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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
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Prealgebra 2 Self-Paced

Prealgebra 2
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Introduction to Algebra A Self-Paced

Introduction to Algebra A
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Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
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Introduction to Algebra B Self-Paced

Introduction to Algebra B
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Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

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Intermediate Algebra
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Tuesday, Jun 10 - Aug 26

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Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

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

MATHCOUNTS/AMC 8 Advanced
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WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
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Programming

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Physics 1: Mechanics
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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
Prove n is square-free given divisibility condition
CatalanThinker   1
N 2 minutes ago by CatalanThinker
Source: 1995 Indian Mathematical Olympiad
Let \( n \) be a positive integer such that \( n \) divides the sum
\[
1 + \sum_{i=1}^{n-1} i^{n-1}.
\]Prove that \( n \) is square-free.
1 reply
CatalanThinker
an hour ago
CatalanThinker
2 minutes ago
Inspired by SXJX (12)2022 Q1167
sqing   0
7 minutes ago
Source: Own
Let $ a,b,c>0 $. Prove that$$\frac{kabc-1} {abc(a+b+c+8(2k-1))}\leq \frac{1}{16 }$$Where $ k>\frac{1}{2}.$
0 replies
1 viewing
sqing
7 minutes ago
0 replies
What is thiss
EeEeRUT   5
N 10 minutes ago by MathLuis
Source: Thailand MO 2025 P6
Find all function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$,such that the inequality $$f(x) + f\left(\frac{y}{x}\right) \leqslant \frac{x^3}{y^2} + \frac{y}{x^3}$$holds for all positive reals $x,y$ and for every positive real $x$, there exist positive reals $y$, such that the equality holds.
5 replies
EeEeRUT
May 14, 2025
MathLuis
10 minutes ago
Thailand geometry
EeEeRUT   4
N 29 minutes ago by MathLuis
Source: Thailand MO 2025 P7
Let $ABC$ be a triangle with $AB < AC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects $BC$ at $D$. The angle bisector of $\angle BAC$ intersect $BC$ at $E$. Suppose that the perpendicular bisector of $AE$ intersect $AB, AC$ at $P,Q$, respectively. Show that $$\sqrt{\frac{BP}{CQ}} = \frac{AC \cdot BD}{AB \cdot CD}$$
4 replies
EeEeRUT
May 14, 2025
MathLuis
29 minutes ago
Docked 4 points Help
sadas123   8
N 2 hours ago by bjump
In school we had this beginners like middle school contest, but we had to right down our solution kind of like usajmo except no proofs. It was also graded out of 7 but I got 4 Points docked for this question. what was my problem??? But I kind of had to rush the solution on this question because there was another problem before this that was like 1000x times harder.

Question:The solutions to the equation x^3-13x^2+ax−48=0 are all positive whole numbers. What is $a$?


Solution: We can see that we can use Vieta's formulas to find that the product of the roots is $48$, and the sum of the roots is $13$. So we need to find a combination of integers that multiply to $48$ and add up to $13$. Let's call the roots of the equation p, q, and r. From Vieta's, we get that $p+q+r=-13$ and $pqr = -48$. Looking at the factors of $48$, which is $2^4*3$, we try to split the numbers in a way that gives us the correct sum and product. Trying 3, -2, and -8, we see that they add up to $-13$ and multiply to $-48$, so they work. That means the roots of the polynomial are -3, -2, and -8, and the factorization is $(x-3)(x-2)(x-8)$. Multiplying it out, we get $x^3-13x^2+46x-48$, so we find that a = 46.
8 replies
sadas123
Yesterday at 4:06 PM
bjump
2 hours ago
Cyclic Quad
worthawholebean   131
N 3 hours ago by mathwiz_1207
Source: USAMO 2008 Problem 2
Let $ ABC$ be an acute, scalene triangle, and let $ M$, $ N$, and $ P$ be the midpoints of $ \overline{BC}$, $ \overline{CA}$, and $ \overline{AB}$, respectively. Let the perpendicular bisectors of $ \overline{AB}$ and $ \overline{AC}$ intersect ray $ AM$ in points $ D$ and $ E$ respectively, and let lines $ BD$ and $ CE$ intersect in point $ F$, inside of triangle $ ABC$. Prove that points $ A$, $ N$, $ F$, and $ P$ all lie on one circle.
131 replies
worthawholebean
May 1, 2008
mathwiz_1207
3 hours ago
Jane street swag package? USA(J)MO
arfekete   40
N 4 hours ago by Pengu14
Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:

- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package

I will reply with my info as an example.
40 replies
arfekete
May 7, 2025
Pengu14
4 hours ago
OTIS or MathWOOT 2
math_on_top   4
N 4 hours ago by Pengu14
Hey AoPS community I took MathWOOT 1 this year and scored an 8 on AIME (last year I got a 6). My goal is to make it to MOP next year through USAMO. It's gonna be a lot of work, but do you think that I should do MathWOOT 2 or OTIS? Personally, I felt that MathWOOT 1 taught me a lot but was more focused on computational - not sure how to split computation vs olympiad prep. So, for those who can address this question:

(1) How much compuational vs olympiad
(2) OTIS or MathWOOT 2 and why
4 replies
math_on_top
Yesterday at 9:56 PM
Pengu14
4 hours ago
An FE. Who woulda thunk it?
nikenissan   118
N 5 hours ago by maromex
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
118 replies
nikenissan
Apr 15, 2021
maromex
5 hours ago
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   21
N 6 hours ago by Penguin117
[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: Follow us on Instagram @indy.integirls, join our Discord, follow us on TikTok @indy.integirls, and 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!
21 replies
Indy_Integirls
May 11, 2025
Penguin117
6 hours ago
Erecting Rectangles
franchester   103
N Yesterday at 9:10 PM by torch
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that \[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.
103 replies
franchester
Apr 15, 2021
torch
Yesterday at 9:10 PM
9 best high school math competitions hosted by a college/university
ethan2011   21
N Yesterday at 7:38 PM by linjiah
I only included college-hosted comps since MAA comps are very differently formatted, and IMO would easily beat the rest on quality since mathematicians around the world give questions, and so many problems are shortlisted, so IMO does release the IMO shortlist for people to practice. I also did not include the not as prestigious ones(like BRUMO, CUBRMC, and others), since most comps with very high quality questions are more prestigious(I did include other if you really think those questions are really good).
21 replies
1 viewing
ethan2011
Apr 12, 2025
linjiah
Yesterday at 7:38 PM
d_k-eja Vu
ihatemath123   50
N Yesterday at 5:57 PM by MathematicalArceus
Source: 2024 USAMO Problem 1
Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n!$ in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
Proposed by Luke Robitaille.
50 replies
ihatemath123
Mar 20, 2024
MathematicalArceus
Yesterday at 5:57 PM
Essentially, how to get good at olympiad math?
gulab_jamun   3
N Yesterday at 5:29 PM by sus_rbo
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!
3 replies
gulab_jamun
Yesterday at 1:53 AM
sus_rbo
Yesterday at 5:29 PM
Unique NT Function
IndoMathXdZ   33
N Apr 10, 2025 by N3bula
Source: IMO SL 2018 N6
Let $f : \{ 1, 2, 3, \dots \} \to \{ 2, 3, \dots \}$ be a function such that $f(m + n) | f(m) + f(n) $ for all pairs $m,n$ of positive integers. Prove that there exists a positive integer $c > 1$ which divides all values of $f$.
33 replies
IndoMathXdZ
Jul 17, 2019
N3bula
Apr 10, 2025
Unique NT Function
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO SL 2018 N6
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IndoMathXdZ
694 posts
#1 • 3 Y
Y by megarnie, Kingsbane2139, Adventure10
Let $f : \{ 1, 2, 3, \dots \} \to \{ 2, 3, \dots \}$ be a function such that $f(m + n) | f(m) + f(n) $ for all pairs $m,n$ of positive integers. Prove that there exists a positive integer $c > 1$ which divides all values of $f$.
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plagueis
157 posts
#2 • 6 Y
Y by Aryan-23, IMUKAT, megarnie, Adventure10, MS_Kekas, khina
Proposed by Ariel García, Mexico
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tastymath75025
3223 posts
#3 • 10 Y
Y by cjquines0, smurfcc, mijail, Mop2018, MiraclesINmaths, Zyz1010, hakN, Adventure10, bhan2025, pokpokben
Great problem :) We’ll split into two cases.
Case 1: $f$ is unbounded. In that case, let $n_1<n_2<...$ be an infinite sequence of integers such that $f(n_i) > f(n_i-1), f(n_i-2), \dots, f(1)$ for each $i$.

For any positive integer $k$ take large $i$ so that $f(n_i-k) \mid f(n_i-k-1) + f(1)$. Note that $f(n_i)\mid  f(n_i-k) + f(k) < 2f(n_i)\implies f(n_i-k)=f(n_i)-f(k)$ and similarly $f(n_i-k-1) = f(n_i) - f(k+1)$. Therefore we have that $f(n_i) - f(k) \mid f(n_i)-f(k+1) + f(1)$, hence $f(n_i)-f(k) \mid f(1)+f(k)-f(k+1)$. As $i$ increases $f(n_i)-f(k)$ gets arbitrarily large, implying $f(k+1)=f(k)+f(1)$ for all $k$, so $f(n)=nf(1)$ and $f(1)>1$ divides each $f(n)$, and we’re done.

Case 2: $f$ is bounded by some $M$. For any $p$ and $n>m$, note that $p\mid f(n),f(m)$ implies $p\mid f(n-m)$ because $f(n)\mid f(m)+f(n-m)$. Consider primes $p$ that divide infinitely many $f(n)$ and let $S$ be the set of all such primes. Let $u$ be the smallest positive integer with $p\mid f(u)$; it follows that for any $n$ with $p\mid f(n)$, we have $p\mid f(n-u), f(n-2u), \dots, f(n \pmod u)$, which contradicts the minimality of $u$ unless $u\mid n$. Notice this allows us to associate each prime $p$ with an “order” $u_p=u$ such that $p\mid f(n)\implies u_p\mid n$. Suppose for contradiction that $u_p\neq 1$ for each $p\in S$.

Now clearly $S$ is finite, and if $m_k= k\text{lcm}_{p\in S} u_p + 1$, then $u_p\nmid m_k$ for each $p\in S$ and each $k$, so no $p\in S$ can divide $f(m_k)$. But the sequence $f(m_1), f(m_2), \dots$ is bounded and infinite, so it takes some value greater than one infinitely many times, meaning there is a prime $q$ which divides infinitely many of the $f(m_i)$. But this would imply $q\in S$, contradiction. Therefore it follows that some $p\in S$ has $u_p=1$. But now by definition there are arbitrarily large $n$ with $p\mid f(n)$, and $p\mid f(n), f(1)\implies p\mid f(n-1)$, so this implies $p\mid f(n)$ for all $n$, and we’re done.
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Kayak
1298 posts
#4 • 3 Y
Y by AlastorMoody, aops29, Adventure10
This was Indian TST D2 P3. Here's my solution to this at TST:

Case-1 $\text{Im}(f)$ is bounded.
Proof: Call a prime $p$ to be Soumil if there's infinitely many integers $i$ such that $p|f(i)$. For a soumil prime $p$, define it's lavil to be the set $T_p : \overset{\text{def}}{=} \{ i \in \mathbb{N} : p | f(i) \}$. Note that if $a, b \in T_p$ then $|a-b| \in T_p$, thus (say by Euclidean algorithm), there's an integer $g_p \in \mathbb{N}$ such that $T_p = g_p \mathbb{N}$.

Let $S$ be the set of all soumil primes. Now I claim atleast one of $g_p$ where $p \in S$ will be equal to $1$ (which would imply the proof). Assume not. Then define $S : \overset{\text{def}}{=} \prod_{p \in S} g_p$, and consider the set $Q = \{ Sm+1 \}_{m \in \mathbb{N}}$ where $m$ varies over integer. Note that by definition of soumil primes, the union of all lavil sets contains all but finitely many integers. But the elements of $Q$ don't belong to any lavil set, which is a contradiction so done. $\square$
Case-2 $\text{Im}(f)$ is unbounded.

In this case I prove $f(m) = f(1)m$.

Lemma: (Growth of $f$) We say the interval $[i, j]$ is soumil $f(n+1)-f(n) = f(1)$ for each $n \in [i, j]$. For each positive integer $m$, there're infinitely many soumil intervals of the form $[mk, m(k+1)-1]$.

Proof

So pick any integer $j$, and an integer $M > \max(f(1), f(2), \cdots, f(j))+100$, and a soumil interval $[n, 2^M+n]$. Then $f(n+M+j) = f(n)+(M+j)f(1) > M + jf(1) > \max(jf(1), f(j))$, and thus \begin{align*}
f(n+M+j) & | f(n+M) + f(j) = f(n) + Mf(1)+f(j) \\
& | f(n+M+j-1) + f(1) = f(n) + (M+j)f(1) = f(n) + Mf(1) + jf(1)
\end{align*}so subtracting them gives $f(n+M+j) | f(j) - jf(1)$ which gives the conclusion $\square$.
This post has been edited 4 times. Last edited by Kayak, Jul 17, 2019, 9:22 PM
Reason: typoes
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juckter
323 posts
#5 • 32 Y
Y by Wizard_32, Kayak, 62861, danepale, Pathological, djmathman, Nothing000, Ti-Ci, p_square, AlastorMoody, aops29, rashah76, RAMUGAUSS, amar_04, Aryan-23, khina, SnowPanda, Supercali, starchan, tigerzhang, L567, megarnie, OlympusHero, MiraclesINmaths, CyclicISLscelesTrapezoid, ZHEKSHEN, IMUKAT, Mz_T, Kingsbane2139, sabkx, Adventure10, Deadline
This was proposed by me :)

I hope you all enjoyed it/will enjoy it!
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v_Enhance
6877 posts
#6 • 12 Y
Y by v4913, Kobayashi, SerdarBozdag, Anshul_singh, HamstPan38825, ZHEKSHEN, sabkx, Nuterrow, Adventure10, Assassino9931, khina, bin_sherlo
The following solution splices together ideas from both of the official solutions.

Claim: Let $d_n = \gcd(f(n), f(1))$. Then $d_{n+1} \mid d_n$ for all $n$.
Proof. Immediate from $f(n+1) \mid f(n) + f(1)$. $\blacksquare$

Hence, we will assume henceforth that $d_n = \gcd(f(n), f(1)) = 1$ for all $n \ge C$, for some constant $C$ (otherwise $\min_n d_n$ is the required constant).

Claim: If $\gcd(a,b) = 1$, then $\gcd(f(a), f(b)) \mid f(1)$. In particular, $\gcd(f(a), f(b)) = 1$ if $\min(a,b) \ge C$.
Proof. This follows by Euclidean algorithm: if $a > b$ then $f(a) \mid f(a-b) + f(b)$, so $\gcd( f(a), f(b) ) \mid \gcd( f(a-b), f(b) )$. $\blacksquare$

The second claim already implies $f$ is unbounded: if $f$ takes only values in $\{2, \dots, L\}$ then taking any $L+1$ primes larger than $C$ gives a contradiction.

Main induction: Assume $f$ is unbounded. Let $X$ be a local maximum, meaning $Y = f(X) \ge \max \left\{ f(1), \dots, f(X-1) \right\}$. We will assume $X$ and $Y$ are large enough such that $X > C + C f(1)$, $Y > 2 f(1) f(C)$. Then: \begin{align*} 	Y = f(X) \mid f(X-C) + f(C) &\implies f(X-C) = Y - f(C) \\ 	Y-f(C) = f(X-C) \mid f(X-2C) + f(C) &\implies f(X-2C) = Y - 2f(C) \\ 	Y-2f(C) = f(X-2C) \mid f(X-3C) + f(C) &\implies f(X-3C) = Y - 3f(C) \\ 	&\vdotswithin{\implies} \\ 	&\implies f(X-f(1)C) = Y - f(1) f(C). \end{align*}However, all the right-hand sides are supposed to not be divisible by $f(1)$, yet we assumed $\gcd(f(C), f(1)) = 1$, contradiction.
This post has been edited 1 time. Last edited by v_Enhance, Nov 25, 2022, 4:13 AM
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MarkBcc168
1595 posts
#7 • 11 Y
Y by hyx, rashah76, Mathasocean, Aryan-23, Kanep, Kobayashi, Adventure10, Kingsbane2139, Mango247, sabkx, Deadline
Here is a short solution by Gems98.

First, we make the following observation.

Claim: If $d\mid f(m)$ and $d\mid f(n)$ then $d\mid f(m-n)$.

Proof: Obvious from the problem's condition.

We will consider two cases.
Case 1: $f$ is bounded.

Let $p_{1}$ be a prime number. By Dirichlet's theorem, there exists a sequence of prime numbers $p_{2}, p_{3}, \cdots, $ such that
$$p_{n+1} \equiv 1 \pmod{\prod_{i=1}^{n}p_{i}}$$By pigeonhole principle, there exists a increasing sequence $a_{1}, a_{2} ,\cdots$ such that
$$c = f(p_{a_{1}}) = f(p_{a_{2}}) = \cdots$$From $p_{a_{i+1}} \equiv 1 \pmod{p_{a_{i}}}$, we get $c \mid f(1)$ by applying the Claim repeatedly. Therefore, if $c \mid f(n)$ then $c \mid f(n-1)$.

Since the sequence $p_{a_{i}}$ is unbounded, we get $c \mid f(n)$ for every $n \in \mathbb{Z}^+$ which is the desired result.

Case 2: $f$ is unbounded.

We call the natural number $n$ $\textit{good}$ iff $f(n) > \max\{f(1), \cdots, f(n-1)\}$. Since $f$ is unbounded, there exist infinitely many $\textit{good}$ number.

It's easy to show that if $m$ is $\textit{good}$, then $f(m) = f(i) + f(m-i)$ for all $i \leq m-1$. Let $n$ be a positive integer and $m$ be a sufficiently large $\textit{good}$ number. From the condition, we get
\begin{align*}
    f(m-n+1) &\mid f(m-n)+f(1) \\
    f(m) - f(n-1) &\mid f(m) - f(n) + f(1) \\ 
    f(m) - f(n) &\mid f(n) - f(n-1) - f(1)
\end{align*}Since $m$ is sufficiently large, we get $f(n) - f(n-1) - f(1) = 0$ which implies that $f$ is linear.
Therefore, there exists positive integer $c = f(1) > 1$ such that $c\mid f(n)$ for all $n \in \mathbb{Z}^+$.
This post has been edited 1 time. Last edited by MarkBcc168, Jul 17, 2019, 2:42 PM
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mofumofu
179 posts
#8 • 7 Y
Y by test20, danepale, AlastorMoody, Kingsbane2139, Adventure10, Mango247, sabkx
While this problem is a really nice one, I would like to point out some similar (and difficult) problems that have appeared before, namely 2007 N5, IMO 2011, Iran 3rd round 2017, and with a different condition Romania TST 2016.
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juckter
323 posts
#9 • 3 Y
Y by starchan, Adventure10, Mango247
Yeah, there are some other pretty cool NT functional problems around. I definitely got reminded of some of those when I came up with this problem :)

I learned of the existence of the Romania TST problem because I heard this one was discarded from the shortlist due to it, which was a shame :( . Had I known that problem existed beforehand I may have not sent my problem to be considered for IMO.
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yayups
1614 posts
#10 • 3 Y
Y by Adventure10, Mango247, sabkx
We split into cases based on whether $f$ is bounded or not.

First, assume $f$ is unbounded. This means that there is an infinite sequence of \textit{peaks} $1=x_0<x_1<x_2<\cdots$ such that $f(x_n)>f(x)$ for all $x<x_n$. Note that $f(x_n)\mid f(x)+f(x_n-x)$ and $f(x)+f(x_n-x)<2f(x_n)$, so in fact $f(x_n)=f(x)+f(x_n-x)$. For any given $x$, we have for large enough $n$ that $f(x_n-x)\mid f(1)+f(x_n-x-1)$, so
\[f(x_n)-f(x)\mid f(1)+f(x_n)-f(x+1),\]so $f(x_n)-f(x)\mid f(1)+f(x)-f(x+1)$ for arbitrarily large $n$. We see that $f(x_n)$ can be arbitrarily large, so we must have $f(x+1)=f(x)+f(1)$, so $f(x)=xf(1)$. Thus, $f(1)\mid f(n)$ for all $n$.

Now suppose $f$ is bounded. Note that $f(n+1)\mid f(n)+f(1)$, so $\gcd(f(n+1),f(1))\mid\gcd(f(n),f(1))$. Thus,
\[\cdots\mid\gcd(f(3),f(1))\mid\gcd(f(2),f(1))\mid\gcd(f(1),f(1)),\]so either $\gcd(f(n),f(1))=1$ for large enough $n$, or $\gcd(f(n),f(1))>1$ for all $n$, in which case we're done. So WLOG, suppose that $\gcd(f(n),f(1))$ for all $n\ge N$.

We claim that $\gcd(f(x),f(xy+1))=1$ for $x\ge N$. To see this, note that
\[f(xy+1)\mid f(x)+f(x(y-1)+1)\implies\gcd(f(xy+1),f(x))\mid\gcd(f(x(y-1)+1),f(x)).\]Iterating this gives $\gcd(f(xy+1),f(x))\mid\gcd(f(1),f(x))=1$, as desired. Thus, any two terms in
\[f(N),f(N+1),f(N(N+1)+1),f(N(N+1)(N(N+1)+1)+1),\ldots\]are relatively prime, so in particular, any two terms are distinct. But this is an infinite sequence of $f(x)$ for distinct values of $x$, so $f$ is unbounded, which is the desired contradiction.
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mathaddiction
308 posts
#11 • 5 Y
Y by amar_04, jishu2003, a_n, Mango247, ihatemath123
This is the first shortlist #6 that I solve(This is also the few ones that I have attempted), and spend a lot of time (10+ hours), so I decided to post it on the thread.
Let $a_n=f(2^n)$ for all $n\geq 0$. We distinguish two cases:
$\underline{Case I:}$For all $n\geq 0$, $a_n$ has an odd prime factor
note that by taking $m=n=2^k$ we have
$$a_k|2a_{k-1}$$Hence any prime factors of $a_n$ must also be a prime factor of $a_i$ where $i$ is any positive integer less than $i$. Since $a_1$ has only finitely many prime factors, there exists a prime p such that $p|a_n$ for all positive integers $n$. We that $p|f(n)$ for all $n\in\mathbb N$.
Indeed, we proceed by backward induction to show that every integer in the interval $[2^k,2^{k+1}]$ has this property, indeed we have $p|f(2^{k+1})$. Now note that if $p|f(n)$ then since $p|f(1)$ together with
$$f(n)|f(n-1)+f(1)$$we have $p|f(n-1)$ and therefore we complete the inductive step and therefore we're done.
$\underline{Case II:}$There exists $k\in\mathbb N$ such that for all $N\geq k$ we have the fact that $a_n$ is a power of $2$.(We assume that $k$ is the minimal choice)
Using similar arguments as in case I we have that for every number $m$ there exists a number $k$ such that for all $N\geq k$ we have $f(m\cdot 2^k)$ is a power of 2.
Note that if $p|f(m)$ and $p|f(n)$, then we have $p|f(n-m)$ therefore $p|f(gcd(m,n))$ by Euclidean algorithm. That is
$$gcd(f(m),f(n))|f(gcd(m,n))(1)$$Let $q$ be the minimal number such that $a_q$ is even. We claim that if $v_2(n)<q$ then $f(n)$ is odd. Otherwise, suppose $v_2(n)=s$, by $(1)$ we have $$2|gcd(f(n),f(2^k))|f(gcd(f(n),f(2^k))=f(2^s)$$this contradicts the minimality of $q$
We further distinguish two cases
$\underline{Case IIa:}$$q\geq1$
We prove the following
$\underline{CLAIM 1.}$The sequence $\{a_n\}$ is unbounded
$\underline{Proof.}$Suppose $\{a_n\}<M$ for all $n$, then since
$$f(2^n+1)|f(2^n)+f(1)$$We have $f(2^n+1)\leq f(2^n)+f(1)\leq M+f(1)$ hence $f(2^n+1)$ is also bounded, therefore by pigeonhole principle there exists an infinite sequence of numbers $\{b_n\}$ such that
$$f(2^{2^{b_i}}+1)=f(2^{2^{b_j}}+1)$$for all $i,j\in\mathbb N$.Let this number be $c$, then since $(2^{2^{b_i}}+1,2^{2^{b_j}}+1)=1$, from $(1)$ we have
$c|f(1)$, using the same induction method as in Case I we have $c|f(n)$ for all positive integer $n$, contradiction.
$\underline{CLAIM 2.}$ Let $a$ be an integer, denote $v_2(n)$ by $m$, then
$$v_2(f(a))=v_2(f(2^m))$$$\underline{Proof.}$Suppose $a=2^mp$, and $v_2(f(a))=x$, $v_2(f(2^m))=y$
If $x>y$, then we prove by induction that $v_2(f(an+2^m))\leq y$ for all positive integers $n$
Indeed the base cases are trivial, suppose the case n holds, then
$$v_2(f(a(n+1)+2^m))\leq v_2(f(an+2^m)+f(n))\leq y$$so we're done. In particular, let $d$ be the order of $2$ mod $p$, pick n such that $np+1=2^wd$ for all integer $w$, we obtain that $v_2(a_{wd+m})=y$, and since $v_2(a_i)+1\geq v_2(a_{i+1})$ we must have $v_2(a_n)\leq y$ that is, $a_n$ is bounded, which violates CLAIM 1.
If $y>x$, then we prove by induction that $v_2(f(a+n2^m))\leq x$ for all positive integers $n$
Indeed the base cases are trivial, suppose the case n holds, then
$$v_2(f(a+(n+1)2^m))\leq v_2(f(a+n2^m)+f(2^m))\leq x$$so we're done. pick n such that $n+p+1=2^w$ for all integer $w$, we obtain that $v_2(a_{w})=x$, that is, $a_n$ is bounded, which violates CLAIM 1.
Combining the two cases, we proves the claim.
$\underline{CLAIM 3.}$If $a\geq k$, then
$$v_2(f(2^{a+1})=v_2(f(2^a))+1$$$\underline{Proof.}$
Otherwise, since $v_2(f(2^{a+1})\leq v_2(f(2^a))+1$ either
(i)$v_2(f(2^{a+1})<v_2(f(2^a))$ or
(ii)$v_2(f(2^{a+1})=v_2(f(2^a))$
If $(i)$ holds,
$$v_2(f(2^a))>v_2(f(2^{a+1})=v_2(f(2^{a+1})+f(2^a))\geq v_2(f(2^{a+1}+2^a))$$This violates CLAIM 2 since $v_2(2^{a+1}+2^a)=v_2(2^a)$
If $(ii)$ holds, then we let $f(2^a)=2^m$. We prove by induction that
$$f(n2^a)=2^m$$for all integer $n$ not divisible by $4$. Indeed from $(ii)$ we have $f(2\cdot 2^a)=2^m$
Now suppose all smaller cases holds, we have
$$f(n2^a)| f((n-1)2^a)+f(2^a)=2^{m+1}$$From CLAIM 2 we have $v_2f(n2^k)=m$ and therefore $f(n2^k)=2^m$, this completes the inductive step. Now we have, for all integers $b$,
$$f(2^{b+m})|f(2^m)+f((2^b-1)2^m)=2^{a+1}$$that is $a_n$ is bounded, this violates CLAIM 1, so we prove CLAIM 3.
$\underline{CLAIM 4.}$
For all integer $n$ we have
$$f(n2^k)=n2^m$$$\underline{Proof.}$
We proceed by induction. In the following we denote $f(2^k)=2^m$. Note that $CLAIM 3$ implies that for all integer $b$ we have $f(2^{b+k})=2^{m+b}$
The base cases are trivial. Now suppose all smaller cases hold we consider the case $n$,suppose $2^{t-1}\leq n\leq 2^t$, we have
$$f(n2^k)|f((n-1)2^k)+f(2^k)$$$$f(2^t\cdot 2^m)|f(n2^k)+f(2^t-n)2^k)$$applying the inductive hypothesis we have
$$n2^m\leq f(n2^k)\leq n2^m$$This completes the inductive step and we hence have proof CLAIM 4.
$\underline{CLAIM 5.}$
$$f(2^{k-1})=2^{m-1}$$$\underline{Proof.}$
Now we have
$$f(n2^k+2^{k-1})|f(n2^k)+f(2^{k-1})$$$$f((n+1)2^k)|f(n2^k+2^{k-1})+f(2^{k-1})$$for all integer $n$. Let the terms on the left and right side of the first divisibility be $A$ and $B$, and the terms on the left and right side of the second divisibility be $C$ and $D$.
Note that $A\leq B$ and $C\leq D$. If $A=B$ and $C=D$, we add them up and apply $CLAIM 4$ to get
$$f(2^{k-1})=2^{m-1}$$which is what we want to prove.
Otherwise, either one of $2A\leq B$ or $2C\leq D$ must hold.
If the former holds we have
$$2^m+f(n2^k+2^{k-1})\leq f(2^{k-1})$$since $f(n2^k+2^{k-1})\geq \frac{1}{2}n2^{m+1}+2^m$
we have $f(2^{k-1})\geq (n+1)2^{m+1}$ which can't hold for large $n$.
If the latter holds we have
$$(n+1)2^m\leq 2f(2^{k-1})$$which also can't hold for large $n$. So CLAIM 5 is proved.
However, we define $k$ to be the minimal integer such that $f(2^k)$ is a power of $2$, so we obtain a contradiction, which rejects the case $q\geq 1$
$\underline{Case IIb.}$$q=0$
Now we have $f(1)$ is even, using the same reasoning as in Case I, we have that for all $n$, $f(n)$ is even, so we're done.
Hence we reject all case and the proof is thus completed.
This post has been edited 2 times. Last edited by mathaddiction, May 5, 2020, 3:07 AM
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niksic
81 posts
#12
Y by
Hopefully this isn't equivalent to any previous solutions :D Nowhere near as pretty as the first one tho :(

Notice that if $p|f(1)$ and $p|f(n)$, then $p|f(m)$ for all $m\leq n$. So for the sake of contradiction, assume that $(f(1),f(n))=1$ for all $n>N$, where $N$ is some integer.

Claim 1: $f(n)\leq n+C$ where $C$ is a constant

Also notice that $f(2n)|2f(n)$, so by induction we have $f(2^k)|2^kf(1)$, so $f(2^k)|2^k$ for $2^k>N$. Because $f(n+m)|f(n)+f(m)$, we also have $f(n+m)\leq f(n)+f(m)$. We will also use these two inequalities: $f(2^k)\leq 2^k$ for $k\geq \log_2(N)$ and $f(2^k)\leq 2^kf(1)$ for the smaller ones. Writing any $n$ as a sum of powers of two and using the inequality as many times as needed, we get $f(n)\leq n+(1+2+4+...+2^{\log_2(N)})f(1)\leq n+C$ where $C$ is constant.

Claim 2: Let $a_p$ be the smallest value $n$ for which $p|f(n)$. Then we have $p|f(n)\implies a_p|n$
Assume the contrary and let $k$ be the smallest value for which $p|f(k)$ and $a_p\nmid k$. Then $k>a_p$ and we have $f(k)|f(k-a_p)+f(a_p)\implies p|f(k-a_p)$, a contradiction.

We now split into two cases:

Case 1 $f(n)\leq \frac{1}{2}(f(n-1)+f(1))$ for infinitely many values of $n$
Take some big $n$ and $\varepsilon>0$ such that $n+2C\leq n(1+\varepsilon)$. Notice that by our assumption we can make $\varepsilon$ arbitrarily small

Take a big integer $m$ and write it as follows: $f(m)=f(an+b)\leq af(n)+f(b)\leq a\frac{1}{2}(f(n-1)+f(1))+f(b)\leq a\frac{1}{2}(n-1+C+1+C)+b+C$
$\leq \frac{1}{2}an(1+\varepsilon)+b\frac{1}{2}(1+\varepsilon)+n+C\leq (an+b)\frac{1}{2}(1+\varepsilon)+C+n = m\frac{1}{2}(1+\varepsilon)+n+C<\frac{3}{4}m$ for $m$ large enough, say for $m\geq N_0$. Take $M=max(N,N_0)$
(These inequalities are far from pretty but they work :D and obviously $3/4$ is arbitrary)

Now look at primes that are larger then $M$ .Say the number of primes that divide $f(1)f(2)\cdots f(k)$ for some $k$ is $g(k)$. By claim 2, you can notice that all primes larger than $M$ are $a_p$ for some $p$, which means $g(k)\geq \pi(k)-\pi(M)$. Say the largest prime dividing $f(1)f(2)\cdots f(M)$ is $P$. Notice that for $k>\frac{4}{3}P$, $\pi(\frac{3}{4}k)\geq g(k)\geq \pi{k}-\pi{M} \iff \pi(k)-\pi(\frac{3}{4}k)\leq \pi(M)$, the right-hand side is fixed, and from here there are many ways to prove that this is a contradiction.
(I won't be writing any of them because I can't think of an elementary way :( )

Case 2 $f(n)=f(n-1)+f(1)$ for $n\geq L$, where $L$ is some integer

This means $f(m)=f(L)+f(1)(m-L) = mf(1)+(f(L)-Lf(1))$ for $m\geq L$. Remember we had the property of $f(2)$ to be a power of $2$. It is trivial to see that this forces $f(L)-Lf(1)=0$ and $f(1)$ is a power of $2$, meaning $2$ divides all $f(m)$ with $m\geq L$, contradiction
This post has been edited 3 times. Last edited by niksic, Aug 13, 2020, 7:00 PM
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Aryan-23
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#13
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Nice problem !

We distinguish cases based on the boundedness of $f$.

Case 1 : $\operatorname {Im}f$ is bounded.

For all primes $p$, define the set $S_p= \{n : p\mid f(n)\}$. Call a prime $p$ strong , if $S_p$ is infinite and weak otherwise. Its easy to see that $p\vert f(a) , f(b) \implies p\mid f(a-b)$, hence $S_p=x_p \cdot \mathbb Z^{+}$ , for some $x_p$. Suppose that for all strong primes $p$, $x_p\neq 1$. Then note that since $f(n)$ is divisible by only finitely many weak primes, hence this means that the set $ \mathbb Z^+ \setminus { \bigcup_{p 
   \quad         \text{strong}} S_p } $ is finite. Consider, $X= \prod_{p  \quad \text{strong}} x_p$ and look at $X\mathbb Z^++1$. Obviously its infinite and has $f(\bullet)$ not divisible by any strong prime, contradiction. $\blacksquare$. Hence $x_p=1$ for some strong prime $p$, and the conclusion is immediate.

Case 2 : $f$ is unbounded.

The idea is to worry only about the size of $f$ and force "Cauchyness" on some intervals.

We show $f(x)=xf(1)$ for all $x\in \mathbb Z^+$

Call a $x$ to be a peak if $f(x) > \operatorname {max} \{f(1),f(2) ,\dots f(x-1)\}$. Obviously there are infinitely many such peaks due to unboundedness. Its easy to see that due to size issues, $f(x)=f(x-k)+f(k)$ holds for all $x>k$.

We show $f(n+1)-f(n)=f(1)$ for all $n$.
Pick a arbitrary $x$ and $y$ such that $x+y$ is a peak and $f(x+y)$ is huge. Spamming the equality $f(x+y)=f(x)+f(y)=f(x-1)+f(y+1)$, we get :

$$ f(y+1) \vert f(y)+f(1) $$$$\implies f(x+y)-f(x-1) \mid f(x+y)-f(x)+f(1)$$$$\implies f(x+y)-f(x-1) \mid f(x)-f(x-1)-f(1)$$
Taking $f(x+y)$ to be huge we're done. $\blacksquare$
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shalomrav
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For the bounded case, just take a prime $p$ for which all the values of $f(n)$ appeared before $p$. So there is some $n$ smaller than $p$ such that $f(p)=f(n)=c$ and by euclidean algorithm $c$ divides $f(1)$, and by more application of euclidean algorithm (induct down, begin with $p,1$) we get $c$ divides every guy in $f(p-1), f(p-2), f(p-3) ... $ and so. we are done since by assumption every value of $f$ is included in this sequence.
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SnowPanda
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jj_ca888
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Let $P(m, n)$ denote the assertion. First of all, $P(1, n)$ yields\[f(n+1) \mid f(n) + f(1) \implies \gcd(f(n+1), f(1)) \mid \gcd(f(n), f(1))\]always. Now, we only need to consider when there exists $M$ for which $\gcd(f(n), f(1)) = 1$ for all $n \geq M$ since\[\gcd(f(n), f(1)) \mid f(n) \mid \gcd(f(n-1), f(1))\]so the sequence $\gcd(f(i), f(1))_{i = 1}^{\infty}$ is in fact nonincreasing, and if it never hits $1$, then the lower bound $L$ of the sequence divides all $\gcd(f(i), f(1))$, and thus divides all $f(i)$, at which point we would be done.

Next, we will prove that in the case where $f(n), f(1)$ are relatively prime for all sufficiently large $n \geq M$, then $f$ is unbounded. First, note that $P(n, m-n)$ yields $f(n) \mid f(m-n) + f(m)$ hence $\gcd(f(n), f(m)) \mid \gcd(f(m-n), f(n))$ which is analogous to the Euclidean Algorithm, and performing this continuously yields $\gcd(f(n), f(m)) \mid f(\gcd(m, n))$. Thus, for pairs $(m, n)$ that are relatively prime, and $m, n \geq M$, it follows that $\gcd(f(m), f(n)) \mid f(1)$ but since\[\gcd(f(m), f(1)) = \gcd(f(n), f(1)) = 1\]we know $\gcd(f(m), f(n)) = 1$. It easily follows that $f$ is then unbounded; if it were upper bounded by $U$, then just pick some $N > U$ numbers $a_1, \ldots , a_N \geq M$ for which their $f$ values are all pairwise relatively prime; no two $f(a_i)$ could be the same, hence the range of $f$ needs to hit at least $N > U$ numbers, thus $U$ could not have been the upper bound, a contradiction.

Lastly, we finish by considering the case where $f$ is unbounded. Consider numbers $k$ for which $f(k) > f(1), \ldots , f(k-1)$. There are clearly infinitely many of these $k$, else $f$ is upper bounded by the largest $f(k)$. Pick arbitrarily large $K$ for which $f(K)$ is large, and greater than all $f(i)$ for $i < K$. Then for any positive $x + y = K$, note $f(x) + f(y) < 2f(K)$ but still must be a whole multiple of $K$, hence $f(x) + f(y) = K$. Note that since $f(x + 1) \mid f(x) + f(1)$,\[f(K) - f(y-1) \mid f(K) - f(y) + f(1) \implies f(K) - f(y-1) \mid f(y-1) - f(y) + f(1)\]where clearly $|f(y-1) - f(y) + f(1)| < |f(K) - f(y-1)|$ from picking $K$ such that $f(K)$ is much larger than every one of $f(y), f(y-1), f(1)$. Finally, this implies that $f(y) = f(y-1) + f(1)$ for all integers $y$, hence $f$ is linear and then induction easily shows that $f(y) = yf(1)$, so indeed $f(1) \mid f(y)$ for all $y$, finishing this case as well.

After exhausting all cases, we have shown $f(1)$ does indeed always divide all $f(n)$.
This post has been edited 11 times. Last edited by jj_ca888, Jun 29, 2021, 9:10 PM
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L567
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Cool problem! :D. I did the bounded case first and then died on the unbounded case for a longg time proving useless stuff like $f(n) > n$ before I realised simple stuff does work :P

Note that every value of $f$ has some prime dividing it since $f(n) > 1$. Let $p$ be an arbitrary prime dividing some value of $f$. Let $S_p$ denote the set of numbers $n$ such that $p | f(n)$. Note that $a,b \in S_p \implies |a-b| \in S_p$ as well, by the problem statement. Let $a_1 < a_2 < a_3, \cdots$ be the elements of $S_p$. Now, we must have $a_{k+1} - a_k = a_1$. This is because if $a_{k+1} - a_k = z < a_1$, then $z < a_1 \in S_p$, not possible. If $a_{k+1} - a_k = z > a_1$, then $a_{k+1} - a_1 \in S_p$ and is between $a_{k}$ and $a_{k+1}$, not possible. So, we have that $S_p$ is just multiples of $a_1$.

Let $q$ be a prime dividing $f(1)$. If the problem is false, then $S_q$ is finite, say with maximal element $M$. Pick a prime $r > M$. Note that $f(r)$ cannot contain any prime divisor that divides $f(n)$ for $n < r$ since then we would need $n | r$, not possible since $n \neq 1$ and $r$ is prime. Since this is true for infinitely many $r$, we have that there are infinitely many primes dividing some value of $f$. In particular, $f$ is unbounded.

I will in fact show that if $f$ is unbounded, then it is linear. Call a number $n$ a peak if $f(n) > f(k)$ for all $1 \le k \le n-1$. So $f(n) | f(k) + f(n-k)$ but since $f(n)$ is greater than both of them, we must have $f(n) = f(k) + f(n-k)$ for all $k < n$. Since $f$ is unbounded, there are infinitely many peaks. Let $m$ be one such peak.

Observe that $f(m) - f(k) = f(m-k) | f(m-k-1) + f(1) = f(m) - f(k+1) + f(1) \implies f(m) - f(k) | f(k) + f(1) - f(k+1)$. Since we can make $m$ sufficiently large, we must have that the RHS is $0$, so $f(k+1) = f(k) + f(1)$ which gives $f(n) = nf(1)$ and so $c = f(1)$ itself divides all values of $f$. Therefore, we are done. $\blacksquare$
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rama1728
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#18
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IndoMathXdZ wrote:
Let $f : \{ 1, 2, 3, \dots \} \to \{ 2, 3, \dots \}$ be a function such that $f(m + n) | f(m) + f(n) $ for all pairs $m,n$ of positive integers. Prove that there exists a positive integer $c > 1$ which divides all values of $f$.

Solved with MathLuis :showoff:

Let \(P(m,n)\) denote the assertion of the given functional equation. We split the problem into two cases.

Case 1. \(\text{Im}(f)\) is bounded.
From the divisibility condition \[f(n+1)\mid f(n)+f(1)\]we have that \[\gcd(f(n+1),f(1))\mid\gcd(f(n)+f(1),f(1))=\gcd(f(n),f(1))\]Also, by Pigeon Hole, there exists a prime \(p\) such that \(p\mid f(n)\) for infinitely many \(n\), let the set of such \(n\) be denoted as \(\mathbb{S}_p\) and let \(k\) be their greatest common divisor. By the Euclidean algorithm, we can show that \(\mathbb{S}_p\) contains all the multiples of \(k\). Now, we choose a subset \(\mathcal{S}\) of \(\mathbb{S}_q\) (for some prime \(q\)) so that \(\mathcal{S}\) contains the least element of \(\mathbb{S}_q\) and all the images of the elements are equal. Let the least element be \(n_0\), then note that \(\gcd(f(n_0),f(1))=\gcd(f(m),f(1))\) for all \(m\geq n_0\), because of the divisibility condition we got earlier. Now, if we choose the prime \(q\) so that it divides \(f(1)\), then we get that \(q\) divides \(f(i)\) and \(f(i+1)\) for some \(i\). And since the only \(k\) that divides \(i\) and \(i+1\) is \(1\), \(q\) divides \(f(i)\) for all natural numbers \(i\) as desired.

Case 2. \(\text{Im}(f)\) is unbounded.
Call a positive integer \(n\) gawaskar, if \(f(n)>f(i)\) for all \(i<n\). Note that there are infinitely many gawaskar numbers and \(f(n)=f(i)+f(n-i)\) for all \(i<n\). Now, let \(x+y\) be a gawaskar number such that \(f(x+y+1)\) is huge. Then, by the problem's condition, \[f(x+y+1)\mid f(x+y)+f(1)=f(x)+f(y)+f(1)\]and \[f(x+y+1)\mid f(x)+f(y+1)\]implying that \[f(x+y+1)\mid f(y+1)-f(y)-f(1)\]Implying that \(f(y+1)=f(y)+1\). Since there are infinitely many gawaskar numbers, we have that \(f(x+1)-f(x)=f(1)\) for all \(x\in\mathbb{N}\). This implies that \(f(x)=xf(1)\), and since \(1\neq f(1)\mid f(x)=xf(1)\), we are done.

Remarks
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AwesomeYRY
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#19
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Solved with help from Max Lu

If $p$ divides $f(1)$, then first there's $n_p$ values of f divisible by p, then everything afterwards isn't divisible by p. Thus, there exists some $N$ s.t. for all $n\geq N$ you have $\gcd(f(1),f(n))=1$.

If $p\nmid f(1)$, then let the first time that $q$ appears in $f(i)$ be $s_q$. Then, $p\mid f(n) \Longrightarrow s_q\mid n$. Combining these two results, for $p\geq N$, $f(p)$ is relatively prime to all previous numbers.

Thus, if $f$ is bounded, you find a contradiction because you can't keep generating relatively prime numbers (finite number of prime divisors).

Otherwise, if $f$ is bounded then we claim that $f$ is linear (and therefore of the form $f(n) = n\cdot f(1)$. There must be arbitrarily long chains of $f(n+1) = f(n)+f(1)$, otherwise if all chains were of length $\leq C$, then the value would go up by $Cf(1)$ and then get halved, which cannot become unbounded. Thus, there exists a chain with arbitrarily large $k$ of the form $[b-k,b]$. Next, $f(b)\mid f(b-k) + f(k) = f(b) - kf(1) + f(k) \leq f(b)$ so $f(k) = kf(1)$ and we're done. $\blacksquare$.
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Inconsistent
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#20 • 1 Y
Y by Nuterrow
Assume otherwise.

Notice that $f(m+n) \leq f(m)+f(n)$ so we can induct to $f(a_1+a_2+\ldots+a_n) \leq f(a_1)+f(a_2)+\ldots+f(a_n)$.

We say a prime $p$ is alive if for any positive integer $N$, there exists $x \geq N$ such that $p \mid f(x)$. We say a prime is dead otherwise.

Notice that for every $p \mid f(1)$ we have $p \nmid f(k)$ for some $k$, so $p$ is dead by induction because $p \nmid f(i) \Longrightarrow p \nmid f(i+1)$ since $f(i+1) \mid f(1)+f(i)$.

We say a dead prime $p$ dies on a day $N$ if for any $x \geq N$ we have $p \nmid f(x)$ and such $N$ is minimal. By definition, every prime has a death day.

Note that $f(2n) \mid 2f(n)$ so for an odd prime $p$ if $p \nmid f(n)$ then $p \nmid f(2n)$. Therefore no new primes appear in the sequence $f(1), f(2), f(4), \ldots, f(2^i), \ldots$ and eventually after all primes that divide $f(1)$ die, $f(2^n)$ is a power of $2$ for all $n \geq C$ for some $C$, and none of them are equal to $1$. This implies $2$ is alive, so $2 \nmid f(1)$. Therefore since $v_2(f(2^{i+1}) \leq 1+v_2(f(2^{i}))$ we know $v_2(f(2^i)) \leq i$. A corollary is that $f(2^i) \leq 2^i$ for all $i \geq C$.

Assume at some integer $T$ we have $v_2(f(2^T)) < T$ then by induction $v_2(f(2^i))<i$ for all integers $i \geq T$. So for all $i \geq \max(T, C)$ we have $f(2^i) \leq \frac{2^i}{2}$

Let $S = \sum_{i = 0}^{\max(T, C)} \left(f(2^i)-2^i\right)$, which is a finite positive integer.

Then for all $i \geq \max(T, C)$ we have $f(i) \leq \sum_{2^s \text{ in binary representation of i}} f(2^s) \leq \frac{i}{2}+S$

Let $R$ be the day when all primes that divide $f(1)$ die.

Lemma: If $p \mid f(q)$ where $p, q$ are primes and $q \geq N$ for some constant $N$ then $p \nmid f(i)$ for all $i < q$.

Proof: Choose $N = R$. Then assume $p \mid f(i)$ for some $i < q$, so pick the minimal such $i$ which must be greater than $1$. Clearly $i \nmid q$ so let $s$ be $q$ reduced modulo $i$. Then $p \nmid f(s)$, and by induction on $p \nmid f(k) \Longrightarrow p \nmid f(k+i)$ using $f(k+i) \mid f(i)+f(k)$, we prove $p \nmid f(q)$ which is a contradiction, so we are done.

Therefore the set of primes between $R$ and $x$ for arbitrary $x$ has an injective mapping to the set of primes at most $\frac{x}{2}+S$. So $\pi\left(x\right) \leq \pi\left(\frac{x}{2}\right)+(R+S)$ for constant $R+S$. This is clearly false by the Prime Number Theorem, giving contradiction.

Therefore $T$ doesn't exist, so $v_2(f(2^i)) = i$ for all $i$, implying $f(2^i) = 2^i$ for all $i \geq C$.

Now let $M = \sum_{i = 0}^C \left(f(2^i)-2^i\right)$. Clearly by summing over the binary representation again, $f(i) \leq i+m$ for all $i$. By taking the smallest power of 2 greater than $i$, $2^k$, we know that $2^k \leq f(i)+f(2^k-i)$ so $f(i) \geq i-M$ since $f(2^k-i) \leq 2^k-i+M$.

For $i, j > 2M$ we must have $f(i+j) \mid f(i)+f(j)$ but if $f(i+j) \neq f(i)+f(j)$ then $i+j-M \leq f(i+j) \leq \frac{f(i)+f(j)}{2} \leq \frac{i+j}{2}+M$, which is a contradiction. So $f(i+j) = f(i)+f(j)$. So letting $j = ki$ and inducting on positive integers $k$ gives us that $f(ki) = kf(i)$ for $i > 2M$. However this means if $p$ is any prime that divides $f(1)$, then $p \mid f(pi) = pf(i)$ for arbitrarily large $i$. This means $p$ is alive, so by contradiction we are done.
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bora_olmez
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#21
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I didn't like this as much as most others.
Solution
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NoctNight
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#22 • 2 Y
Y by PRMOisTheHardestExam, Mango247
Solution
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IAmTheHazard
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#23
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Really long solution, but this problem feels like you just keep throwing stuff at it and keep track of what you've proved and you eventually grind it down.

As with everyone else, split into cases where $f$ is bounded and unbounded.

In general, note that if $n \mid f(a),f(b)$ for $a>b$, then $n \mid f(a) \mid f(b)+f(a-b) \implies n \mid f(a-b)$. Call an integer $i$ with $n \mid f(i)$ $n$-cool. I claim that if there exist $n$-cool integers, then they are all consecutive multiples of some positive integer $k$, and $k$ is one of them (i.e. they are $\{k, 2k, \ldots\}$ or $\{k, 2k, \ldots, ak\}$ for some $a \geq 1$). I claim that setting $k$ equal to the least $n$-cool integer works, which is by induction. Suppose that the $c$ smallest $n$-cool integers are $\{k,\ldots,ck\}$, and the $c+1$-th smallest $n$-cool integer exists. Then if $x$ is this $c+1$-th smallest $n$-cool integer, $x-k$ must be $n$-cool as well, and also must be in $S$ as it's smaller than $x$, hence $x-k=ck \implies x=(c+1)k$, completing the induction.

Suppose $f$ is bounded, so there exists some maximal $M$ such that there are infinitely many $n$ with $f(n)=M$, and call such $n$ fat. The key idea is that for any fat $n$, we have $f(a)+f(b) \in \{M,2M\}$ for all $a+b=n$ with $a,b$ sufficiently large (so $f(a),f(b) \leq M$), where we only achieve $2M$ if $a$ and $b$ are fat as well.
First, we use the above idea but loosened a bit: let $m$ be the least integer such that $M \mid f(m)$ (which exists, since it's at most the smallest fat number). Then if $n$ is fat and $n-m$ is large, $M \mid f(n-m) \implies f(n-m)=M$. On the other hand, if some $n>k>n-m$ exists such that $f(k)=M$, then $M=f(n) \mid f(k)+f(n-k) \implies M \mid f(n-k)$: impossible by definition, as $n-k<m$. This is enough to imply that the gap between large enough consecutive fat numbers is constant (and equal to $m$).
Now, pick positive integers $a,b$, such that $a,b,b-a$ are large enough, and $b$ is fat, so $b+m$ is fat as well. Then $f(b-a)=f(b+m-a)=M-f(a)$. By fixing $a$ and letting $b$ vary, this means that $f$ is eventually periodic (with minimal period $m$).
Pick $x$ large and let $1<k=f(mx+1)$, which is constant for large enough $x$. Then by our "general fact", $k$ divides $mx+1,mx+m+1$, which implies that it divides $\gcd(mx+1,mx+m+1)=\gcd(mx+1,m)=1$. Then because there are infinitely many $k$-cool integers, and $1$ is one of them, every positive integer is $k$-cool, completing the proof. $\blacksquare$

Now to tackle the case where $f$ is unbounded. Let $M$ be the minimum value attained by $f$, and call an $n$ such that $f(n)=M$ skinny. Further, let $n$ be powerful if $f(n)\geq f(n-1),f(n-2),\ldots,f(1)$, and overpowered if the inequality is strict, so there are infinitely many overpowered (and thus powerful) $n$. It's easy to see that for all overpowered $n$, we have $f(a)+f(b)=f(n)$ for all $a+b=n$, and if $n$ is powerful (but not overpowered) this equality holds so long as both $a$ and $b$ aren't powerful either, and if one is, then both are.
Pick some large overpowered $n$ with $f(n)=x$ and $x>9999M$. Then we have $f(m)=x-M \iff f(n-m)=M$ for $m<n$. On the other hand, by our general claim, the set of $a$ such that $x-M \mid f(a)$ is equal to consecutive multiples of some positive integer $k$, including $k$in particular, the least $m$ with $f(m)=x-M$ is $k$ (important!). On the other hand, if $a<n$, then $2(x-M)\geq x$, so any such $a$ must have $f(a)=x-M$. Hence for $m<n$, $f(m)=x-M \iff m=ck$ where $c$ ranges from $1$ to some arbitrary positive integer (possibly $1$ as well). This means that the skinny integers less than $n$ are evenly spaced out with a gap of $k$ as well. But since $k$ must equal the least $m$ satisfying $f(m)=x-M$ (which is overpowered), we can repeat this argument with a different (larger) choice of overpowered $n$, giving us a different value of $k$, unless $c \leq 1$ always (in which case the gap isn't defined). This means that there is exactly one skinny number $K$, which in turn readily implies that if $n>K$ is overpowered, then $n-K$ is overpowered as well, and they are consecutive (i.e. there is no overpowered number between them). Thus there exists some $1 \leq r \leq K$ such that the overpowered integers which are at least $r$ are precisely those of the form $Kx+r$ for $r \geq 0$. Note that if $f(r)=C$, then $f(Kx+r)=Mx+C$.
For any $y$, pick some $x$ such that $Kx+r>y$, so $f(y)+f(Kx+r-y)=f(Kx+r)=Mx+C$. We then also have
$$f(y+K)+f(Kx+r-y)=f(K(x+1)+r)=M(x+1)+C,$$so $f(y+K)=f(y)+M$ for all $y$.
We are now ready to finish. Pick some $x,y$ and consider the equation $x+(y+cK)=(x+y+cK)$ for all $c \geq 0$. Then,
$$f(x+y+cK) \mid f(x)+f(y+cK) \implies f(x+y)+cM \mid f(x)+f(y)+cM \implies f(x+y)+cM \mid f(x)+f(y)-f(x+y).$$Since $f(x+y)+cM$ is unbounded, $f(x)+f(y)=f(x+y)$, hence $f$ satisfies Cauchy and is linear. This clearly implies that $f(1)>1$ divides all values of $f$, so we're done. $\blacksquare$
This post has been edited 3 times. Last edited by IAmTheHazard, Jan 25, 2023, 10:31 PM
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blackbluecar
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#24
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Assume $f$ is unbounded. We have a sequence $m_1<m_2< \cdots$ where $f(m_i)>f(n)$ for all positive integers $n<m_i$. First, by size stuff, we see that $f(m_i) = f(m_i-k)+f(k)$ for every positive integer $k<m_i$. Likewise $f(m_i) = f(m_i-k-1)+f(k+1)$. So, $f(m_i)-f(k)$ divides $f(m_i)-f(k+1)+f(1)$. Thus, $f(m_i)-f(k)$ divides $f(k)-f(k+1)+f(1)$. But $f(m_i)-f(k)$ gets arbitrarily large. So, $f(k)-f(k+1)+f(1)=0$. Thus, $f(n)=n \cdot f(1)$

Assume $f$ is bounded. First, we claim that there exist positive integers $a$, $b$, and $p$ where $\gcd(a,b)=1$ and $p$ divides $f(a)$ and $f(b)$. This follows from the fact that $\text{Im}(f)$ is bounded. If there are $k$ primes dividing elements in $\text{Im}(f)$ then just consider a set of $k+1$ pairwise relatively prime positive integers, by PHP we get what we want. Now, let $A=\{a_1<a_2< \cdots \}$ be the set of positive integers where $f(a_i)$ is divisible by $p$. We claim $a_k = c \cdot k$ for some positive integer $c$. Indeed, if $m>n$ are positive integers then $f(a_m)$ divides $f(a_m-a_n)+f(a_n)$. Since $p$ divides $f(a_m)$ and $f(a_n)$, it must follow that $p$ divides $f(a_m-a_n)$. So, $a_m-a_n \in A$. Now, we will proceed by induction. Suppose $a_i = c \cdot i$ for every positive integer $i<k$, we will show $a_k = c \cdot k$. Assume $a_k < c \cdot k$. Then, $a_{k}-a_{k-1}<k = a_1$. But this contradicts minimality since $a_k-a_{k-1} \in A$. Now, assume $a_k > c \cdot k$. Then $a_k>a_k-a_1>a_{k-1}$. But because $a_k-a_1\in A$, this implies there is an element in $A$ between two consecutive elements. Contradiction.

So, we have positive integers $\gcd(a,b)=1$ where $p$ divides $f(a)$ and $f(b)$. But, this implies that $a = k \cdot A$ and $b = k \cdot B$ for some $A$ and $B$. So, $k=1$. Implying that $p$ divides $f(n)$ for all positive integers $n$.
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Anzoteh
126 posts
#25
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Lol I ended up turning this into an analysis problem smh (also thanks to some theorem a few have pointed out).

Sketch
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VicKmath7
1390 posts
#26 • 1 Y
Y by Maths_Girl
Solution
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Leo.Euler
577 posts
#27
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Assume for contradiction that there exists large $N$ such that for $n \ge N$, $\gcd(f(1), f(n))=1$.

Claim: $f$ is unbounded.
Proof. First, we note that $\gcd(f(1), f(n+1)) \mid \gcd(f(1), f(n))$ for all $n$, and inductively we have \[ \gcd(f(1), f(a)) \mid \gcd(f(1), f(b)) \]whenever $a > b$. Furthermore, for such $a, b$, it follows by the Euclidean algorithm that \[\gcd(f(a), f(b)) \mid \gcd(f(a-b), f(b)), \]and we can accordingly condense this to \[ \gcd(f(a), f(b)) \mid \gcd(f(1), f(n)) \mid f(1) \]for some $n$. By our assumption, when both $a$ and $b$ are larger than $N$, $\gcd(f(a), f(b)) \mid 1$, which forces $\gcd(f(a), f(b))=1$, implying that $f$ is unbounded. :yoda:

We assume the claim henceforth. Say that a positive integer $n$ is Skywalker if and only if $f(n) \ge f(k)$ for each $k < n$; clearly there is an infinitude of such numbers. By definition $f(n)=f(k)+f(n-k)$ for all $k<n$ whenever $n$ is Skywalker. Let $n$ be an arbitrarily large Skywalker, and $m<n$ be arbitrary. Fixing $m$ and variating $n$, the original functional equation implies \[ f(n-m+1) \mid f(n-m)+f(1), \]which we easily reduce to \[ f(n) - f(m) \mid f(m)-f(m-1)-f(1), \]so by size reasons we have $f(m)-f(m-1)-f(1)=0$, which rewrites as $f(m)=f(m-1)+f(1)$. Inductively we have $f(i)=if(1)$ for all $i \ge 1$, but this is a contradiction to our initial assumption that $\gcd(f(1), f(n))=1$ for sufficiently large $n$. We conclude. :starwars:

Remark: This proof consists of three main components, namely
  • showing that $f$ is unbounded by following the Euclidean algorithm,
  • defining Skywalker numbers using the unboundedness of $f$ and observing their infinitude, and
  • solving the functional equation by fixing some $m<n$ (where $n$ is a large Skywalker) and variating $n$.
In fact, the final two steps are motivated by the solution to 2019 N4, in which the functional divisibility is taken to advantage by fixing the dividend and varying the divisor enough to conclude that the dividend is 0.
This post has been edited 1 time. Last edited by Leo.Euler, Aug 14, 2023, 7:42 PM
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huashiliao2020
1292 posts
#28
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I was suggested to solve this by Leo.Euler;
here's a long and inefficient solution, where I just threw things at the wall until it stuck:

Suppose FTSOC that there exists a number $L$ (we take the smallest) that does not share the common gcd of the numbers smaller than it; let the earlier common gcd be g. Let $d_n=\gcd(f(n),f(1))$. Note that $$d_{n+1}=\gcd(f(n+1),f(1))\mid\gcd(f(n)+f(1),f(1))=d_n\stackrel{\text{induction}}{\rightarrow}d_i\mid d_j\forall i>j\implies\forall l\ge L,d_l\mid d_L,$$and by assumption $\gcd(d_L,g)=1\implies\gcd(d_l,g)=1$. We also extract $$\forall a\equiv1\pmod b,a,b\ge L,\gcd(f(a),f(b))\mid\gcd(f(a-b)+f(b),f(b))=\gcd(f(a-b),f(b))=...=\gcd(f(1),f(k))\mid f(1)$$by iterating. By assumption, this is one (otherwise $1<\gcd(f(a),f(b),f(1))\mid\gcd(f(a),f(1))$); in particular, f is unbounded.

Since f is unbounded, there are an infinite number of times when a new largest value appears in the sequence (namely $f(n)>f(k)\forall k<n$); namely $f(n)>\max(f(k),f(n-k))>\frac{f(k)+f(n-k)}2\implies f(n)=f(k)+f(n-k)$ for infinite n. Fix $m$ and vary n with this property; we have $$f(n)-f(m)=f(n-m)\mid f(n-m-1)+f(1)=f(n)-f(m+1)+f(1)\leftrightarrow f(n)-f(m)\mid f(m)-f(m+1)+f(1),$$which for sufficiently large f(n) s.t. LHS>RHS, we must have RHS=0; in particular, $f(m+1)=f(m)+f(1)\implies f(m)=mf(1)$, contradiction since f(1)>1 and f(1) is a common gcd, so our inital FTSOC was wrong.
This post has been edited 8 times. Last edited by huashiliao2020, Sep 5, 2023, 3:53 AM
Reason: better latex
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math_comb01
662 posts
#29
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Fun problem!
Sketch:
Case 1: IM(f) is unbounded
The idea is to look at a "peak" and do some sort of downward induction to get $f(x)=xf(1)$.
Case 2:Im(f) is bounded
Here, the idea is to take the set of primes dividing $f$ infinitely many times, then prove one of them divides all by contradiction argument.
This post has been edited 3 times. Last edited by math_comb01, Mar 30, 2024, 11:17 AM
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Iveela
116 posts
#30
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We begin by solving the case where $d = f(1)$ is the minimum value attained by $f$. In particular, we prove that $d$ divides all values of $f$. Assume the contrary. Then, obviously $f$ is nonconstant. Notice that $d \mid f(x)$ implies $d \mid f(x-1)$ since $d \mid f(x) \mid f(x-1)+f(1) = f(x-1) + d$. Consequently $d$ doesn't divide $f(k)$ for all sufficiently large values of $k$.
Now suppose $x \in \mathbb{N}$ satisfies $f(x) \leq f(x+1)$. Then, the condition implies that $f(x+1) \mid f(x) + f(1) \leq 2f(x+1)$. Since equality holds only if $d=f(x)=f(x+1)$, this tells us that either $f(x+1)=f(x) + d$ or $f(x+1)=d$. Consider the smallest positive integer $k$ such that $f(k) \neq kd$. Suppose that a positive integer $x$ satisfies $d \neq f(x) \leq f(x+1) \leq \dots \leq f(x+k)$. Then, the condition implies that $f(x) + kd = f(x+k) \mid f(x) + f(k) \implies f(x + k) \mid f(k) - kd$ and hence $f(x+k) \leq |f(k)-kd|$ which is a constant number.
On the other hand, suppose that $x \in \mathbb{N}$ satisfies $f(x) > f(x+1)$. Then, we have $f(x+1) \mid f(x) + d \implies f(x+1) \leq \frac{f(x)+d}{2}$. It is clear that these observations are enough to conclude that $f$ is bounded. (Actually, this implies that if $f$ is unbounded we must have $f(x)=dx$.)
Let $\alpha$ be the largest value that appears infinitely many times in $f$. Let $\alpha_1, \alpha_2, \dots$ be the numbers mapped to $\alpha$ by $f$. Consider two sufficiently large positive integers $i > j$ such that $i-j$ is also very large. Then, we have $f(\alpha_j) \geq f(\alpha_j-1)$ and hence $f(\alpha_j-1) = \alpha - d$. The condition implies that $f(\alpha_i) \mid f(\alpha_j-1) + f(\alpha_i-\alpha_j+1)$. Since $i-j$ is very large, we have $f(\alpha_i-\alpha_j+1) \leq \alpha \implies f(\alpha_j - 1) + f(\alpha_i - \alpha_j +1 ) < 2 \alpha$. Hence, $f(\alpha_j-1) + f(\alpha_i-\alpha_j+1) = \alpha \implies f(\alpha_i-\alpha_j+1) = d$, a contradiction. (We actually dont use the fact that $f(1)$ is the minimum here.) $\square$

Now we solve the general case. Let $c$ be the smallest positive integer such that $f(c) \leq f(k)$ for all $k \in \mathbb{N}$. Consider the function $g : \mathbb{N} \to \mathbb{N}$ such that $g(x) = f(cx)$. Observe that $g(m+n) \mid g(m) + g(n)$ for all $m, n \in \mathbb{N}$ and $g(1) \leq g(k)$ for all $k \in \mathbb{N}$. It follows that $g(1) \mid g(k) \Leftrightarrow f(c) \mid f(ck)$ for all positive integers $k$.
Case 1: The sequence $f(c), f(2c), \dots$ is bounded.
It suffices to show that $f$ is bounded. Assume the contrary. Then there exists a positive integer $s$ such that the sequence $f(s), f(c+s), f(2c+s), \dots$ is unbounded. Let $f(ck+s)$ be a sufficiently large term of this sequence. Our condition implies that $f(ck+s) \mid f(ck) + f(s)$ which is bounded by a constant number, a contradiction. $\square$.
Case 2: The sequence $f(c), f(2c), \dots$ is unbounded.
Note that $f(ck)=kf(c)$. We claim that $f(x+c) = f(x) + f(c)$. Using the same logic as Case 1, we see that the sequences $f(s), f(c+s), f(2c+s), \dots$ are unbounded for all positive integers $s$. Consider positive integers $k$ such that $f(ck+s) > f(ck_0+s)$ for all $k > k_0$. Suppose that $f(ck) > f(ck+s)$. Consider the largest positive integer $\alpha \leq k$ such that $f(c\alpha) \leq f(ck+s)$. It is clear that $f(ck+s) - f(c\alpha) < f(c)$ The condition implies that $f(ck+s) \mid f(c\alpha) + f(c(k-\alpha)+s)<2f(ck+s)$. Hence, $f(c(k- \alpha)+s) < f(c)$ a contradiction. Then, $f(ck+s) \mid f(c\alpha) + f(c(k-\alpha)+s) < 2f(ck+s)$ implies that $f(c(k-\alpha)+s) + \alpha f(c) = f(ck+s)$.
Now we prove that $f(x+y)=f(x)+f(y)$. Let $x=\alpha c+ x_0$ and $y=\beta c + y_0$. Then, $f(x+y)=(\alpha+\beta)c+f(x_0+y_0) \mid f(x) + f(y) = c(\alpha+\beta) + f(x_0) + f(y_0)$ and hence $c(\alpha+\beta)c + f(x_0 + y_0) \mid f(x_0) + f(y_0) - f(x_0+y_0)$ for all positive integers $\alpha$ and $\beta$. This implies $f(x_0) + f(y_0) = f(x_0+y_0) \implies f(x) + f(y) = f(x+y)$. Hence, $f(x) \equiv kx, \ k \in \mathbb{N}$. $\blacksquare$
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idkk
118 posts
#31
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Claim: if $c|f(a),c|f(b)$ then $c|f(gcd(a,b))$

Proof: Its easy to see $c|f(a),c|f(b) \implies c|f(|b-a|)$ use euclidian division algorithim to prove the claim .


Case 1: $f$ is bounded above or in other words $f$ attains finite values.


Let $f(M)=t$ be the largest value in its range which repeats infinitely often. say $a$ is the smallest natural number such that $f(a)=t$ .
we know then that $f(x)=t \implies a|x$


Claim: $f(x)=t \iff a|x$ for large $x$

assuming $n$ is large we have $f(n)|f(a)+f(n-a)$ if $f(n)=t$ we know $f(n-a)=t$ using this the claim is easy to see .


Claim: For large $x$ , $f(x)=t$ or $\frac{t}{2}$

Consider the sequence $t=f(ax_1)=f(a(x_1+1))=......$ where $x_1$ is large covering all large solutions for $f(x)=t$


Now for any large $m$ not divisible by $a$ pick some large $n$ such that $m+n=ax_1$


Now its easy to see $I=f(m)=f(m+a)=f(m+2a)=........$ we know then $|$ divides $t$.

Pick any other large $m_1$ such that $a|m+m_1$ and choose some large $n_1$ such that $a|m_1+n_1$ to similarly get $a|f(m_1)$ and repeat similarly
from here observe we are writing $t$ as a sum of its divisors only way to do is $\frac{t}{2}+\frac{t}{2}$


Now once we have the main problem proved for large $x$ , smaller $x$'s automatically follow pretty easily .


Case 2: $f$ is unbounded .






Its easy to observe

$f(2^d)|2f(2^{d-1})|4f(2^{d-2})......|2^df(1)$

now i define the function $f_2(x)=$ largest odd factor of any natural no $x$ , observe $f_2(2^{d+1})|f_2(2^{d})$

Case 1: $f_2(f(2^d)) > 1$ always .


Prove that the common odd factor divides any number of form $f(K 2^t)$ where $K$ is a odd no. By induction.

Case 2: The sequence $f(1),f(2),f(4),.....$ eventually turns into powers of 2 .


Say $f(2^d)$ is the smallest power of $2$ in the sequence .

Claim: All multiples of $f(2^d)$ come in the sequence $f(2^d),f(2^{d+1}),f(2^d*3),f^(2^{d+2}),.....$

appolozie my laziness prove alll terms basically are multiples of $f(2^d)$ and do $f^(2^d+2^dt)|f(2^d)+f(2^dt)$ , its easy prove that the sequence wont be bounded . which will give u this claim .


Now observe $f^(2^dx+r)|f(2^dx)+f(1)$ where $f(1)$ is odd

Now as $f(1)$ has to be odd[can be proved with more that if it wasent the case all terms would be even]. , we know there exist ifninitely many $x$ such that $f(2^dx)+f(1)$ is a power of $f(1)$ , by which the result isnt hard to see[ too lazy to add the details :P]
This post has been edited 5 times. Last edited by idkk, May 28, 2024, 9:20 AM
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Akacool
14 posts
#32 • 1 Y
Y by Titibuuu
Lets prove that $f$ is bounded. If we assume that $f$ is unbounded we can say that there are infinite amount of $N$ that $f(N)$ is the prefix max. So it becomes that for any $x < N$
$f(x) + f(N - x) = f(N)$. Now if we take large N and some x to be $f(N - x) | f(N - x - 1) + f(1)$ and it becomes that $f(N) - f(x) | f(N) - f(x + 1) + f(1)$ which in turn means that
$f(x + 1) = f(x) + f(1)$ so $f(x) = xf(1)$.
Now that $f$ is bounded we can assume that there is a set of values of $f$ lets call $A$ that they appear infinitely many times in $f$. It is clear that the $\gcd(f(n), f(1))$ divides all $f$ less than $n$.
If we use euclidean alg on $f$ we can easily see that the $\gcd(f(x), f(y) | f(\gcd(x, y))$. So it means that every occurrence of a element of $S$ is divisible by some "primitive occurrence".
And if that occurrence is $1$ we will get that that element of $S$ is divisible by $f(1)$ in turn all of $f$ will be divisible. So every "primitive" occurrence divides every occurrence of elements of $S$. Every $f(x)$ if $x$ is large will be an element of $S$ which means that every $x$ will be divisible by some "primitive" occurrence but we can take an $X$ that isn't divisible by any so thus a contradiction.
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HamstPan38825
8867 posts
#33
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woah this is like the most perfect number theory FE ever made

We split into two cases.

Case 1: $f$ is bounded, say by $N$. Then there exists a nonempty set of positive integers $S$ such that for any $k \in S$, $k \mid f(n)$ for infinitely many values of $n$. Otherwise, every $k$ only divides $a_k$ values of $n$, and we can only define $f$ for less than $a_1+a_2+\cdots+a_N$ values of $n$.

For each $n \in S$, let $m_n$ be minimal such that $n \mid f(m_n)$, and let $a_1$ be the minimum among the $m_n$. Furthermore, let $a_2, a_3, \dots$ be the values with this $n \mid f(a_i)$ in increasing order.

Claim: For any $q \in \mathbb N$, $n \mid f(qa_1)$.

Proof: Observe we have
\begin{align*}
f(a_i) \mid f(a_i - a_1) + f(a_1) &\implies n \mid f(a_i - a_1) \\
f(a_i - a_1) \mid f(a_i - 2a_1) + f(a_1) &\implies n \mid f(a_i- 2a_1) \\
&\vdots
\end{align*}so we inductively have $n \mid f(a_i \bmod a_1)$. If $a_1 \nmid a_i$, this is less than $a_1$, hence we must have $a_1 \mid a_i$.

By definition, the sequence $\{a_i\}$ has infinitely many terms, so for any $q$, we can find a $a_\ell > qa_1$, and the above process implies $n \mid f(qa_i)$, as needed. $\blacksquare$

Claim: $a_1 = 1$.

Proof: Otherwise, any divisor $d$ of $f(1)$ divides only finitely many $f(n)$. In particular, for sufficiently large primes $p$, $f(p) \mid f(1) p$ implies $f(p) = p$. But this contradicts $f$ bounded. $\blacksquare$

These two claims solve the bounded case.

Case 2: $f$ is unbounded. Then by definition, there exists a sequence $n_1, n_2, n_3, \dots$ such that $f(n_i) > f(k)$ for all $k < n_i$. For these $n_i$, we must have $f(n_i) = f(k) + f(n_i-k)$ as $2f(n_i) > f(k) + f(n_i-k)$ for all $k$. In particular, we have the following claim:

Claim: $f(k+1)-f(k) = c$ for some constant $c$.

Proof. We will show that $f(k) - f(k-1) = f(k+1) - f(k)$ for each $k$. Take some $n_i > k$; by the previous property, \[f(k) + f(n_i-k) = f(n_i) = f(k+1) + f(n_i-k-1) \implies f(k+1) - f(k) = f(n_i-k) - f(n_i-k-1).\]Let $a_{k-1} = f(k) - f(k-1)$ and $a_k = f(k+1) - f(k)$. We have
\[f(n_i-1) \mid \gcd(f(k-1) + f(n_1 - k), f(k) + f(n_1-k-1)) = \gcd(f(k-1)+f(n_1-k), a_{k-1}-a_k).\]In particular, $f(n_i - 1) \mid a_{k-1} - a_k$. But $\{f(n_i)\}$ is unbounded, hence $\{f(n_i - 1)\} = \{f(n_i) - f(1)\}$ is unbounded, so $a_{k-1} = a_k$, as needed. $\blacksquare$

Therefore $f$ is linear in $n$. Let $f(1) = b$ and $f(2) = a+b$; if $a$ and $b$ are relatively prime, then $a+b \mid 2b$, which is a clear contradiction as $\gcd(a+b, 2b) = \gcd(a+b, 2)$. Hence $a$ and $b$ share a common factor $d$, and this $d$ divides $f(n)$ for every $n$.
This post has been edited 1 time. Last edited by HamstPan38825, Aug 21, 2024, 9:17 PM
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N3bula
279 posts
#34
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Claim: If $f(n)\geq \alpha n$ for some fixed $\alpha$ we get the result.

Proof:
Let $\alpha$ be the largest value such that for all $n$ we have that $f(n)\geq \alpha n$. Clearly from the definition we can find values such that for any $\epsilon$ we get that $\frac{f(n)}{\alpha n} \leq 1+\epsilon$, now take a value of $r$ such that $\frac{f(r)}{\alpha r} < 2$. I claim that we have for that value of $r$ that $f(nr)=nf(r)$. Suppose that $f((n-1)r)=(n-1)f(r)$, we must have that $f(nr)\mid nf(r)$ if $f(nr)\neq nf(r)$ we have that $f(nr)\leq \frac{nf(r)}{2}$, thus $\frac{f(nr)}{\alpha nr} \leq \frac{f(r)}{2\alpha r} < 1$ however this is a contradiction. Thus for all $n$ we have that $f(nr)=nf(r)$. Now suppose that for some $p\mid f(1)$ that $p\nmid f(k)$. Clearly we have that $p \nmid f(k+1)$ and thus for all $m\geq k$ we get $p\nmid f(m)$. Thus we must have unless the result holds true that $f(r)$ is coprime to $f(1)$. Now let $p$ be a prime such that $p\mid f(1)$. Clearly there exists infinetly many $k$ such that $f(n) \mid p^k-1$. Thus we can choose abitrarily large $n$ such that $f(nr)=p^k-1$. Now consider $f(f(1)nr+f(1)) \mid f(1)f(nr)+f(1)$ as $f(nr)=p^k-1$, we get that $f(f(1)nr+f(1)) \mid f(1)p^k$ as we can make $n$ abitrarily large, as $f(n)\geq \alpha n$ we can get that $p \mid f(f(1)nr+f(1))$ for all sufficiently large $n$ this suffices as this means that $p\mid f(n)$ for all $n$.

Now suppose no such lower bound exists.
I will prove that such a $c$ still exists. Clearly we must have that if $gcd(a, b)=1$ that $gcd(f(a), f(b))=1$. Thus if we let the first $n$ primes be $p_1$, $p_2$, $\dots$, $p_n$. We know that for all $n$ we have $f(n)\leq f(1)n$. Now let $k$ be a value such that $\frac{f(k)}{k}< \frac{1}{10^{1000}}$. We get that above some point from that we must have for all $m>N$ that $\frac{f(m)}{m}<\frac{1}{10^{1000}}$. Let the largest prime divisor amongst $f(p_1)$, $f(p_2)$, $\dots$, $f(p_n)$, such that $p_n\leq N < p_{n+1}$ be $q$. Suppose $q$ is the $i$th prime, now let $u$ be a value such that $u \geq n, i$. Take the first $u$ primes. Clearly a prime divisor of size at least $p_u$ divides one of $f(p_{n+1}), f(p_{n+2}), \dots, f(p_u)$ which contradicts the size condition thus implying the result.
This post has been edited 2 times. Last edited by N3bula, Apr 10, 2025, 10:35 AM
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