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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.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
-----------------------------
13. Blind speculation: discouraged
Along these lines, if you do see a question that you don't have an answer to, we discourage "blindly guessing" as it leads to spreading of baseless rumors. For example, if you see some user posting "why are there fewer qualifiers than usual this year?", you should not reply "the MAA must have been worried about online cheating so they took fewer people!!". Was sich überhaupt sagen lässt, lässt sich klar sagen; und wovon man nicht reden kann, darüber muss man schweigen.
-----------------------------
14. Discussion of cheating: strongly discouraged
If you have evidence or reasonable suspicion of cheating, please report this to your Competition Manager or to the AMC HQ; these forums cannot help you.
Otherwise, please avoid public discussion of cheating. That is: no discussion of methods of cheating, no speculation about how cheating affects cutoffs, and so on --- it is not helpful to anyone, and it creates a sour atmosphere. A longer explanation is given in Seriously, please stop discussing how to cheat.
-----------------------------
15. Cutoff jokes: never allowed
Whenever the cutoffs for any major contest are released, it is very obvious when they are official. In the past, this has been achieved by the numbers being posted on the official AMC website (here) or through a post from the AMCDirector account.

You must never post fake cutoffs, even as a joke. You should also refrain from posting cutoffs that you've heard of via email, etc., because it is better to wait for the obvious official announcement. A longer explanation is given in A Treatise on Cutoff Trolling.
-----------------------------
16. Meanness: never allowed
Being mean is worse than being immature and unproductive. If another user does something which you think is inappropriate, use the Report button to bring the post to moderator attention, or if you really must reply, do so in a way that is tactful and constructive rather than inflammatory.
-----------------------------

Finally, we remind you all to sit back and enjoy the problems. :D

-----------------------------
(EDIT 2024-09-13: AoPS has asked to me to add the following item.)

Advertising paid program or service: never allowed

Per the AoPS Terms of Service (rule 5h), general advertisements are not allowed.

While we do allow advertisements of official contests (at the MAA and MATHCOUNTS level) and those run by college students with at least one successful year, any and all advertisements of a paid service or program is not allowed and will be deleted.
0 replies
v_Enhance
Jun 12, 2020
0 replies
k i Stop looking for the "right" training
v_Enhance   50
N Oct 16, 2017 by blawho12
Source: Contest advice
EDIT 2019-02-01: https://blog.evanchen.cc/2019/01/31/math-contest-platitudes-v3/ is the updated version of this.

EDIT 2021-06-09: see also https://web.evanchen.cc/faq-contest.html.

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
My Unsolved FE on R+
ZeltaQN2008   2
N 2 minutes ago by ZeltaQN2008
Source: IDK
Give $a>0$. Find all funcitions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(xf(y)+a)=yf(x+y+a)$$
2 replies
ZeltaQN2008
an hour ago
ZeltaQN2008
2 minutes ago
Insects walk
Giahuytls2326   1
N 23 minutes ago by removablesingularity
Source: somewhere in the internet
A 100 × 100 chessboard is divided into unit squares. Each square has an arrow pointing up, down, left, or right. The board square is surrounded by a wall, except to the right of the top right corner square. An insect is placed in one of the squares.

Every second, the insect moves one unit in the direction of the arrow in its square. As the insect moves, the arrow of the square it just left rotates 90° clockwise.

If the specified movement cannot be performed, then the insect will not move for that second, but the arrow in the square it is standing on will still rotate. Is it possible that the insect never leaves the board?
1 reply
Giahuytls2326
May 3, 2025
removablesingularity
23 minutes ago
Inequality
MathsII-enjoy   3
N 32 minutes ago by MathsII-enjoy
A interesting problem generalized :-D
3 replies
MathsII-enjoy
Saturday at 1:59 PM
MathsII-enjoy
32 minutes ago
something...
SunnyEvan   1
N 33 minutes ago by SunnyEvan
Source: unknown
Try to prove : $$ \sum csc^{20} \frac{2^{i} \pi}{7} csc^{23} \frac{2^{j}\pi }{7} csc^{2023} \frac{2^{k} \pi}{7} $$is a rational number.
Where $ (i,j,k)=(1,2,3) $ and other permutations.
1 reply
SunnyEvan
Today at 1:13 AM
SunnyEvan
33 minutes ago
9 Did I make the right choice?
Martin2001   33
N Today at 2:11 AM by happypi31415
If you were in 8th grade, would you rather go to MOP or mc nats? I chose to study the former more and got in so was wondering if that was valid given that I'll never make mc nats.
33 replies
Martin2001
Apr 29, 2025
happypi31415
Today at 2:11 AM
late ross acceptance
basketball314159   4
N Today at 2:07 AM by NoSignOfTheta
Hi everyone,

I got rejected from Ross a couple weeks ago and then today got an acceptance letter, but I didn't to my knowledge get put on a waitlist. Did anyone experience something similar?

Also, for those who have done Ross, honest thoughts? I did HCSSiM and absolutely loved it, but I realize every math program is different.

Thanks!
4 replies
basketball314159
Yesterday at 12:30 AM
NoSignOfTheta
Today at 2:07 AM
Mathcounts state
happymoose666   36
N Today at 1:47 AM by happypi31415
Hi everyone,
I just have a question. I live in PA and I sadly didn't make it to nationals this year. Is PA a competitive state? I'm new into mathcounts and not sure
36 replies
happymoose666
Mar 24, 2025
happypi31415
Today at 1:47 AM
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   10
N Today at 12:47 AM by qwerty123456asdfgzxcvb
Hi everyone!

After the resounding success of the first three years of PVMT, the Poolesville High School Math Team is excited to announce the fourth annual Poolesville High School Math Tournament (PVMT)! The PVMT team includes a MOPper and multiple USA(J)MO and AIME qualifiers!

PVMT is open to all 6th-9th graders in the country (including rising 10th graders). Students will compete in teams of up to 4 people, and each participant will take three subject tests as well as the team round. The contest is completely free, and will be held virtually on June 7, 2025, from 10:00 AM to 4:00 PM (EST).

Additionally, thanks to our sponsors, we will be awarding approximately $10K+ worth of prizes (including gift cards, Citadel merch, AoPS coupons, Wolfram licenses) to top teams and individuals. More details regarding the actual prizes will be released as we get closer to the competition date.

Further, newly for this year we might run some interesting mini-events, which we will announce closer to the competition date, such as potentially a puzzle hunt and integration bee!

If you would like to register for the competition, the registration form can be found at https://pvmt.org/register.html or https://tinyurl.com/PVMT25.

Additionally, more information about PVMT can be found at https://pvmt.org

If you have any questions not answered in the below FAQ, feel free to ask in this thread or email us at falconsdomath@gmail.com!

We look forward to your participation!

FAQ
10 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
qwerty123456asdfgzxcvb
Today at 12:47 AM
USAMO Medals
YauYauFilter   9
N Yesterday at 11:35 PM by Alex-131
YauYauFilter
Apr 24, 2025
Alex-131
Yesterday at 11:35 PM
everyone will get zero marx on this
Th3Numb3rThr33   48
N Yesterday at 9:50 PM by Blast_S1
Source: JMO 2018 Problem 6
Karl starts with $n$ cards labeled $1,2,3,\dots,n$ lined up in a random order on his desk. He calls a pair $(a,b)$ of these cards swapped if $a>b$ and the card labeled $a$ is to the left of the card labeled $b$. For instance, in the sequence of cards $3,1,4,2$, there are three swapped pairs of cards, $(3,1)$, $(3,2)$, and $(4,2)$.

He picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had $i$ card to its left, then it now has $i$ cards to its right. He then picks up the card labeled $2$ and reinserts it in the same manner, and so on until he has picked up and put back each of the cards $1,2,\dots,n$ exactly once in that order. (For example, the process starting at $3,1,4,2$ would be $3,1,4,2\to 3,4,1,2\to 2,3,4,1\to 2,4,3,1\to 2,3,4,1$.)

Show that, no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.
48 replies
Th3Numb3rThr33
Apr 19, 2018
Blast_S1
Yesterday at 9:50 PM
Find the radius of circle O
TheMaskedMagician   3
N Yesterday at 8:38 PM by fruitmonster97
Source: 1976 AHSME Problem 18
IMAGE

In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$; point $D$ is interior to the circle; and $DB$ intersects the circle at $C$. If $BC=DC=3$, $OD=2$, and $AB=6$, then the radius of the circle is

$\textbf{(A) }3+\sqrt{3}\qquad\textbf{(B) }15/\pi\qquad\textbf{(C) }9/2\qquad\textbf{(D) }2\sqrt{6}\qquad \textbf{(E) }\sqrt{22}$
3 replies
TheMaskedMagician
May 18, 2014
fruitmonster97
Yesterday at 8:38 PM
MathPath
PatTheKing806   13
N Yesterday at 8:05 PM by Nora2021
Is anybody else going to MathPath?

I haven't gotten in. its been 3+ weeks since they said my application was done.
13 replies
PatTheKing806
Mar 24, 2025
Nora2021
Yesterday at 8:05 PM
ARMl Local 2025 Final Results
PaulDreyer   27
N Yesterday at 5:50 PM by llddmmtt1
Results, problems, and solutions are here. Congratulations to SFBA ARML / AlphaStar: Foxes and Friends and Leading Aces Academy for placing 1st and 2nd overall and to Liam Reddy (Utah Rookies) for their perfect score on the individual round and for being the only student with a perfect score to answer the tiebreaker correctly.
27 replies
PaulDreyer
Saturday at 5:25 PM
llddmmtt1
Yesterday at 5:50 PM
MathILy 2025 Decisions Thread
mysterynotfound   28
N Yesterday at 5:49 PM by mysterynotfound
Discuss your decisions here!
also share any relevant details about your decisions if you want
28 replies
mysterynotfound
Apr 21, 2025
mysterynotfound
Yesterday at 5:49 PM
IMO 2018 Problem 5
orthocentre   79
N Apr 17, 2025 by cursed_tangent1434
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
79 replies
orthocentre
Jul 10, 2018
cursed_tangent1434
Apr 17, 2025
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.
Martin.s
1536 posts
#76 • 1 Y
Y by buddyram
First of all, it \(n=2\) is not possible since we must have \(a_{2^m} + 2a_{2^{m+1}} = 0\) for each \(m\) which inductively gives \(a_1 = (-2)^ma_{2^m},\) which means that \(2^m|a_1\) for every \(m\) — this is impossible since \(a_1\) must be non-zero.

We will show that it applies to each \(n \geq 3.\) It is enough for each \(n\) to find a function \(f_n : \mathbb{N} \to \mathbb{N},\) which from now on I will denote with \(f,\) with the properties
(a) \(f(rs) = f(r)f(s)\) for each \(r,s \in \mathbb{N}\)
(b) \(f(1) + 2f(2) + \cdots + nf(n) = 0\).

Indeed, if \(f\) satisfies (a) and (b), then setting \(a_m = f(m)\) for each \(m\), we have \(\displaystyle{a_k + 2a_{2k} + \cdots + na_{nk} = f(k) + 2f(2k) + \cdots + nf(nk) = f(k)(1 + 2f(2) + \cdots +nf(n)) = 0.}\)

We can therefore consider \(n \geq 3.\)

We know from Bertrand's theorem that for every \(n \geq 2,\) there exists a prime \(p\) with \(n/2 < p \leq n.\) Let \(p\) be the greatest prime less than or equal to \(q\) and \(n,\) where \(q\) is the largest prime that is less than \(n.\) Then we have \(n/4 < q < p \leq n.\)

If \(f(m) = a^{m_p}b^{m_q},\) posit where \(m_p, m_q\) are their powers in the prime factorization, and \(a, b\) are nonzero integers which will be determined later. We see that (a) is satisfied, and it is enough to choose them appropriately \(a, b\) so that (b) is also satisfied.

Case 1: \(q > n/2.\) Then we want \(n-2 + pf(p) + qf(q) = 0.\) Because \(p, q\) are coprime, there are \(x, y\) such \(px + qy = 1.\) We define \(a = -(n-2)x, b = -(n-2)y,\) and we are done.

Case 2: \(n/3 < q \leq n/2.\) Then we want \(n-3 + pf(p) + qf(q) + 2qf(2q) = 0.\)

Case 2a: If \(q=2,\) then it must be \(n=3\) or \(n=4\) (since if \(n \geq 5,\) then \(p \geq 5\) and \(q \geq 3\)). The case \(n=3\) is rejected after \(q > n/2.\) For \(n=4,\) we have \(p=3, q=2,\) and want \(1 + 3f(3) + 2f(2) + 4f(2)^2 = 0,\) and it is enough to choose \(a = f(3) = -1\) and \(b = f(2) = -1.\)

Case 2b: If \(q > 2,\) then we want \(n-3 + pf(p) + 3qf(q) = 0,\) and because \(p, 3q\) are coprime, we can find suitable \(a, b\) as in case 1. (For the \(a, b\) non-zeros to be, it is enough to check that \(n-3 \neq 0,\) which is true since if \(n=3,\) we would have \(q=2\) and would not be in case 2.)

Case 3: \(n/4 < q \leq n/3.\) Then we want \(n-4 + pf(p) + qf(q) + 2qf(2q) + 3qf(3q) = 0.\)

Case 3a: If \(q \leq 3,\) then you must \(n \leq 6.\) (Since if \(n \geq 7,\) then \(p \geq 7\) and \(q \geq 5 > 3.\)) But if \(n \leq 6,\) then \(q \leq n/3 \leq 2.\) So \(q = 2,\) and \(n=6.\) But this is inappropriate since for \(n=6,\) we have \(q=3.\) So here we have nothing to control.

Case 3b: If \(q > 3,\) then we want \(n-4 + pf(p) + 6qf(q) = 0,\) and because \(p, 6q\) are coprime, we can find suitable \(a, b\) as in case 1. (For the \(a, b\) non-zeros to be, it is enough to check that \(n \neq 4,\) which is true since if \(n=4,\) we would have \(q=2\) and would not be in case 3.)
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Markas
105 posts
#77
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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.
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Mathandski
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#78
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My solution was incorrect. Thanks for report!
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MathLuis
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#79
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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
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numbertheory97
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#80
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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
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ihatemath123
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#82
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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
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bjump
1013 posts
#83
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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.
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lelouchvigeo
181 posts
#84 • 1 Y
Y by alexanderhamilton124
Easy?
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ihategeo_1969
231 posts
#85
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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.
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mathwiz_1207
96 posts
#86
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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
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VideoCake
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#87
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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.
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Maximilian113
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#88
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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
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clarkculus
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#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.
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chenghaohu
71 posts
#90
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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.
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cursed_tangent1434
616 posts
#91
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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.
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