Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Advertising paid program or service: never allowed

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

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

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

Original 2013 post
50 replies
v_Enhance
Feb 15, 2013
blawho12
Oct 16, 2017
pairs (m, n) such that a fractional expression is an integer
cielblue   0
8 minutes ago
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
0 replies
cielblue
8 minutes ago
0 replies
the same prime factors
andria   6
N 30 minutes ago by MathLuis
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
6 replies
andria
Sep 6, 2015
MathLuis
30 minutes ago
Inspired by RMO 2006
sqing   1
N an hour ago by SomeonecoolLovesMaths
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
1 reply
sqing
5 hours ago
SomeonecoolLovesMaths
an hour ago
Problem 4 of RMO 2006 (Regional Mathematical Olympiad-India)
makar   7
N an hour ago by SomeonecoolLovesMaths
Source: Combinatorics (Box Principle)
A $ 6\times 6$ square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always two congruent rectangles.
7 replies
makar
Sep 13, 2009
SomeonecoolLovesMaths
an hour ago
Perfect squares: 2011 USAJMO #1
v_Enhance   227
N an hour ago by ray66
Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.
227 replies
v_Enhance
Apr 28, 2011
ray66
an hour ago
Mustang Math Recruitment is Open!
MustangMathTournament   0
2 hours ago
The Interest Form for joining Mustang Math is open!

Hello all!

We're Mustang Math, and we are currently recruiting for the 2025-2026 year! If you are a high school or college student and are passionate about promoting an interest in competition math to younger students, you should strongly consider filling out the following form: https://link.mustangmath.com/join. Every member in MM truly has the potential to make a huge impact, no matter your experience!

About Mustang Math

Mustang Math is a nonprofit organization of high school and college volunteers that is dedicated to providing middle schoolers access to challenging, interesting, fun, and collaborative math competitions and resources. Having reached over 4000 U.S. competitors and 1150 international competitors in our first six years, we are excited to expand our team to offer our events to even more mathematically inclined students.

PROJECTS
We have worked on various math-related projects. Our annual team math competition, Mustang Math Tournament (MMT) recently ran. We hosted 8 in-person competitions based in Washington, NorCal, SoCal, Illinois, Georgia, Massachusetts, Nevada and New Jersey, as well as an online competition run nationally. In total, we had almost 900 competitors, and the students had glowing reviews of the event. MMT International will once again be running later in August, and with it, we anticipate our contest to reach over a thousand students.

In our classes, we teach students math in fun and engaging math lessons and help them discover the beauty of mathematics. Our aspiring tech team is working on a variety of unique projects like our website and custom test platform. We also have a newsletter, which, combined with our social media presence, helps to keep the mathematics community engaged with cool puzzles, tidbits, and information about the math world! Our design team ensures all our merch and material is aesthetically pleasing.

Some highlights of this past year include 1000+ students in our classes, AMC10 mock with 150+ participants, our monthly newsletter to a subscriber base of 6000+, creating 8 designs for 800 pieces of physical merchandise, as well as improving our custom website (mustangmath.com, 20k visits) and test-taking platform (comp.mt, 6500+ users).

Why Join Mustang Math?

As a non-profit organization on the rise, there are numerous opportunities for volunteers to share ideas and suggest projects that they are interested in. Through our organizational structure, members who are committed have the opportunity to become a part of the leadership team. Overall, working in the Mustang Math team is both a fun and fulfilling experience where volunteers are able to pursue their passion all while learning how to take initiative and work with peers. We welcome everyone interested in joining!

More Information

To learn more, visit https://link.mustangmath.com/RecruitmentInfo. If you have any questions or concerns, please email us at contact@mustangmath.com.

https://link.mustangmath.com/join
0 replies
MustangMathTournament
2 hours ago
0 replies
Zsigmondy's theorem
V0305   2
N 2 hours ago by Andyluo
Is Zsigmondy's theorem allowed on the IMO, and is it allowed on the AMC series of proof competitions (e.g. USAJMO, USA TSTST)?
2 replies
V0305
2 hours ago
Andyluo
2 hours ago
Moving P(o)in(t)s
bobthegod78   71
N 2 hours ago by ray66
Source: USAJMO 2021/4
Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area 2021?

(A lattice point is a point $(x, y)$ in the coordinate plane where $x$ and $y$ are both integers, not necessarily positive.)
71 replies
bobthegod78
Apr 15, 2021
ray66
2 hours ago
Jane street swag package? USA(J)MO
arfekete   45
N 4 hours ago by vsarg
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.
45 replies
arfekete
May 7, 2025
vsarg
4 hours ago
Proof Writing Help
gulab_jamun   3
N 6 hours ago by maromex
Ok so like, i'm working on proofs, and im prolly gonna use this page for any questions. My question as of now is what can I cite? Like for example, if for a question I use Evan Chen's fact 5, in my proof do I have to prove fact 5 all over again or can i say "this result follows from Evan Chen's fact 5"?
3 replies
gulab_jamun
Yesterday at 2:45 PM
maromex
6 hours ago
MAN IS KID
DrMath   136
N Today at 11:00 AM by lakshya2009
Source: USAMO 2017 P3, Evan Chen
Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$. Ray $AI$ meets $\overline{BC}$ at $D$ and meets $\Omega$ again at $M$; the circle with diameter $\overline{DM}$ cuts $\Omega$ again at $K$. Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\overline{IS}$. The circumcircles of $\triangle KID$ and $\triangle MAN$ intersect at points $L_1$ and $L_2$. Prove that $\Omega$ passes through the midpoint of either $\overline{IL_1}$ or $\overline{IL_2}$.

Proposed by Evan Chen
136 replies
DrMath
Apr 19, 2017
lakshya2009
Today at 11:00 AM
have you done DCX-Russian?
GoodMorning   83
N Today at 7:30 AM by ray66
Source: 2023 USAJMO Problem 3
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Proposed by Holden Mui
83 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:30 AM
Titu Factoring Troll
GoodMorning   77
N Today at 7:19 AM by ray66
Source: 2023 USAJMO Problem 1
Find all triples of positive integers $(x,y,z)$ that satisfy the equation
$$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.$$
77 replies
GoodMorning
Mar 23, 2023
ray66
Today at 7:19 AM
Need help with combi problems
JARP091   4
N Today at 4:19 AM by JARP091
I want to create a problem set of some of the hardest combi problems that are yet to appear in any contest. Can anyone help me out? Also can anyone give me some tips to create combi problems.
4 replies
JARP091
Yesterday at 5:43 PM
JARP091
Today at 4:19 AM
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   82
N May 21, 2025 by iyappana
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
82 replies
EthanWYX2009
Jul 16, 2024
iyappana
May 21, 2025
gcd (a^n+b,b^n+a) is constant
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 IMO P2
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Assassino9931
1364 posts
#77 • 1 Y
Y by cubres
straight wrote:
Assassino9931 wrote:
Did anyone mention that this problem appeared in the IMSC China International camp 2024 and benefitted some students??? :)

(P.S. I am not blaming anyone that a setup has been made; in fact I can assure that no such thing has happened, I am just giving a fun fact here.)

@above I also realized this and I think it's very unfair indeed. It is completely the same idea with completely the same construction. If I recall proposer of that problem was Navid Safaei (forgive me if I'm wrong) . So far I haven't heard of anyone claiming to have known this problem before the IMO though

No, the origin of the problem I put is Ukraine 2019 8.8/9.7 by Arsenii Nikolaiev (who was actually Observer B at IMO 2024 if I am not mistaken, so in particular not part of the leaders/observer A problem voting). And, well, I did hear about at least one student benefitting from that, but prefer keep the identity of the country confidential.
This post has been edited 1 time. Last edited by Assassino9931, Mar 31, 2025, 10:53 AM
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Cali.Math
128 posts
#78 • 1 Y
Y by cubres
We uploaded our solution https://calimath.org/pdf/IMO2024-2.pdf on youtube https://youtu.be/daboPS8Dtyk.
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Assassino9931
1364 posts
#79 • 3 Y
Y by pavel kozlov, VicKmath7, cubres
Here is the (primitive root)-styled solution, in spirit of what VicKmath7 wrote above.

Let $p \geq 3$ be a prime. Consider firstly the following: when is there an integer $n$ such that $a^n + b \equiv 0 \pmod p$? If $g$ is a primitive root mod $p$ (where $p$ is a prime not dividing $a$ or $b$!), then writing $a \equiv g^A$, $b \equiv g^B$ transfers to the equivalent $g^{B}(g^{nA-B}+1) \equiv 0 \pmod p$, i.e. $nA - B \equiv \frac{p-1}{2} \pmod {p-1}$. Similarly, for $b^n+a$ to be divisible by $p$ we must have $nB - A \equiv \frac{p-1}{2} \pmod {p-1}$. Now if it is the case that $A+B\equiv \frac{p-1}{2} \pmod {p-1}$, then taking $n\equiv -1 \pmod {p-1}$ would work not only for $a^n+b$, but also for $b^n + a$.

But note that $A + B \equiv \frac{p-1}{2} \pmod {p-1}$ if and only if $g^{A+B} \equiv -1 \pmod p$, i.e. $p$ divides $ab+1$. Therefore if we initially take $p$ to be an odd prime divisor of $ab+1$ (note that such does not divide $a$ or $b$), then there are infinitely many $n$, for which the required greatest common divisor is divisible by $p$. However, it cannot be the case that $p$ divides $a^n+b$ and $b^n+a$ for all large $n$ -- otherwise, $p$ would divide $a^{n+1} + ab \equiv a^{n+1} - 1$, so $p$ would divide $a-1$ (due to $a^{n+1} \equiv a^{n+2} \equiv 1 \pmod p$ and $\gcd(a,ab+1) = 1$), similarly $p$ would divide $b-1$, but now $ab+1 \equiv 0 \pmod p$ implies that $p$ must also divide $a+1$ and hence $(a+1) - (a-1) = 2$, contradicting $p\geq 3$.

Therefore, if $a$ and $b$ are such that $ab+1$ has an odd prime divisor, then they cannot satisfy the problem conditions. Finally, suppose $ab+1$ is a power of $2$. Fortunately, we only need that $ab+1$ is divisible by $4$ (unless $a=b=1$, which satisfies the problem conditions). Indeed, we may assume $a\equiv 3 \pmod 4$ and $b \equiv 1 \pmod 4$ and now if $n$ is even, then $a^n + b$ is divisible by $2$, but not by $4$ (and $b^n + a$ is divisible by $4$), while if $n$ is odd, then both $a^n + b$ and $b^n + a$ are divisible by $4$, so $\gcd(a^n+b,b^n+a)$ is infinitely often divisible by $4$ and not divisible by $4$, contradiction.
This post has been edited 3 times. Last edited by Assassino9931, Sep 18, 2024, 8:22 AM
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L13832
268 posts
#80 • 2 Y
Y by radian_51, cubres
What an amazing problem, kudos to the problem proposer :)
$\textbf{Answer:}$ $a=b=1$
Let $p$ be a prime such that $p\mid g$, so $p\mid b^{n-1}(a^n+b)-b^n-a\implies p\mid ab-1$ if $p\nmid a$.
Finally checking if $ab+1\mid g$ and choosing $n\equiv -1\pmod{\phi(ab+1)}$(this is possible because $\gcd(ab+1,a)=\gcd(b,ab+1)=1$) we have
$$a^n+b\equiv a^{-1}+b\equiv \frac{ab+1}{a}\equiv 0\pmod{ab+1}$$$$b^n+1\equiv a+b^{-1}\equiv \frac{ab+1}{b}\equiv 0\pmod{ab+1}$$Motivation
The last part comes from the fact that $a,b$ can be inverted modulo ${ab+1}$. Similarly we obtain $b^n+a\equiv 0\pmod{ab+1}$ which gives us $ab+1\mid g\implies p\mid ab+1$
Now we take $n \equiv 0 \pmod{p-1}$, we get that $p \mid g  = \gcd(a^n + b, b^n + a)$, so by FLT, $p \mid a + 1, p \mid b + 1 \implies p \mid 2\implies p=2$.
If $ab+1=2$ then $\boxed{a=b=1}$
Otherwise $ab+1\equiv 0\pmod{4}\implies a,b\equiv \pm 1\pmod{4}$.
WLOG $a \equiv -1 \pmod{4}$ and $b \equiv 1 \pmod{4}$. If $n$ is odd then we obtain that $a^n + b$ and $b^n + a$ are both divisible by $4$, and therefore $4 \mid g$. But having $n$ even we get $a^n + b \equiv 2 \pmod{4}$ and $4 \nmid a^n + b$, a contradiction, so we are done! :yoda:
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Golden_Verse
5 posts
#81 • 1 Y
Y by cubres
Answer
Solution
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Vedoral
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#82 • 1 Y
Y by cubres
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AshAuktober
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#83 • 1 Y
Y by cubres
Note that $ab+1 \mid g \mid a-b$, so indeed we must have $a  =b $, from where the only working pair is $(1, 1).
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zaidova
88 posts
#84 • 1 Y
Y by cubres
For all n, which are positive integers ($a=b=1$) is the only solution.
This post has been edited 3 times. Last edited by zaidova, Jan 3, 2025, 7:43 PM
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cursed_tangent1434
640 posts
#85 • 1 Y
Y by ihategeo_1969
I just can't believe I didn't manage to solve this problem in contest. The idea of considering $ab+1$ is not that random either, especially since it is almost natural to consider $a+b$ at which point you decide that the considered expression had better be symmetric but also relatively prime to both $a$ and $b$, which leads to the considered form.

The entirety of the problem is the following key claim.

Claim : For any such pair $(a,b)$ we must have $ab+1$ a power of two.

Proof : Consider a prime $p\mid ab+1$. Then, since $\gcd(ab+1,a)=\gcd(ab+1,b)=1$ it follows that $p\nmid a,b$ and so letting $n=k(p-1)-1$ for sufficiently large positive integers $k$ we have,
\[a^n+b \equiv \frac{1}{a}+b \equiv \frac{ab+1}{a} \equiv 0 \pmod{p}\]and
\[a+b^n \equiv a+\frac{1}{b} \equiv \frac{ab+1}{b} \equiv 0 \pmod{p}\]which implies that $p \mid \gcd(a^n+b,b^n+a)$. But then, if the $\gcd$ is eventually constant, we have that $p \mid \gcd(a^n+b,b^n+a)$ for all sufficiently large positive integers $n$. But then, considering $n=k(p-1)$ for sufficiently large positive integers $k$ we have that,
\[0 \equiv a^n+b \equiv b+1 \pmod{p}\]And similarly, $a+1 \equiv 0 \pmod{p}$. But then, since $p\mid ab+1$ and $p \mid a+1$ it follows that $p \mid b-1$ which in conjunction with $p\mid b+2$ implies that $p=2$. This means that there cannot exist any odd prime $p$ dividing $ab+1$ proving the claim.

Now, if $ab+1>2$ and
\[ab+1=2^r\]for some $r \ge 2$, it follows that $ab \equiv 3 \pmod{4}$. Thus, $a\equiv 1 \pmod{4}$ and $b \equiv 3 \pmod{4}$ (or vice versa). But note that this means for even $n$,
\[a^n + b \equiv 1+3 \equiv 0 \pmod{4}\]but
\[a+b^n \equiv 1+1 \equiv 2 \pmod{4} \]Thus, $\nu_2(\gcd(a^n+b,b^n+a))=1$. But note that for odd $n$ we have,
\[a^n+b \equiv 1 + 3 \equiv 0 \pmod{4}\]and
\[a+b^n \equiv 1 + 3 \equiv 0 \pmod{4}\]which implies that $4\mid \gcd(a^n+b,b^n+a)$. But, if the $\gcd$ is eventually constant this is a clear contradiction, which implies that we must have $ab+1=2$ and thus, $a=b=1$ as desired.
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quantam13
113 posts
#86
Y by
Cute solution. The main idea is to consider the sequence modulo $ab+1$, which can be motivated by "$n=-1$".
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dstanz5
243 posts
#89
Y by
quantam13 wrote:
Cute solution. The main idea is to consider the sequence modulo $ab+1$, which can be motivated by "$n=-1$".

Sorry for necroposting, I am going through the recent problems and I wish to know more about why $ab+1$ is the key expression here. "$n = -1$" would give $gcd(\frac{1}{a} + b, \frac{1}{b} + a)$ but I don't know what this gives.
EDIT: Oh I see, if you multiply them by a and b respectively they both give $ab+1$. Thanks to khina and vEnhance
This post has been edited 1 time. Last edited by dstanz5, Mar 31, 2025, 7:51 AM
Reason: :(
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Ilikeminecraft
658 posts
#90
Y by
I claim that $(a, b) = (1, 1)$ is the only valid answer.

We take $((a^c + b, b^c + a), (a^{c + 1} + b, b^{c + 1} + a))$ first. Note that $((a, b), (c, d)) = ((a, c), (b, d)).$ Thus,
\begin{align*}
  ((a^c + b, b^c + a), (a^{c + 1} + b, b^{c + 1} + a)) & = ((a^c + b, a^{c + 1} + b), (b^c + a, b^{c + 1} + a)) \\
  (a^c + b, a^{c + 1} + b) & = (a^c + b, a^{c + 1} - a^c) \\
  & = (a^c + b, a - 1)(a^c + b, a^c) \text{ since } (a^c, a - 1) = 1\\
  & = (a^{c - 1} + b, a - 1)(b, a^c) \\
  & = (b + 1, a - 1)(b, a^c) \\
  ((a^c + b, b^c + a), (a^{c + 1} + b, b^{c + 1} + a)) & = ((b + 1, a - 1)(b, a^c), (b - 1, a + 1)(a, b^c)) \\
  & \mid ((b + 1, a-1),(b - 1, a + 1)(a, b^c))((b, a^c),(b - 1, a + 1)(a, b^c)) \\
  & \mid ((b + 1, a-1),(b - 1, a + 1))((b, a^c), (a, b^c)) \\
  & = 2(a, b)
\end{align*}Thus, we have that $(a^n + b, b^n + a) \mid 2(a, b)$ when it becomes constant. Now, let $q = ab + 1.$ By picking an $n \equiv-1\pmod {\phi(q)},$ we have that $q\mid a + \frac1b, b + \frac1a \implies q\mid(a^n + b, b^n + a) \implies q \mid 2(a, b).$ However, we have that $q = ab + 1\implies (q, a) = (q, b) = 1.$ Thus, either $q = 1,$ or 2. If $q = 1,$ then $a$ or $b$ aren't positive. Thus, $q = 2\implies \boxed{a = b = 1}.$
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santhoshn
5 posts
#91
Y by
1, 1 Answer
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GingerMan
5 posts
#92
Y by
Answer is $(a,b)=(1,1)$ only, which works.
Take any prime divisor $p$ of $ab+1$ and let $t=\nu_p(ab+1)$. Then
\begin{align*}
  a^n+b &\equiv a^n - \frac 1a \equiv \frac{a^{n+1}-1}{a} \pmod{p^t}\\
  b^n+a &\equiv b^n - \frac 1b \equiv \frac{b^{n+1}-1}{b} \pmod{p^t}
\end{align*}So $p^t\mid \gcd(a^n+b,b^n+a)$ if and only if $\operatorname{lcm}(x,y) \mid n+1$, where $x$, $y$ are the orders of $a$, $b$ mod $p^t$, respectively. If $p^t>2$ and $(a,b)\neq (1,1)$, then $\operatorname{lcm}(x,y)>1$, so the sequence is not eventually constant.
Thus $p^t=2$, implying $ab+1=2$, which still gives $(a,b)=(1,1)$.
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iyappana
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#93
Y by
Answer is 1,1
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