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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

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

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MathWOOT Level 1
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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
mathemetics
Pangbowen   0
2 minutes ago
Let a,b,c≥0 and a+b+c=7. Prove that : a/b+b/c+c/a+abc≥ab+bc+ca-2
0 replies
Pangbowen
2 minutes ago
0 replies
Inspired by Austria 2025
sqing   3
N 4 minutes ago by Pangbowen
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
3 replies
sqing
Today at 2:01 AM
Pangbowen
4 minutes ago
Property of a function
Ritangshu   1
N 18 minutes ago by Natrium
Let \( f(x, y) = xy \), where \( x \geq 0 \) and \( y \geq 0 \).
Prove that the function \( f \) satisfies the following property:

\[
f\left( \lambda x + (1 - \lambda)x',\; \lambda y + (1 - \lambda)y' \right) > \min\{f(x, y),\; f(x', y')\}
\]
for all \( (x, y) \ne (x', y') \) and for all \( \lambda \in (0, 1) \).

1 reply
Ritangshu
May 3, 2025
Natrium
18 minutes ago
max value
Bet667   2
N 39 minutes ago by Natrium
Let $a,b$ be a real numbers such that $a^2+ab+b^2\ge a^3+b^3.$Then find maximum value of $a+b$
2 replies
Bet667
an hour ago
Natrium
39 minutes ago
[MAIN ROUND STARTS MAY 17] OMMC Year 5
DottedCaculator   34
N Today at 1:09 AM by DottedCaculator
Hello to all creative problem solvers,

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

Check out the fifth annual iteration of the

Online Monmouth Math Competition!

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

Our website: https://www.ommcofficial.org/
Our Discord (6000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal.vercel.app/

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

How hard is it?

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

How are the problems?

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

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

Solo teams?

Test Policy

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

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

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

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]


34 replies
DottedCaculator
Apr 26, 2025
DottedCaculator
Today at 1:09 AM
ARMl Local 2025 Final Results
PaulDreyer   32
N Today at 12:54 AM by deduck
Results, problems, and solutions are here. Congratulations to SFBA ARML / AlphaStar: Foxes and Friends and Leading Aces Academy for placing 1st and 2nd overall and to Liam Reddy (Utah Rookies) for their perfect score on the individual round and for being the only student with a perfect score to answer the tiebreaker correctly.
32 replies
PaulDreyer
May 3, 2025
deduck
Today at 12:54 AM
MathILy 2025 Decisions Thread
mysterynotfound   33
N Yesterday at 10:37 PM by Vivaandax
Discuss your decisions here!
also share any relevant details about your decisions if you want
33 replies
mysterynotfound
Apr 21, 2025
Vivaandax
Yesterday at 10:37 PM
Conditions for Integer
worthawholebean   22
N Yesterday at 10:18 PM by daijobu
Source: AIME 2008II Problem 15
Find the largest integer $ n$ satisfying the following conditions:
(i) $ n^2$ can be expressed as the difference of two consecutive cubes;
(ii) $ 2n+79$ is a perfect square.
22 replies
worthawholebean
Apr 3, 2008
daijobu
Yesterday at 10:18 PM
A box contains 5 chips numbered 1, 2, 3, 4, 5
CobbleHead   23
N Yesterday at 8:46 PM by superhuman233
Source: 2018 AMC 10B #6
A box contains $5$ chips, numbered $1$, $2$, $3$, $4$, and $5$. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$. What is the probability that $3$ draws are required?

$\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}$
23 replies
CobbleHead
Feb 16, 2018
superhuman233
Yesterday at 8:46 PM
[$10K+ IN PRIZES] Poolesville Math Tournament (PVMT) 2025
qwerty123456asdfgzxcvb   11
N Yesterday at 8:31 PM by Ruegerbyrd
Hi everyone!

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

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

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

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

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

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

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

We look forward to your participation!

FAQ
11 replies
qwerty123456asdfgzxcvb
Apr 5, 2025
Ruegerbyrd
Yesterday at 8:31 PM
ranttttt
alcumusftwgrind   39
N Yesterday at 8:29 PM by Ruegerbyrd
rant
39 replies
alcumusftwgrind
Apr 30, 2025
Ruegerbyrd
Yesterday at 8:29 PM
Centroids form Equilateral Triangle
Generic_Username   22
N Yesterday at 7:13 PM by Tetra_scheme
Source: 2019 AMC 12B #25
Let $ABCD$ be a convex quadrilateral with $BC=2$ and $CD=6.$ Suppose that the centroids of $\triangle ABC,\triangle BCD,$ and $\triangle ACD$ form the vertices of an equilateral triangle. What is the maximum possible value of the area of $ABCD$?

$\textbf{(A) } 27 \qquad\textbf{(B) } 16\sqrt3 \qquad\textbf{(C) } 12+10\sqrt3 \qquad\textbf{(D) } 9+12\sqrt3 \qquad\textbf{(E) } 30$
22 replies
Generic_Username
Feb 14, 2019
Tetra_scheme
Yesterday at 7:13 PM
mathpath: how much do recommendations matter
mm999aops   22
N Yesterday at 5:38 PM by ZMB038
See question^

I'm hoping only the QT matters : P
22 replies
mm999aops
Feb 3, 2023
ZMB038
Yesterday at 5:38 PM
Is anyone else going to MathPath 2024?
megahertz13   74
N Yesterday at 5:35 PM by ZMB038
Is anyone else going to MathPath 2024? We should connect.
74 replies
megahertz13
Feb 19, 2024
ZMB038
Yesterday at 5:35 PM
IMO ShortList 1998, number theory problem 5
orl   64
N Apr 25, 2025 by Ilikeminecraft
Source: IMO ShortList 1998, number theory problem 5
Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.
64 replies
orl
Oct 22, 2004
Ilikeminecraft
Apr 25, 2025
IMO ShortList 1998, number theory problem 5
G H J
Source: IMO ShortList 1998, number theory problem 5
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AlanLG
241 posts
#56 • 1 Y
Y by cubres
nice :blush: $$\boxed{\text{The answer are all powers of 2.} }$$ Claim. $n$ must be a power of $2$
take a odd divisor $k>1$ of $n$, then $2^k-1\mid 2^n-1\mid m^2+9$ by Fermat Christmas Theorem all prime divisors of $m^2+9$ are $1\pmod 4$ or $3$, but $2^k-1\not\equiv 3\pmod 3$ as $k$ odd, nor $2^k-1\equiv 1 \pmod 4$, a contradiction.

Claim. all powers of $2$ work.
Let $n=2^k$, write $$2^{2^k}-1=(2+1)(2^2+1)(2^{2^2}+1)\cdots (2^{2^{k-1}}+1)$$Note that each term except for the first one, by Fermat Christmas Theorem have only $1\pmod 4$ prime divisors, so choose $g$ a primitive root $\pmod {p^\theta}$ then by Chinese Remainder Theorem exists $m$ such that $$m\equiv 3 g^{\frac{\phi({p^\theta})}{4}}\pmod {p^\theta} \hspace{0.4cm}\forall\hspace{0.2cm} p\mid 2^{2^j}+1 \hspace{0.5cm}\text{and} \hspace{0.5cm} m\equiv 0\pmod 3$$then $p^\theta\mid m^2+9$ , and $3\mid m^2+9$, as $\text{Fermat numbers}$ are relative primes we would have that $2^{2^n}-1\mid m^2+9$, as desired.
Z K Y
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joshualiu315
2534 posts
#57 • 1 Y
Y by cubres
The answer is $\boxed{\text{powers of 2}}$.

Let $d$ be the largest odd prime divisor of $n$. For the sake of contradiction, assume $d>1$. Fermat's Christmas Theorem states that the factors of $m^2+9$ are either $1 \pmod{4}$ or $3$. Since $2^d-1$ divides $2^n-1$, all the factors of $2^d-1$ are either $1 \pmod{4}$ or $3$. Then, as

\[2^d-1 \equiv 3 \pmod{4},\]
it must contain a factor that is $3 \pmod{4}$, contradicting our assumption that $d>1$.

To prove the powers of $2$ work, induction with difference of squares easily works.
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shendrew7
795 posts
#58 • 1 Y
Y by cubres
Our answer is $\boxed{n=2^b, b \ge 0}$. The most convenient method to show each works is by setting $m=3a$. Our condition is reduced to proving there exists such an $a$ with
\[2^{2^b}-1 = 3\left(2^2+1\right)\left(2^{2^2}+1\right) \ldots \left(2^{2^{b-1}}-1\right) \mid 3(a^2+1) \mid (3a)^2+9.\]
Since the Fermat primes are pairwise relatively prime, and we have the solution $a \equiv 2^{2^{t-1}} \pmod{2^{2^t}+1}$ for each $t$, there must exist such an $a$.

Otherwise, suppose $p>1$ is an odd prime divisor of $n$. Then neither 2 or 3 are factors of $2^p-1 \equiv 3 \pmod 4$, so Fermat's Christmas Theorem tells us we must have $p \equiv 1 \pmod 4$ for all prime divisors of $n$, contradiction. $\blacksquare$
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peppapig_
281 posts
#60 • 1 Y
Y by cubres
Since it's not specified that $m$ has to be an integer, the answer is obviously all positive integers $n$! (Just kidding! Real solution below)

I claim that the answer is all $n$ that can be expressed in the form of $2^k$ for some nonnegative integer $k$.

First, note that by Fermat's Christmas Theorem, if a prime $p$ divides $m^2+3^2$, then it is either $1$ mod $4$ or it divides $\gcd(m,3)$. Using this, I now claim that $n$ cannot have any prime factor $p$ that is larger than $2$. FTSOC, assume that $p\mid n$, where $p>2$. This then implies that
\[2^p-1\mid 2^n-1,\]and since $p>2$, we have that $2^p-1$ must be $3$ mod $4$. Additionally, since $p$ is a prime $>2$, this means that $p$ is odd, implying that
\[2^p-1 \equiv 2-1\equiv 1\mod 3,\]so if $2^p-1$ is $3$ mod $4$, some other prime $q\neq 3$ divides $2^p-1$, a contradiction to the Christmas Theorem statement. Therefore $n$ cannot have any prime factor $p>2$, meaning that $n$ must be in the form of $2^k$ in order for $2^n-1$ to have a multiple in the form of $m^2+9$.

I now claim that for all $n=2^k$, $2^n-1$ has a multiple of the form $m^2+9$. I will prove this using induction. Suppose that $2^{2^k}-1$ for $k\geq 1$ has a multiple of the form $m^2+9$. This implies that $-9$ is a quadratic residue mod $2^{2^k}-1$. Now, note that
\[2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1),\]and that $\gcd(2^{2^k}-1,2^{2^k}+1)=1$. Therefore, since we already know that $-9$ is a quadratic residue mod $2^{2^k}-1$, we just need to prove that $-9$ is also a quadratic residue mod $2^{2^k}-1$. Since the two modulos are relatively prime, by CRT, this will also prove that $-9$ is a quadratic residue mod $2^{2^{k+1}}-1$, meaning that a multiple of $2^{2^{k+1}}-1$ in the form of $m^2+9$.

Using this, notice that,
\[(2^{2^{k-1}})^2 \equiv -1 \mod (2^{2^{k}}+1) \iff (3*2^{2^{k-1}})^2 \equiv -9 \mod (2^{2^{k}}+1),\]which proves that $-9$ is indeed a quadratic residue mod $2^{2^{k}}+1$. Therefore, if there exists a multiple of $2^{2^{k}}+1$ in the form of $m_1^2+9$, then there exists a multiple of $2^{2^{k+1}}+1$ in the form of $m_2^2+9$!

Finally, to complete our induction, we need to cover the base case of $n=2$ and the external case $n=1$. The latter is covered by $m=0$ and the former also by $m=0$. Therefore, there exists a multiple of $2^n-1$ in the form of $m^2+9$ if and only if $n$ is a power of $2$, finishing the problem.
This post has been edited 2 times. Last edited by peppapig_, Mar 14, 2024, 12:48 AM
Reason: Parentheses
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SenorSloth
37 posts
#61 • 2 Y
Y by OronSH, cubres
We claim that the answer is $n=2^x$ for some nonnegative integer $x$.

We start by proving that all other numbers fail. For any odd integer $n>1$, we have that $3\nmid 2^n-1$. We also have that $2^n-1\equiv 3\pmod{4}$, which implies that $2^n-1$ must have some prime divisor $p\equiv 3\pmod{4}$, and $p\neq3$. Since we require $2^n-1\mid m^2+9$, this prime must also divide $m^2+9$. However, we can then apply Fermat's Christmas theorem on $m^2+9$ to show that any prime $p$ dividing $m^2+3^2$ is either $p\equiv 1\pmod{4}$ or $p=3$, contradiction. Any even $n$ that is not a power of $2$ will have some odd factor $x>1$, and since $2^x-1\mid 2^n-1$ we get the same contradiction.

Now we prove that $n=2^k$ works. $k=0$ gives $2^n-1=1$, which clearly works. We will now prove that for positive $k$, $2^{2^k}-1$ is the product of $3$ and some (not necessarily distinct) primes that are $1\pmod{4}$. We will use induction to do this, with base case $k=1$ and $2^{2^k}-1=3$. For the inductive step, we notice that $2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1)$, so we just need to prove that $2^{2^k}+1$ only has prime factors that are $1\pmod 4$. This is true by applying Fermat's Christmas Theorem, so the induction is complete.

Now we just need to show there exists a working $m$. We can factor $2^n-1$ into powers of distinct primes, and consider each prime power separately. For the factor of $3$, just select $m\equiv0\pmod{3}$. For the other primes, which must have $p\equiv 1\pmod{4}$, we need to show that there exists a value of $m\pmod{p^k}$ such that $m^2+9\equiv\left(\frac m3 \right)^2+1\equiv 0 \pmod{p^k}$. It is well-known that there exists a primitive root $g$ with order $\phi(p^k)=(p-1)(p^{k-1})$. Since we know $\frac{p-1}{4}$ is an integer, we can let $\frac m3\equiv g^{\frac{(p-1)(p^{k-1})}{4}}\pmod{p^k}$ so that $\left(\frac m3 \right)^2\equiv -1 \pmod{p}$. Then we just multiply by $3$ to get a working value for $m\pmod{p^k}$. Then by applying CRT on all of the separate prime powers, we know that there will exist a value of $m$ such that $m^2+9$ is divisible by all of the prime powers together, and we are done.
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blueberryfaygo_55
340 posts
#62 • 2 Y
Y by megarnie, cubres
The answer is all $n =2^k$ where $k$ is a nonnegative integer. We first introduce the following lemma.

Lemma. For $n>1$, every prime divisor of $2^n-1$ is either $3$ or congruent to $1 \pmod 4$ if and only if $n$ is a power of $2$.
Proof. First suppose that $n=2^k$, and let $p$ be a prime divisor of $2^n-1$. We have $$2^{2^k} \equiv 1 \pmod p$$so letting $\mathrm{ord}_p(2) = r$, it follows that $$\begin{cases} r \mid 2^k \\ r \mid p-1 \end{cases}$$so $r=2^l$ for $l \geq 1$. If $l=1$, then $4 \equiv 1 \pmod p$ or $p=3$; otherwise, $r \equiv 0 \pmod 4$, so we must have $4 \mid p-1$ or $p \equiv 1 \pmod 4$.
Now, we show that if every prime divisor $p$ of $2^n-1$ is either $3$ or congruent to $1 \pmod 4$, then $n$ must be a power of $2$. For the sake of contradiction, suppose $n$ is not a power of $2$. Then, $n = 2^u \cdot w$ where $u$ is a nonnegative integer and $w$ is an odd integer greater than $1$. We know that $$2^w - 1 \mid 2^n - 1$$but $2^w - 1 \equiv (-1)^w - 1 \equiv 1 \not \equiv 0 \pmod 3$, so every prime dividing $2^w-1$ is congruent to $1 \pmod 4$. It is then clear that $$2^w - 1 \equiv 1 \pmod 4$$so $w=1$, a contradiction, giving the lemma. $\blacksquare$

Returning to the problem, suppose $q$ is a prime divisor of $2^n-1$. The given condition implies that $$m^2 \equiv -9 \pmod q$$so either $q=3$ or $$\left(\dfrac{-9}{q}\right) = \left(\dfrac{9}{q}\right) \left(\dfrac{-1}{q}\right) = \left(\dfrac{-1}{q}\right) = -1.$$It follows that $q \equiv 1 \pmod 4$, but $q$ is an arbitrary prime dividing $2^n-1$, so every prime dividing $2^n-1$ is congruent to $1 \pmod 4$. Applying the lemma finishes, since if $n$ is a power of $2$, we have $-9$ as a quadratic residue modulo every prime divisor of $n$ (or divisible by the prime divisor $3$), and the construction of $m$ follows from using the Chinese Remainder Theorem on the prime divisors of $n$. $\blacksquare$
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Eka01
204 posts
#63 • 1 Y
Y by cubres
Been a while since I wrote solutions.
We claim that the answer is all $n$ of the form $\boxed {n=2^k}$.
First note by Fermat's Two Squares Theorem that any prime factor of $m^2+9$ is either equal to $3$ or congruent to $1(mod \ 4)$.
Now if $n$ is odd, it is trivial to observe that $2^n -1 \equiv 3(mod \ 4)$ for $n \geq 3$ so it follows that the required $n$ must be of the form $2^k$. We now show that these indeed work.
Proceed by induction.
We assume that $ 2^{2^k} -1 | m^2 +9$. Now we need to show that $2^{2^{k+1}} -1 | (m+a)^2 +9$ or that $a^2 +2am$ is divisible by $2^{2^k} -1$ and $2^{2^k} +1$. Since they are both coprime, the result follows by $CRT$ (This probably requires some care but good enough for a comeback I suppose).
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onyqz
195 posts
#64 • 1 Y
Y by cubres
great problem
solution
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SomeonesPenguin
128 posts
#65 • 1 Y
Y by cubres
The answer is $n=2^k$ where $k$ is a positive integer. We will use the following lemma:

If $p\equiv 3\pmod 4$ and $p\mid x^2+y^2$, then $p\mid x$ and $p\mid y$.

In particular, this means that $2^n-1$ has at most one prime factor that is $3$ modulo $4$, which is $3$. Suppose now that there is some prime number $p\mid n$, $p\equiv 3\pmod 4$. Clearly \[2^p-1\mid 2^n-1\mid m^2+9\]Notice now that $2^p-1\equiv 3\pmod 4$ so it must have a prime factor that is $3$ modulo $4$. This also can't be $3$ since $3\nmid 2^p-1$ (because $p$ is odd), hence we get a contradiction.

Therefore, $n=2^k$ for some positive integer $k$. We prove that all such $n$ work.

Note that we basically don't care about the $3$ in $2^{2^k}-1$ since from LTE $\nu_3\left(2^{2^k}-1\right)=1$ so we can just take $3\mid m$.

Now I claim that the only prime factor that is $3$ modulo $4$ of $2^{2^k}-1$ is $3$.
Proof: Note that \[2^{2^k}-1=(2+1)\left(2^2+1\right)\dots\left(2^{2^{k-1}}+1\right)\]Besides the first factor, all of them are of the form $x^2+1$ so they can't have a prime factor $p\equiv 3\pmod 4$. $\square$

For any other prime number $p\mid 2^{2^k}-1$ (note that from the above we have $p\equiv 1\pmod 4$) we have $\gcd(3,p)=1$ so $3\mid m$ doesn't affect us. Now note \[\left(\dfrac{-9}{q}\right)=\left(\dfrac{-1}{q}\right) \cdot \left(\dfrac{9}{q}\right)=1\]So there is some $m$ such that $p\mid m^2+9$. Therefore, by CRT we can find the desired $m$. $\blacksquare$
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EpicBird08
1751 posts
#67 • 1 Y
Y by cubres
Does my proof of sufficiency work?

The answer is $n = 2^k$. Assume $n > 1$ since $n = 1$ trivially holds.

Necessity: Suppose that an odd number $d > 1$ divided $n,$ so that $2^d - 1$ divides $m^2 + 9$ and so $m^2 \equiv -9 \pmod{2^d - 1}.$ Since $d$ is odd, we have that $2^d - 1$ is not divisible by $3,$ and so we can further reduce this to $m^2 \equiv -1 \pmod{2^d - 1},$ i.e. $2^d - 1$ divides $m^2 + 1.$ We know by orders that $m^2 + 1$ only has prime divisors equivalent to $1 \pmod{4},$ so all divisors of $m^2 + 1$ are equivalent to $1 \pmod{4}.$ But $2^d - 1 \equiv 3 \pmod{4}$ for $d > 1,$ yielding a contradiction.

Sufficiency: Let $n = 2^k.$ We will construct such a number $m$ inductively, with the base case $k = 1$ holding with $m = 0.$ Now assume that we have a number $m$ such that $$m^2 \equiv -9 \pmod{2^{2^k} -1}.$$Since $2^{2^{k+1}} - 1 = (2^{2^k} - 1)(2^{2^k} + 1)$ and $\gcd(2^{2^k} - 1, 2^{2^k} + 1) = 1,$ we just need to find a number $m$ such that $m^2 \equiv -9 \pmod{2^{2^k} + 1}$, from which we are done by the Chinese Remainder Theorem. Clearly $2^{2^k} + 1$ is not divisible by $3,$ so we can reduce this to $m^2 \equiv -1 \pmod{2^{2^k} + 1}.$

Factor $2^{2^k} + 1 = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$ where the $p_i$ are primes and $e_i \in \mathbb{N}_0$ such that $p_i \equiv 1 \pmod{4}$ for all $i.$ By the Chinese Remainder Theorem, it suffices to show that $m^2 \equiv -1$ has a solution modulo $p_i^{e_i}$ for all $i.$ Notice that $p_i^{e_i}$ has a primitive root, and $\phi(p_i^{e_i}) = (p_i - 1)p_i^{e_i - 1}$ is divisible by $4,$ which readily implies that $-1$ is a quadratic residue. Combining all these congruences gives us our desired $m.$

Therefore, the only $n$ such that $2^n - 1$ has a multiple of the form $m^2 + 9$ are powers of two, which we have confirmed to work.
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smileapple
1010 posts
#68 • 1 Y
Y by cubres
Suppose that $k\mid n$ for some odd $k>2$. If $m^2\equiv-9\pmod{2^n-1}$ for some $m$, then $m^2\equiv-9\pmod{2^k-1}$. Since $2^k\equiv2\pmod3$, we have $(\frac{m}3)^2\equiv-1\pmod{2^k-1}$. Since $2^k-1\equiv3\pmod4$, it follows that $-1$ is a quadratic residue modulo $p$ for some prime divisor $p$ such that $p\equiv3\pmod4$, a contradiction. Hence, if $n$ is not a power of $2$, then $2^n-1$ has no multiple of the form $m^2+9$.

Suppose that $-9$ is a quadratic residue modulo $2^{2^r}-1$ for some nonnegative integer $r$. note that $(3\cdot 2^{2^{r-1}})^2\equiv-9\pmod{2^{2^r}+1}$. Then $-9$ is a quadratic residue modulo $(2^{2^r}-1)(2^{2^r}+1)=2^{2^{r+1}}-1$. By induction, it follows that $-9$ is a quadratic residue modulo $2^{2^r}-1$ for all $r$, so that $2^{2^r}-1$ has a multiple of the form $m^2+9$.

We thus conclude that the solution set for $n$ is given by $\boxed{\{2^r\mid r\in\mathbb{Z},r\ge0\}}.$ $\blacksquare$
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cursed_tangent1434
620 posts
#69 • 1 Y
Y by cubres
Mid problem, but I think it's quite instructive. Doesn't help if you misread and spend a non-trivial amount of time trying to solve the problem with 'divisor' instead of 'multiple'.

We claim that the answer is all positive integers $n$ of the form $n=2^k$ for some non-negative integer $k$. When $n=1$ the result is clear so we deal with $n>1$ in what follows. First we shall show that all the claimed solutions indeed work.

Clearly $3 \mid 2^{2^k}-1$ and in particular, Lifting the Exponent Lemma tells us that $\nu_3(2^{2^{k}}-1)=\nu_3(3)+\nu_3(2^{k-1})=1$. Thus, we can write
\[2^n-1=3 \cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a^r}\]for distinct odd primes $p_1, p_2, \cdots , p_r$. We first claim the following.

Claim : All primes $p_1,p_2,\dots , p_r$ are $1 \pmod{4}$.

Proof : Simply note that since
\[2^{2^k}-1=(2-1)(2+1)(2^2+1)\dots (2^{2^{k-1}}+1)\]Since each index of $2$ beyond the second factor is even, none of these factors are divisible by 3. Further, if $p \equiv 3 \pmod{4}$ (and $p>3$) divides one of these terms, then $p \mid 2^{2^i}+1$ for some $i \ge 1$. However, this is a clear contradiction by Fermat's Christmas Theorem implying that the left-hand expression has no $3 \pmod{4}$ primes factors as claimed.


We let $3 \mid m$. Further, for each prime $p_i \mid 2^{2^k}-1$ we have,
\[(-9)^{\frac{p_i-1}{2}} \equiv (-1)^{\frac{p_i-1}{2}}\cdot 9^{\frac{p_i-1}{2}} \equiv (-1)^{\frac{p_i-1}{2}} \equiv 1 \pmod{p_i}\]since $4 \mid p_i-1$ as noted above. Thus, $-9$ is a QR $\pmod{p_i}$. Thus, for each prime $p_i$ there exists some positive integer $x_i$ for which $p_i \mid x_i^2+9$. We now resort to induction.

Say there exists a positive integer $x_{im}$ such that $p^m \mid x_{im}^2+9$ for some $m \ge 1$. We note that,
\[(x_{im}+kp^m)^2+9\]is injective $\pmod{p^{m+1}}$ and in the set $\{0,p^m,2\cdot p^m,\dots (p-1)\cdot p^m\}$ as $k$ ranges from $0$ to $p-1$. Thus, one of these values for $k$ gives a new positive integer $x_{i(m+1)}=x_{im}+kp^m$ such that $p^{m+1}\mid x_{i(m+1)}^2+9$.

Now, applying the inductive described above to $p_i$, we conclude that there must exist some positive integer $y_i$ such that $p_i^{a_i} \mid y_i^2 +9$ for all $1 \le i \le r$. We apply CRT on $3 , p_1,p_2,\dots , p_r$ to construct a suitable value for $m$ and finish.

To see why no other $n$ work, say $p \mid n$ and $2<p<n$. Thus,
\[2^p-1\mid 2^n-1 \mid m^2+9\]Since $p>2$ the left-hand side is $3 \pmod{4}$ and thus, some prime factor of $2^p-1$ , $q \ne 9 \equiv 3 \pmod{4}$ exists. However, by Fermat's Christmas Theorem this is a clear contradiction as we must have $q \mid 9$.
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AshAuktober
1004 posts
#70 • 1 Y
Y by cubres
Odd $n$ don't work so $n$ must be a power of 2. Such $n$ can be shown to work using CRT combined with LTE to show $\nu_3$ can't be too large.

Also why is misreading this so real lmfao.
This post has been edited 1 time. Last edited by AshAuktober, Apr 10, 2025, 12:18 PM
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ATM_
21 posts
#71 • 1 Y
Y by cubres
Suppose $n$ has an odd prime divisor d , such that : $d>1$
We have : $2^d-1|2^n-1$
Hence : $2^d-1|a^2+9$
$d>1\implies 2^d-1>1$ , thus: $2^d-1$ has an odd prime divisor p
By Fermat Christmas theorem: $p \equiv 1[4]$
Hence : $p|a^2+9\implies a^2\iff -9[p]$
Using quadratic recopricity : $\left(\dfrac{-9}{p}\right)=\left(\dfrac{-1}{p}\right).\left(\dfrac{3}{p}\right)^2=-1.\left(\dfrac{3}{p}\right)^2$
Because : $p\iff 3[4]$

If: $p\neq 3$ , then:$ \left(\dfrac{3}{p}\right)=\pm 1\implies \left(\dfrac{3}{p}\right)^2=1$
So :b$ \left(\dfrac{-9}{p}\right)=-1$ (not a quadratic residu mod p)
Absurde , Hence: $p=3$

So: $3|2^d-1\implies 2^d\equiv 1[3]$
Hence : $o_p(2)|d\implies 2|d$
Absurde ,cause d is odd
Thus n has no odd divisors
Which means $n$ is a power of 2
Let : $n=2^k/k\in \mathbb{N}$

Proof by induction:
For $n=1,2^n-1=1|a^2+9$
True
Suppose :$\exists a\in \mathbb{N}:a^2\equiv -9[2^{2^{k}}-1]$

We habe : $2^{2^{k+1}}-1=(2^k-1)(2^k+1)$
For some positive integer $a_1=3×2^{2^{k-1}},with :k\in \mathbb{N}*$
We have : $a_1^2+9=9×2^{2^k}+9=9(2^{2^k}+1)$
Thus : $2^{2^k}+1|a_1^2+9$
And : $a^2\equiv -9[2^{2^k}-1]$
Because : $pgcd(2^{2^k}-1,2^{2^k}+1)=gcd(2^{2^k}+1,2)=1$
By CRT: $\exists b\in \mathbb{N}:b\equiv a_1[2^{2^k}+1]$ and $b\equiv a[2^{2^k}-1]$
$\implies b^2\equiv a^2\equiv -9[2^{2^k}-1]$ et $b^2\equiv a_1^2\equiv -9[2^{2^k}+1]$
So : $\exists b\in \mathbb{N}:b^2\equiv -9[2^{2^{k+1}}-1]$

Hence : $\exists a\in \mathbb{N}:2^n-1|a^2+9$ iff $n$ is a power of 2
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Ilikeminecraft
616 posts
#72
Y by
We are solving when $2^n - 1\mid m^2 + 9.$ I claim that an integer $d$ has a multiple of the form $m^2 + 9$ if and only if none of $d$'s prime factors, other than $3$, are $3\pmod4$, and $\nu_3(d) \leq 2.$

We first show that all $d$ that has a multiple of the form $m^2 + 9$ must be of that form. By Fermat's Christmas theorem, we have that all of the prime divisors of $m^2 + 9$ must be $1\pmod4,$ or is 2. Now, assume that $\nu_3(d) \geq 3.$ Thus, any multiple of $d$ divides 27. However, $\nu_3(m^2 + 9) \leq2.$ Thus, this direction is now proved.

Now, we show that all $d$ of this form has a multiple that can be written as $m^2 + 9.$ By using the Jacobi Symbol, we can use quadratic reprocity to show that there exists an $m$ such that $m^2 \equiv -1\pmod d,$ and thus we are done.

Now, we are looking for when $2^{n} - 1$ satisfies the conditions of $d.$ For all the prime factors to be $1\pmod4$ or $3$ we claim that $n$ is a power of 2. This can be seen by using orders.

Since $\nu_3(2^{k - 1}) = 0,$ by $LTE,$ our expression has at most one power of 3. Thus, the answer is $n = \boxed{2^k}$ for $k\in\mathbb N$
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