough. 2) $\nu_p(\frac{a_{n+1}}{a_1}) = \nu_p(\frac{a_n}{a_{n+1}})$ $\Rightarrow$ ${\nu_p(a_{n+1}) = \frac{\nu_p(a_n) + \nu_p(a_1)}{2}}$, which is also enough. 3) $\nu_p(\frac{a_{n}}{a_1}) = \nu_p(\frac{a_n}{a_{n+1}})$ $\Rightarrow$ $\nu_p(a_{n+1}) = \nu_p(a_1)$, but this is false $\Rightarrow$ $\nu_p(a_{n+1}) \ge \nu_p(a_n)$ and $\nu_p(a_n)$ is upper bounded by $\nu_p(a_1)$, so we will get to a constant eventually. Since we apply this to finitely many primes, at some point $\nu_p(a_n)$ will get fixed for all $p \mid a_1$ $\Rightarrow$ the sequence will satisfy $a_{n+1} \mid a_n$ for all n $\Rightarrow$ it will eventually be constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathandski
757 posts
#78
Y by
My solution was incorrect. Thanks for report!
This post has been edited 1 time. Last edited by Mathandski, Jan 13, 2025, 6:25 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathLuis
1524 posts
#79
Y by
Let the whole sum be $S_n$ then for $n \ge N$ we will consider $S_{n+1}-S_n$, so we have that:
\[S_{n+1}-S_n \in \mathbb Z \implies \frac{a_{n+1}}{a_1}+\frac{a_n}{a_{n+1}}-\frac{a_n}{a_1}=\frac{a_{n+1}-a_n}{a_1}+\frac{a_n}{a_{n+1}}=\frac{a_{n+1}^2-a_na_{n+1}+a_na_1}{a_1a_{n+1}} \in \mathbb Z \]Therefore $a_1a_{n+1} \mid a_{n+1}^2-a_na_{n+1}+a_na_1$ so $a_{n+1} \mid a_na_1$
Now if $p \not \; \mid a_1$ then $\nu_p(a_{n+1}) \le \nu_p(a_n)$ so it becomes a decreasing sequence so it will be eventually constant.
Now notice that if we had for some $p$ prime that $p \mid a_1,a_n$ then $p \mid a_{n+1}$ so we can let $a_i=p \cdot b_i$ and realice we have the same condition for the new sequence, therefore by repeating until stuck we have that $\gcd(a_1,a_n) \mid a_{n+1}$ and from here we can also easly get $a_{n+1} \mid \text{lcm}(a_1,a_n)$, so now for a prime $p \mid a_1$ we have that $\text{min} \{\nu_p(a_1), \nu_p(a_n) \} \le \nu_p(a_{n+1}) \le \text{max} \{\nu_p(a_1), \nu_p(a_n) \}$
This holds for all $n \ge N$ so if we ever had $\nu_p(a_n)=\nu_p(a_1)$ then we would get $\nu_p(a_j)=\nu_p(a_1)$ for all $j \ge n$
And in the other 2 cases we get that either both $\nu_p(a_n), \nu_p(a_{n+1})$ are less than $\nu_p(a_1)$ or we get that $\nu_p(a_n) \ge \nu_p(a_{n+1})$ so either way for any such $p$ we get that the sequence of $\nu_p$'s stabilizes and converges at some point.
Therefore for some $M$ we have that $a_n$ is constant for all $n \ge M$ (since then we would get that $a_{n+1} \mid a_n$ and so on, so we would mess with decreasing-ness) thus we are done.
This post has been edited 2 times. Last edited by MathLuis, Sep 12, 2024, 9:13 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
numbertheory97
42 posts
#80
Y by
We prove the following quick claim, which allows us to focus on a finite number of primes.

Claim: Only finitely many primes divide $a_1, a_2, \dots$.

Proof. Actually, the only primes that divide elements of the sequence are the divisors of $a_1, a_2, \dots, a_{N - 1}$. Suppose $p \nmid a_1a_2 \cdots a_{N - 1}$ and $p \mid a_M$ for some $M \geq N$. Then \[\nu_p\left(\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{M - 1}}{a_M} + \frac{a_M}{a_1}\right) = \nu_p\left(\frac{a_{M - 1}}{a_M}\right) < 0\]since $\nu_p(a_M) > 0$, a contradiction since the expression is parentheses is an integer. Since finitely many primes are divisors of $a_1a_2 \dots a_{N - 1}$, we're done. $\square$

Now let $p$ one of these primes; it suffices to show that the sequence $\nu_p(a_1), \nu_p(a_2), \dots$ is eventually constant.

Claim: For any integer $n \geq N$, we have \[\nu_p(a_1) \leq \nu_p(a_n) \leq \nu_p(a_N)\]if $\nu_p(a_1) \leq \nu_p(a_N)$ and \[\nu_p(a_N) \leq \nu_p(a_n) \leq \nu_p(a_1)\]otherwise.

Proof. Observe that
\begin{align*}
    \left(\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_n}{a_{n + 1}} + \frac{a_{n + 1}}{a_1}\right) - \left(\frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots + \frac{a_{n - 1}}{a_n} + \frac{a_n}{a_1}\right) \\
    = \frac{a_n}{a_{n + 1}} + \frac{a_{n + 1}}{a_1} - \frac{a_n}{a_1}
\end{align*}is an integer, so $a_n/a_{n + 1} + a_{n + 1}/a_1$ and $a_n/a_1$ have the same denominator. Thus \[\nu_p\left(\frac{a_n}{a_{n + 1}} + \frac{a_{n + 1}}{a_1}\right) \geq 0 \iff \nu_p\left(\frac{a_n}{a_1}\right) \geq 0\]and \[\nu_p\left(\frac{a_n}{a_{n + 1}} + \frac{a_{n + 1}}{a_1}\right) < 0 \iff \nu_p\left(\frac{a_n}{a_1}\right) < 0 \iff \nu_p\left(\frac{a_n}{a_{n + 1}} + \frac{a_{n + 1}}{a_1}\right) = \nu_p\left(\frac{a_n}{a_1}\right) < 0.\]
Let $k = \nu_p(a_n/a_{n + 1} + a_{n + 1}/a_1)$. Suppose first that $\nu_p(a_n/a_1) \geq 0$, or alternately $\nu_p(a_n) \geq \nu_p(a_1)$. This implies that $k \geq 0$ as well, so if $\nu_p(a_{n + 1}) > \nu_p(a_n)$ or $\nu_p(a_{n + 1}) < \nu_p(a_1)$ we get $k < 0$, a contradiction. Thus $\nu_p(a_1) \leq \nu_p(a_{n + 1}) \leq \nu_p(a_n)$.

On the other hand, suppose $\nu_p(a_n) < \nu_p(a_1)$. Then $k = \nu_p(a_n/a_1) = \nu_p(a_n) - \nu_p(a_1)$, but \[k = \min\left(\nu_p\left(\frac{a_n}{a_{n + 1}}\right), \nu_p\left(\frac{a_{n + 1}}{a_1}\right)\right)\]unless $\nu_p(a_n/a_{n + 1}) = \nu_p(a_{n + 1}/a_1)$, which is clearly impossible since this implies $k = \frac12(\nu_p(a_n) - \nu_p(a_1))$. Thus $\nu_p(a_{n + 1}) \in \{\nu_p(a_1), \nu_p(a_n)\}$.

In either case, by starting at $n = N$ and inducting upward, we obtain the desired bounds on $\nu_p(a_n)$. $\square$

The claim implies that for $n \geq N$, the sequence $\nu_p(a_n)$ is monotonic and bounded between $\nu_p(a_1)$ and $\nu_p(a_n)$, so it clearly has a limiting value. Since we're only examining finitely many primes, there is some integer $K$ for which $\nu_p(a_K) = \nu_p(a_{K + 1}) = \dots$, and thus the sequence is constant beginning with $a_K$. This completes the proof. $\square$

Remark. I would have lost partial credit if I had actually done this problem in contest, since it didn't occur to me until reading hints after solving that simply showing $\nu_p(a_n)$ converges for each prime doesn't quite finish.
This post has been edited 1 time. Last edited by numbertheory97, Sep 24, 2024, 12:27 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3446 posts
#82
Y by
Claim: There are finitely many primes that divide an integer in the sequence.
Proof: For $i \geq N$ and for any prime $p$, the integer condition implies that
\[ \nu_p \left( \tfrac{a_{i-1}}{a_i} \right) \geq \min \left(\nu_p \left( \tfrac{a_1}{a_2} \right), \nu_p \left( \tfrac{a_2}{a_3} \right), \ldots, \nu_p \left( \tfrac{a_{i-2}}{a_{i-1}} \right), \nu_p \left( \tfrac{a_{i}}{a_{1}} \right) \right). \]In particular, if we set $p$ to a prime that does not divide any of $a_1, \dots, a_{i-1}$, it follows that $a_N, a_{N+1}, \dots$ cannot contain new prime divisors that don't divide any of $a_1, \dots, a_{N-1}$.

From hereon, let $i$ be any integer index greater than $N$ and fix some arbitrary prime $p$.

Claim: If $\nu ( a_{i-1} )\geq \nu ( a_1 )$, it follows that $\nu ( a_i ) \geq \nu ( a_1 )$.
Proof: Suppose FTSOC that $\nu ( a_i )< \nu ( a_1 )$. Subtracting the assertion for $i-1$ from $i$ implies that $- \tfrac{a_{i-1}}{a_1} + \tfrac{a_{i-1}}{a_i} + \tfrac{a_i}{a_1}$ is an integer. But by assumption, $\nu ( \tfrac{a_{i-1}}{a_1} )$ and $\nu ( \tfrac{a_{i-1}}{a_i} )$ are positive while $\nu ( \tfrac{a_i}{a_1} )$ is negative. This is a contradiction.

Claim: If $\nu (a_{i-1} ) < \nu ( a_i )$, we have $\nu (a_i) = \nu (a_1)$.
Proof: Once again, we only use that $- \tfrac{a_{i-1}}{a_1} + \tfrac{a_{i-1}}{a_i} + \tfrac{a_i}{a_1}$ is an integer. By assumption, $\nu (\tfrac{a_{i-1}}{a_i})$ is negative – moreover, by assumption, $\nu ( \tfrac{a_{i-1}}{a_1} )$ and $\nu ( \tfrac{a_i}{a_1} )$ are not equal. Therefore, to satisfy the integer condition, the smaller of the two, $\nu ( \tfrac{a_{i-1}}{a_1} )$, must equal $\nu (\tfrac{a_{i-1}}{a_i})$. This implies the claim.

If $\nu_p (a_{i-1} ) < \nu_p ( a_i )$ for some $i$, due to the first and second claims combined, $\nu_p (a_k) = \nu_p (a_1)$ for all $k \geq i$. Otherwise, we're already done. Applying this on our finite set of primes finishes the problem.
This post has been edited 1 time. Last edited by ihatemath123, Nov 27, 2024, 9:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bjump
1023 posts
#83
Y by
Note that for integral $C \ge 0$
$$\frac{a_1}{a_2} + \frac{a_2}{a_3}+ \cdots  +\frac{a_{N+C-1}}{a_{N+C}} + \frac{a_{N+C}}{a_1}  \in \mathbb Z$$$$\frac{a_1}{a_2} + \frac{a_2}{a_3}+ \cdots  +\frac{a_{N+C}}{a_{N+C+1}} + \frac{a_{N+C+1}}{a_1}  \in \mathbb Z$$Subtracting the first expression from the second expression gives:
$$\frac{a_{N+C+1}}{a_1} + \frac{a_{N+C}}{a_{N+C+1}} -\frac{a_{N+C}}{a_1} \in \mathbb Z$$Suppose for some prime $p$, $\nu_p (a_1) = 0$ this implies $\nu_p(a_{N+C}) \ge  \nu_p(a_{N+C+1})$ implying that the sequence is eventually constant. Now if $\nu_p (a_1) \ge 1$ then if $\nu_p(a_1) >\nu_p (a_{N+C+1}) > \nu_p (a_{N+C})$ We have
$$\nu_p (a_{N+C}) - \nu_p (a_{N+C+1}) = \nu_p(a_{N+C+1} - a_{N+C}) -\nu_p (a_1)$$$$\nu_p (a_{N+C}) - \nu_p (a_{N+C+1}) = \nu_p(a_{N+C}) -\nu_p (a_1)$$$$\nu_p (a_{N+C+1}) = \nu_p (a_1)$$If $\nu_p (a_{N+C})= \nu_p (a_1)$ suppose for the sake of contradiction that $\nu_p (a_{N+C+1}) \neq \nu_p (a_1)$, we have:
$$\nu_p (a_{N+C}) - \nu_p ( a_{N+C+1}) = \nu_p(a_{N+C+1}) - \nu_p (a_1)$$$$\nu_p (a_{N+C})+\nu_p (a_1)  = 2 \nu_p(a_{N+C+1}) $$$$\nu_p(a_1)  = \nu_p(a_{N+C+1})$$A contradiction.
If $\nu_p(a_1) > \nu_p (a_{N+C}) > \nu_p (a_{N+C+1})$ we have the middle fraction is an integer and it is impossible for $\nu_p(a_{N+C+1} - a_{N+C}) =\nu_p( a_{N+C+1}) \ge \nu_p(a_1) $ to be true.

Now suppose $\nu_p (a_{N+C}) > \nu_p (a_1)$ we have that
$$\frac{a_{N+C+1}}{a_1}+\frac{a_{N+C}}{a_{N+C+1}} \in \mathbb Z$$If $\nu_p (a_1) < \nu_p( a_{N+C+1})$ we have $\nu_p(a_{N+C+1} \le \nu_p (a_{N+C})$. Otherwise $\nu_p (a_{N+C+1})  \le  \nu_p (a_i)$.

Therefore the sequence will be stuck at a constant with $\nu_p$ less than $\nu_p (a_1)$, $\nu_p(a_1)$ if the sequnce changes at all. It is impossible for the sequence to stay strictly above $\nu_p (a_1)$ due to our last argument. Thus $(a_n)$ is eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
181 posts
#84 • 1 Y
Y by alexanderhamilton124
Easy?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihategeo_1969
235 posts
#85
Y by
No way I still haven't done this.

See that \[\frac{a_{n+1}}{a_1}+\frac{a_n}{a_{n+1}}=b_n \iff a_{n+1}a_1b_n-a_na_1=a_{n+1}(a_{n+1}-a_n) \text{ for large }n \left(\clubsuit \right)\]Say a prime $p \mid a_1$ if it exists (or else $a_{n+1} \mid a_n \implies a_{n+1} \leq a_n$ so by discrete IVT we are done).

See that there finitely many such primes $p$ and say $\nu_p(a_1)=C>0$.

Claim: Either the sequence $(\nu_p(a_n))_{n \geq 1}$ is eventually just $C$ or eventually $\nu_p(a_{n+1}) \leq \nu_p(a_n)$.
Proof: Say $\nu_p(a_{n+1})>\nu_p(a_n)$. See that applying $\nu_p$ in $\clubsuit$ we get \begin{align*}
& C+\nu_p(a_{n+1}b_n-a_n)=\nu_p(a_{n+1})+\nu_p(a_{n+1}-a_n) \iff C+\nu_p(a_n)=\nu_p(a_{n+1})+\nu_p(a_n) \iff \nu_p(a_{n+1})=C
\end{align*}Now see that if $\nu_p(a_{n+2}) \geq \nu_p(a_{n+1})$ then $\nu_p(a_{n+2})=C$. So assume the contrary. Again applying $\nu_p$ in $\clubsuit$ but rearranged and shifting $n \mapsto n+1$; we get \[\nu_p(a_{n+2})+C+\nu_p(b_{n+1})=\nu_p(a_{n+2}^2-a_{n+2}a_{n+1}+a_{n+1}a_1)=2\nu_p(a_{n+2}) \implies \nu_p(a_{n+2}) \geq C\]Which is a contradiction. $\square$

This claim obviously finishes.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathwiz_1207
97 posts
#86
Y by
We will prove that the sequence $\{v_p(a_n)\}$ is constant after a finite number of terms. Note that the condition is equivalent to
\[\frac{a_{n + 1}}{a_n} + \frac{a_n}{a_{n+1}} - \frac{a_n}{a_1} \in \mathbb{Z} \leftrightarrow \frac{(a_{n+1} - a_n)(a_{n + 1} - a_1)}{a_{n + 1}a_1} \in \mathbb{Z}\]Let $n$ be such that $n \geq N$ in what follows.


If $v_p(a_{n+1}) < v_p(a_n)$, we must have $v_p(a_{n+1}) \geq v_p(a_1)$. If $v_p(a_{n + 1}) < v_p(a_1)$, then
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_{n+1}) + v_p(a_{n+1}) - v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


If $v_p(a_{n+1}) > v_p(a_n)$, we must have $v_p(a_{n + 1}) = v_p(a_1)$. If $v_p(a_{n + 1} < v_p(a_1)$, we have
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_n) + v_p(a_{n+1}) - v_p(a_{n + 1}) - v_p(a_1) < 0\]a contradiction. Similarly, if $v_p(a_{n+1}) > v_p(a_1)$, we have
\[v_p(a_{n+1} - a_n) + v_p(a_{n + 1} - a_1) - v_p(a_{n+1}) - v_p(a_1) = v_p(a_n) + v_p(a_1) - v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


If $v_p(a_{n + 1}) = v_p(a_n)$, we must have $v_p(a_{n+1}) \geq v_p(a_1)$. If $v_p(a_{n+1}) < v_p(a_1)$, we have
\[v_p(a_{n+1}^2 - a_na_{n + 1} + a_1a_n) - v_p(a_{n + 1}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_1) < 0\]a contradiction.


Now, let $b_n = v_p(a_{n})$. Then, $b_n \geq b_1$ for all $n \geq N + 1$, since $b_1 \leq b_{n + 1} < b_n$, $b_n < b_{n + 1} = b_1$ or $b_1 \leq b_{n + 1} = b_n$. This implies that after a finite number of terms, either $\{b_n\}$ is $b_1$ or it is constant, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 17, 2025, 9:43 PM
Reason: formatting
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VideoCake
9 posts
#87
Y by
Solved together with raflikk! :)

Solution. Denote given expression by \(S_n\), and notice that \(S_{n+1} - S_n\) has to be an integer, so
\[S_{n+1} - S_n = \frac{a_{n+1} - a_{n}}{a_{1}} - \frac{a_{n}}{a_{n+1}}\]meaning that \(a_1a_{n+1} \mid a_{n+1}(a_{n+1} - a_{n}) - a_na_1\). This implies \(a_1 \mid a_{n+1}(a_{n+1} - a_{n})\) and \(a_{n+1} \mid a_na_1\). Suppose that a prime \(p\) divides \(a_1\) and \(a_n\). Then,
\[p \mid a_1 \mid a_{n+1}(a_{n+1} - a_n) \implies p \mid a_{n+1}^2\]which means that \(p \mid a_{n+1}\). Thus, \(p \mid a_i\) for all \(i \geq n\). Now we repeat the following step:

Assume that there exists a positive integer \(C\) such that all terms \(a_i\) with \(i \geq C\) are integers, and assume that \(a_1\) is an integer. Pick a prime \(p\) such that \(p \mid a_1\) and \(p \mid a_i\) (with \(i \geq C\)). Since all \(a_j\) with \(j \geq i \geq C\) are integers, we know that \(p \mid a_j\) for all \(j \geq i\). Now we divide every term in the sequence by \(p\). All ratios are still the same. (We allow some terms in the sequence to be non-integers after this step). Note how all \(a_j\) with \(j \geq i\) are still integers, so we pick our new constant \(C\) to be equal to \(i\), and note how \(a_1\) is still an integer.

Eventually, it is not possible to perform the step by picking a prime \(p\), as \(a_1\) only has a finite amount of divisors. Then, \(\gcd(a_1, a_i) = 1\) for all \(i \geq C\). Lastly, this means that for every integer \(n \geq C\), we have:
\[a_{n+1} \mid a_1a_n \implies a_{n+1} \mid a_n \implies a_{n+1} \leq a_n\]We divided all terms in the sequence with the same primes, so \(a_{n+1} \leq a_n\) also holds in the original sequence, so this sequence has to be eventually constant, we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#88
Y by
Note that for all $k \geq N,$ $$\frac{a_{k}}{a_{k+1}}+\frac{a_{k+1}}{a_1} - \frac{a_{k+1}}{a_1} \in \mathbb Z \implies \frac{a_k-a_{k+1}}{a_k}+\frac{a_{k+1}}{a_1} - \frac{a_k}{a_1} \in \mathbb Z.$$Thus $$S_k=(a_{k+1}-a_k) \left( \frac{1}{a_1} - \frac{1}{a_{k+1}} \right) \in \mathbb Z.$$Now fix a prime $p.$ If $v_p(a_N) < v_p(a_{N+1}),$ we have that $$v_p(a_N) = v_p(a_{N+1}-a_N).$$So if $v_p\left( \frac{1}{a_1}\right) \neq v_p \left( \frac{1}{a_{N+1}} \right),$ it follows that $$v_p \left( \frac{1}{a_1} - \frac{1}{a_{N+1}} \right) \leq v_p \left( \frac{1}{a_{N+1}} \right) < v_p \left(\frac{1}{a_N}\right),$$therefore adding this with above yields $v_p(S_k) < 0,$ contradiction.

Thus $v_p\left( \frac{1}{a_1} \right) = v_p \left( \frac{1}{a_{N+1}} \right) \implies v_p(a_1) = v_p(a_{N+1}).$ Now by similar logic to above, $v_p(a_{N+2}) \leq v_p(a_{N+1})$ as it is impossible for $v_p(a_{N+2}) > v_p(a_{N+1})$ and $v_p (a_1) = v_p (a_{N+2})$ to happen at the same time. But if $v_p(a_{N+2}) < v_p(a_{N+1}),$ we see that $$v_p(S_{N+1}) = v_p(a_{N+2}) + v_p \left( \frac{1}{a_1}-\frac{1}{a_{N+2}} \right) = v_p \left(\frac{a_{N+2}}{a_1}-1 \right) < 0,$$contradiction. Thus $v_p(a_{N+2}) = v_p(a_1),$ and applying this yields that the sequence $v_p(a_n)$ is eventually constant.

Now suppose that for the sake of a contradiction $v_p(a_n)$ is not eventually constant, then by above for all $k \geq N$ we have $v_p(a_k) \geq v_p(a_{k+1}).$ But this is a contradiction, as there are a finite number of possible values $v_p(a_k)$ can take, as $a_k$ are positive integers. Since $p$ is not special it follows that $\{ a_n \}$ is eventually constant. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
clarkculus
233 posts
#89 • 1 Y
Y by centslordm
Let the given expression be $f(n)$. Define $g(n)=f(n+1)-f(n)=\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}-\frac{a_n}{a_1}$, which is an integer for $n\ge N$. I claim the sequence $\nu_p(a_{N}),\nu_p(a_{N+1}),\dots$ is eventually weakly decreasing via casework.

1) $\nu_p(a_{N+1})>\nu_p(a_N)$: $g(N)=\frac{a_N}{a_{N+1}}+\frac{a_{N+1}}{a_1}-\frac{a_N}{a_1}\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_N}{a_{N+1}}\right)=\nu_p\left(\frac{a_{N+1}-a_N}{a_1}\right)\implies\nu_p(a_N)-\nu_p(a_{N+1})=\nu_p(a_{N+1}-a_N)-\nu_p(a_{1})\]so $\nu_p(a_{N+1})=\nu_p(a_1)$. Now, if $k\ge1$ and $\nu_p(a_{N+k})=\nu_p(a_1)$, $g(N+k)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+k}}{a_{N+k+1}}+\frac{a_{N+k+1}}{a_1}-\frac{a_{N+k}}{a_1}\right)\ge0\]Since $\nu_p(a_{N+k}/a_1)=0$, if $\nu_p(a_{N+k+1})>\nu_p(a_1)$, then $\nu_p(a_{N+k}/a_{N+k+1})<0$ with the second term an integer, contradiction, and if $\nu_p(a_{N+k+1})<\nu_p(a_1)$, then $\nu_p(a_{N+k+1}/a_1)<0$ with the first term an integer, contradiction. Therefore, $\nu_p(a_{N+k+1})=\nu_p(a_1)$ as well. An inductive argument then shows that $\nu_p(a_1)=\nu_p(a_{N+k})$ for all $k\ge1$.

2) $\nu_p(a_{N+1})<\nu_p(a_N)$: $g(N)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+1}-a_N}{a_1}\right)\ge0\implies\nu_p(a_{N+1})\ge\nu_p(a_{1})\]For $k\ge1$ and assuming $\nu_p(a_{N+k-1})\ge\nu_p(a_{N+k})\ge\nu_p(a_1)$, $g(N+k)\in\mathbb{Z}$ implies
\[\nu_p\left(\frac{a_{N+k}}{a_{N+k+1}}+\frac{a_{N+k+1}}{a_1}-\frac{a_{N+k}}{a_1}\right)\ge0\]so either $\nu_p(a_{N+k}/a_{N+k+1})\ge0$ and $\nu_p(a_{N+k+1}/a_1)\ge0$, which implies $\nu_p(a_{N+k})\ge \nu_p(a_{N+k+1})\ge\nu_p(a_1)$, or $\nu_p(a_{N+k}/a_{N+k+1})=\nu_p(a_{N+k+1}/a_1)$, which implies the same thing. By induction, for all $k\ge1$,
\[\nu_p(a_{N+1})\ge\nu_p(a_{N+2})\ge\dots\ge\nu_p(a_{N+k})\ge\nu_p(a_1)\]
3) $\nu_p(a_{N+1})=\nu_p(a_N)$: Trivial, as the decreasingness of the sequence of $\nu_p(a_n)$ is unchanged.

Partition all prime numbers into two sets $P$ and $Q$ such that $p\in P$ iff $\nu_p(a_1)=0$. For all $p\in P$, $g(N)\in\mathbb{Z}$ implies $\nu_p(a_N/a_{N+1})\ge0$, so $\nu_p(a_N)\ge\nu_p(a_{N+1})$, and a similar induction shows $\nu_p(a_n)$ is weakly decreasing for $n\ge N$. Also, by our claim, for each $q\in Q$, there exists a constant $N_q\ge N$ for which $\nu_q(a_n)$ is weakly decreasing for $n\ge N_q$. Since $Q$ is finite (as $a_1$ has a finite number of prime divisors), the union of $\{N\}$ and the set of all the $N_q$ is finite and thus has a maximal element $M$. Hence, for all primes $p$, $\nu_p(a_n)$ is weakly decreasing for $n\ge M$. This implies the sequence $\nu_p(a_n)$ is eventually constant for all primes $p$, so the sequence $a_n$ is eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
chenghaohu
71 posts
#90
Y by
Let $S_n = \frac{a_1}{a_2} + .... + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}$. Consider the difference between $S_{n+1}$ and $S_n$, for $n \ge N$. We will show that the fact that this difference, $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ must be an integer for all such $n \ge N$ will force $v_p (a_k) = v_p(a_{k+1})$ for all primes $p$ at sufficiently large $k\ge N$.

Consider the following cases:

Case 1: If $v_p(a_n) > v_p(a_{n+1})$.

If $v_p(a_n) > v_p(a_{n+1})$, then for $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ to be an integer, it is necessary that $v_p(a_{n+1}) \ge v_p(a_1).$ Thus considering $\frac{a_{n+2} - a_{n+1}}{a_1} + \frac{a_{n+1}}{a_{n+2}}$, we see that $v_p(a_{n+1}) \ge v_p(a_{n+2}) \ge v_p(a_1)$ resulting from $v_p(a_{n+2}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_{n+2})$. Continuing this iteration, we see that $v_p(a_{n+k})$ are all equal to some number between $v_p(a_{n+1})$ and $v_p(a_1)$, inclusive at the end, so this case indeed provides a constant value for $a_m$ where $m$ is a sufficiently large positive integer.

Case 2: If $v_p(a_n) < v_p(a_{n+1})$

Since $\frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}}$ is an integer, it is necessary that $v_p(a_{n+1} - a_n) - v_p(a_1) = v_p(a_n) - v_p(a_{n+1})$, forcing that $v_p(a_1) = v_p(n+1)$.
Now we can use an inductive process to show that if $v_p(a_z) = v_p(a_1),$ then $v_p(a_{z+1}) = v_p(a_1)$.

Base case: $z = n+1$. This works because it is necessary that $v_p(a_{n+2}) - v_p(a_1) = v_p(a_{n+1}) - v_p(a_{n+2})$ for $\frac{a_{n+2} - a_{n+1}}{a_1} + \frac{a_{n+1}}{a_{n+2}}$ to be an integer. Inductive step is the same idea as the base case.

Case 3: If $v_p(a_n) = v_p(a_{n+1})$.

$v_p(a_k) = v_p(a_{k+1})$ for all $k\ge n$, then we are good. Else it becomes one of Case $1$ or Case $2$, which resolves successfully. Thus this case also provides a constant value for $a_m$ where $m$ is a sufficiently large positive integer.

Since all $3$ possible cases works, we are done with the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
623 posts
#91
Y by
We assume the contrary and work towards a contradiction. First note that, for all $n \ge N$, both
\[\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots+\frac{a_n}{a_1} \text{ and } \frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots + \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1}\]are positive integers. Hence their difference,
\[\frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_1} - \frac{a_n}{a_1} = \frac{a_na_1+a_{n+1}^2 - a_na_{n+1}}{a_{n+1}a_1}\]is also a positive integer. Hence,
\[\frac{a_na_1+a_{n+1}^2 - a_na_{n+1}}{a_{n+1}a_1} - 1 = \frac{a_na_1+a_{n+1}^2-a_na_{n+1}-a_{n+1}a_1}{a_{n+1}a_1} = \frac{(a_{n+1}-a_1)(a_{n+1}-a_n)}{a_{n+1}a_1}\]is a non-negative integer. We can now show our key claim.

Claim : There exists a positive integer $M$ such that for all $n \ge M$ we have $\nu_p(a_n) \le \nu_p(a_1)$ or $\nu_p(a_n)$ is constant for all primes $p$.

Proof : Since the sequence $(a_i)$ is not eventually constant by assumption, consider any prime $p$ for which $\nu_p(a_i)$ is also not eventually constant. Consider the sequence $\nu_p(a_i)$. Since this is a sequence of non-negative integers which is not eventually constant it cannot be non-increasing. Consider a term $a_{n+1}$ ($n+1>N$ and $a_{n+1} \not \in \{a_1,a_n\}$) such that $\nu_p(a_{n+1})>\nu_p(a_n)$. Then, since
\[\frac{(a_{n+1}-a_1)(a_{n+1}-a_n)}{a_{n+1}a_1}\]is a non-negative integer,
\begin{align*}
    \nu_p(a_{n+1}-a_1) + \nu_p(a_{n+1}-a_n) &\ge \nu_p(a_1)+\nu_p(a_{n+1})\\
    \nu_p(a_{n+1}-a_1)+\nu_p(a_n) & \ge \nu_p(a_1)+\nu_p(a_{n+1})
    \end{align*}Now if $\nu_p(a_{n+1})>\nu_p(a_1)$ we would have,
\[\nu_p(a_1)+\nu_p(a_n) \ge \nu_p(a_1) + \nu_p(a_{n+1})\]which contradicts the condition that $\nu_p(a_{n_1})>\nu_p(a_n)$. Thus, $\nu_p(a_{n+1}) \le \nu_p(a_1)$ for all such terms $a_{n+1}$.

Even if the sequence $\nu_p(a_i) > \nu_p(a_1)$ for some $i > N$ this implies that this sequence may not increase beyond this point. Hence, it can only increase after reaching a value below $\nu_p(a_1)$ beyond which point $\nu_p(a_i) \le \nu_p(a_1)$ which implies the claim.

However, further note that the sequence $\nu_p(a_i)$ may not eventually be non-decreasing since the sequence is not eventually constant and is bounded above by $\nu_p(a_1)$. Thus, we can consider a term $a_m$ such that $m>M$ but $\nu_p(a_{m+1}) < \nu_p(a_m)$. But now note that,
\begin{align*}
    \nu_p(a_{m+1}-a_1)+\nu_p(a_{m+1}-a_m) & \ge \nu_p(a_1)+\nu_p(a_{m+1})\\
    \nu_p(a_{m+1}) + \nu_p(a_{m+1}) & \ge \nu_p(a_1) + \nu_p(a_{m+1})
\end{align*}which is a clear contradiction since this implies $\nu_p(a_m) > \nu_p(a_{m+1}) \ge \nu_p(a_1)$ violating the previous claim. But this means that the sequence $\nu_p(a_i)$ does not decrease beyond a certain point which is a contradiction. Hence, the only possibility is for $\nu_p(a_i)$ to be eventually constant for all primes $p$ which implies that $(a_i)$ is indeed eventually constant.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1733 posts
#92
Y by
The condition gives $\left\{\frac{a_i}{a_{i+1}}\right\}=\left\{\frac{a_i-a_{i+1}}{a_1}\right\}$. If $\nu_p(a_n)<\nu_p(a_{n+1})$, we have $\nu_p(a_1)=\nu_p(a_{n+1})$ from $i=n$. Then $i=n+1$ gives $\nu_p(a_{n+2})=\nu_p(a_1)$, and so on to get $\nu_p$ becomes constant. Thus if a prime has $\nu_p$ increase, it happens only once, and it must divide $a_1$, so it happens finitely often. Thus eventually all $\nu_p$s are nonincreasing, and can also only decrease finitely often since finitely many of them are nonzero. Thus they are all eventually constant, as desired.
Z K Y
N Quick Reply
G
H
=
a