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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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
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Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
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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
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Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
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Introduction to Number Theory
Friday, May 9 - Aug 1
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Introduction to Algebra B Self-Paced

Introduction to Algebra B
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Introduction to Geometry
Sunday, May 11 - Nov 9
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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
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Wednesday, Jun 25 - Dec 10
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Intermediate Counting & Probability
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Precalculus
Friday, May 16 - Oct 24
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Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
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Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
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Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
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AMC 10 Problem Series
Friday, May 9 - Aug 1
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Tuesday, May 27 - Aug 12
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Thursday, May 22 - Jul 31

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Sunday, Jun 22 - Sep 21

F=ma Problem Series
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WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
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Programming

Introduction to Programming with Python
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Tuesday, May 13 - Jul 29
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Physics

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Physics 1: Mechanics
Thursday, May 22 - Oct 30
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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 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
2021 KMO 1st round Gauss Combinatoric Problem
kwan2010   3
N 3 minutes ago by MathIQ.
Find the number of functions that satisfy both of the following conditions.
(i) f(1)≤f(2)≤...≤f(9)
(ii) The number of elements in the range of the composition function f∘f is 7.

The answer is Click to reveal hidden text
3 replies
kwan2010
Feb 16, 2025
MathIQ.
3 minutes ago
IMO Shortlist 2014 G6
hajimbrak   30
N 18 minutes ago by awesomeming327.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic.
Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$
Proposed by Ali Zamani, Iran
30 replies
hajimbrak
Jul 11, 2015
awesomeming327.
18 minutes ago
sum of 4 primes with 5 <p <q <r <s <p + 10 is divisible by 60
parmenides51   4
N 26 minutes ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 4
Let $p, q, r$ and $s$ be four prime numbers such that $$5 <p <q <r <s <p + 10.$$Prove that the sum of the four prime numbers is divisible by $60$.

(Walther Janous)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
26 minutes ago
2player game, adding numbers, whoever reaches no >= 2019 wins
parmenides51   2
N 28 minutes ago by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 3
Alice and Bob are playing a year number game.
There will be two game numbers $19$ and $20$ and one starting number from the set $\{9, 10\}$ used. Alice chooses independently her game number and Bob chooses the starting number. The other number is given to Bob. Then Alice adds her game number to the starting number, Bob adds his game number to the result, Alice adds her number of games to the result, etc. The game continues until the number $2019$ is reached or exceeded.
Whoever reaches the number $2019$ wins. If $2019$ is exceeded, the game ends in a draw.
$\bullet$ Show that Bob cannot win.
$\bullet$ What starting number does Bob have to choose to prevent Alice from winning?

(Richard Henner)
2 replies
parmenides51
Dec 18, 2020
MathIQ.
28 minutes ago
4th grader qual JMO
HCM2001   23
N 2 hours ago by nitride
i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier
23 replies
+3 w
HCM2001
Yesterday at 12:53 AM
nitride
2 hours ago
[TEST RELEASED] OMMC Year 5
DottedCaculator   107
N 2 hours ago by ethan2011
Test portal: https://ommc-test-portal-2025.vercel.app/

Hello to all creative problem solvers,

Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.

Our website: https://www.ommcofficial.org/

This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.

How hard is it?

We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!

How are the problems?

You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.

Final Round: May 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.

Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]
107 replies
DottedCaculator
Apr 26, 2025
ethan2011
2 hours ago
for the contest high achievers, can you share your math path?
HCM2001   25
N 3 hours ago by KnowingAnt
Hi all
Just wondering if any orz or high scorers on contests at young age (which are a lot of u guys lol) can share what your math path has been like?
- school math: you probably finish calculus in 5th grade or something lol then what do you do for the rest of the school? concurrent enrollment? college class? none (focus on math competitions)?
- what grade did you get honor roll or higher on AMC 8, AMC 10, AIME qual, USAJMO qual, etc?
- besides aops do you use another program to study? (like Mr Math, Alphastar, etc)?

You're all great inspirations and i appreciate the answers.. you all give me a lot of motivation for this math journey. Thanks
25 replies
HCM2001
Wednesday at 7:50 PM
KnowingAnt
3 hours ago
another diophantine about primes
AwesomeYRY   133
N 3 hours ago by EpicBird08
Source: USAMO 2022/4, JMO 2022/5
Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.
133 replies
AwesomeYRY
Mar 24, 2022
EpicBird08
3 hours ago
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   40
N 4 hours ago by OGMATH
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

----------

[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
40 replies
Indy_Integirls
May 11, 2025
OGMATH
4 hours ago
MAA finally wrote sum good number theory
IAmTheHazard   96
N Yesterday at 4:54 PM by megahertz13
Source: 2021 AIME I P14
For any positive integer $a,$ $\sigma(a)$ denotes the sum of the positive integer divisors of $a.$ Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a.$ Find the sum of the prime factors in the prime factorization of $n.$
96 replies
IAmTheHazard
Mar 11, 2021
megahertz13
Yesterday at 4:54 PM
Integer Functional Equation
msinghal   99
N Yesterday at 3:55 PM by DeathIsAwe
Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that \[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\] for all $x, y \in \mathbb{Z}$ with $x \neq 0$.
99 replies
msinghal
Apr 29, 2014
DeathIsAwe
Yesterday at 3:55 PM
mathpath: how much do recommendations matter
mm999aops   25
N Yesterday at 2:09 PM by ethan2011
See question^

I'm hoping only the QT matters : P
25 replies
mm999aops
Feb 3, 2023
ethan2011
Yesterday at 2:09 PM
Segment has Length Equal to Circumradius
djmathman   74
N Yesterday at 12:19 PM by amirhsz
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
74 replies
djmathman
Apr 30, 2014
amirhsz
Yesterday at 12:19 PM
Essentially, how to get good at olympiad math?
gulab_jamun   11
N Yesterday at 2:13 AM by oinava
Ok, so I'm posting this as an anynonymous user cuz I don't want to get flamed by anyone I know for my goals but I really do want to improve on my math skill.

Basically, I'm alright at computational math (10 AIME, dhr stanford math meet twice) and I hope I can get good enough at olympiad math over the summer to make MOP next year (I will be entering 10th as after next year, it becomes much harder :( )) Essentially, I just want to get good at olympiad math. If someone could, please tell me how to study, like what books (currently thinking of doing EGMO) but I don't know how to get better at the other topics. Also, how would I prepare? Like would I study both proof geometry and proof number theory concurrently or just study each topic one by one?? Would I do mock jmo/amo or js prioritize olympiad problems in each topic. I have the whole summer ahead of me, and intend to dedicate it to olympiad math, so any advice would be really appreciated. Thank you!
11 replies
gulab_jamun
May 18, 2025
oinava
Yesterday at 2:13 AM
IMO 2018 Problem 5
orthocentre   80
N May 7, 2025 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
May 7, 2025
IMO 2018 Problem 5
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Source: IMO 2018
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Markas
150 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
771 posts
#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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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
3449 posts
#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
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#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
183 posts
#84 • 1 Y
Y by alexanderhamilton124
Easy?
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ihategeo_1969
239 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
103 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
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#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
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#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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OronSH
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#92
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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.
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