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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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 events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/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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MATHCOUNTS/AMC 8 Basics
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Programming

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0 replies
jlacosta
Apr 2, 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
Distributing coins in a circle
quacksaysduck   1
N 6 minutes ago by BR1F1SZ
Source: JOM 2025 Mock 1 P4
There are $n$ people arranged in a circle, and $n^{n^n}$ coins are distributed among them, where each person has at least $n^n$ coins. Each person is then assigned a random index number in $\{1,2,...n\}$ such that no two people have the same number. Then every minute, if $i$ is the number of minutes passed, the person with index number congruent to $i$ mod $n$ will give a coin to the person on his left or right. After some time, everyone has the same number of coins.

For what $n$ is this always possible, regardless of the original distribution of coins and index numbers?

(Proposed by Ho Janson)
1 reply
quacksaysduck
Jan 26, 2025
BR1F1SZ
6 minutes ago
Do not try to case bash lol
ItzsleepyXD   3
N 39 minutes ago by reni_wee
Source: Own , Mock Thailand Mathematic Olympiad P3
Let $n,d\geqslant 6$ be a positive integer such that $d\mid 6^{n!}+1$ .
Prove that $d>2n+6$ .
3 replies
ItzsleepyXD
Yesterday at 9:08 AM
reni_wee
39 minutes ago
The product of two p-pods is a p-pod
MellowMelon   10
N 43 minutes ago by Mathandski
Source: USA TST 2011 P3
Let $p$ be a prime. We say that a sequence of integers $\{z_n\}_{n=0}^\infty$ is a $p$-pod if for each $e \geq 0$, there is an $N \geq 0$ such that whenever $m \geq N$, $p^e$ divides the sum
\[\sum_{k=0}^m (-1)^k {m \choose k} z_k.\]
Prove that if both sequences $\{x_n\}_{n=0}^\infty$ and $\{y_n\}_{n=0}^\infty$ are $p$-pods, then the sequence $\{x_ny_n\}_{n=0}^\infty$ is a $p$-pod.
10 replies
MellowMelon
Jul 26, 2011
Mathandski
43 minutes ago
Nordic squares!
mathisreaI   36
N an hour ago by awesomehuman
Source: IMO 2022 Problem 6
Let $n$ be a positive integer. A Nordic square is an $n \times n$ board containing all the integers from $1$ to $n^2$ so that each cell contains exactly one number. Two different cells are considered adjacent if they share a common side. Every cell that is adjacent only to cells containing larger numbers is called a valley. An uphill path is a sequence of one or more cells such that:

(i) the first cell in the sequence is a valley,

(ii) each subsequent cell in the sequence is adjacent to the previous cell, and

(iii) the numbers written in the cells in the sequence are in increasing order.

Find, as a function of $n$, the smallest possible total number of uphill paths in a Nordic square.

Author: Nikola Petrović
36 replies
mathisreaI
Jul 13, 2022
awesomehuman
an hour ago
MOP Emails Out! (not clickbait)
Mathandski   101
N Today at 1:01 PM by pingpongmerrily
What an emotional roller coaster the past 34 days have been.

Congrats to all that qualified!
101 replies
Mathandski
Apr 22, 2025
pingpongmerrily
Today at 1:01 PM
How many approaches you got? (A lot)
IAmTheHazard   86
N Today at 8:27 AM by User141208
Source: USAMO 2023/2
Let $\mathbb{R}^+$ be the set of positive real numbers. Find all functions $f \colon \mathbb{R}^+ \to \mathbb{R}^+$ such that, for all $x,y \in \mathbb{R}^+$,
$$f(xy+f(x))=xf(y)+2.$$
86 replies
1 viewing
IAmTheHazard
Mar 23, 2023
User141208
Today at 8:27 AM
Berkeley mini Math Tournament Online 2025 - June 7
BerkeleyMathTournament   0
Today at 7:38 AM
Berkeley mini Math Tournament is a math competition hosted for middle school students once a year. Students compete in multiple rounds: individual round, team round, puzzle round, and relay round.

BmMT 2025 Online will be held on June 7th, and registration is OPEN! Registration is $8 per student. Our website https://berkeley.mt/events/bmmt-2025-online/ has more details about the event, past tests to practice with, and frequently asked questions. We look forward to building community and inspiring students as they explore the world of math!

3 out of 4 of the rounds are completed with a team, so it’s a great opportunity for students to work together. Beyond getting more comfortable with math and becoming better problem solvers, our team is preparing some fun post-competition activities!

Registration is open to students in grades 8 or below. You do not have to be local to the Bay Area or California to register for BmMT Online. Students may register as a team of 1, but it is beneficial to compete on a team of at least 3 due to our scoring guideline and for the experience.

We hope you consider attending, or if you are a parent or teacher, that you encourage your students to think about attending BmMT. Thank you, and once again find more details/register at our website,https://berkeley.mt.
0 replies
BerkeleyMathTournament
Today at 7:38 AM
0 replies
How to get good at comp math
fossasor   28
N Today at 6:27 AM by Konigsberg
I'm a rising ninth grader who wasn't in the school math league this year, and basically put aside comp math for a year. Unfortunately, that means that now that I'm in high school and having the epiphany about how important comp math actually is, and how much it would help my chances of getting involved in other math-related programs. In addition, I do enjoy math in general, and suspect that things like the AMCs are probably going to be some of the best practice I can get. What this all means is that I'm trying to go from mediocre to orz, 2 years after I probably should have started if I wanted to be any good.

So my question is: how do I get good at comp math?

This year, my scores on AMC 10 (and these are the highest I've ever gotten) were a 73.5 and an 82.5 (AMC 8 was 21/25, but that doesn't matter much). This is not good enough to qualify for AIME, and I probably need to raise my performance on each by at least 10 points. I've been decently good in the past at Number Theory, but I need to work on Geo and Combinatorics, and I'm trying to find the best resources to do that. My biggest flaw is probably not knowing many algorithms like Stars and Bars, and the path is clear here (learn them) but I'm still not sure which ones I need to know.

I'm aware that some of this advice is going to be something like "Practice 5 hours a day and start hardgrinding" or something along those lines. Unfortunately, I have other extracurriculars I need to balance, and for me, time is a limiting resource. My parents are somewhat frowning upon me doing a lot of comp math, which limits my time as well. I have neither the time nor motivation to do more than an hour a day, and in practice, I don't think I can be doing that consistently. As such, I would need to make that time count.

I know this is a very general question, and that aops is chock-full of detailed advice for math competitions. However, I'd appreciate it if anyone here could help me out, or show me the best resources I should use to get started. What mocks are any good, or what textbooks should I use? Where do I get the best practice with the shortest time? Is there some place I can find a list of useful formulas that have appeared in math comps before?

All advice is welcome!

28 replies
fossasor
Apr 10, 2025
Konigsberg
Today at 6:27 AM
ranttttt
alcumusftwgrind   21
N Today at 5:47 AM by nmlikesmath
rant
21 replies
alcumusftwgrind
Yesterday at 11:04 PM
nmlikesmath
Today at 5:47 AM
MasterScholar North Carolina Math Camp
Ruegerbyrd   17
N Today at 4:51 AM by fake123
Is this legit? Worth the cost ($6500)? Program Fees Cover: Tuition, course materials, field trip costs, and housing and meals at Saint Mary's School.

"Themes:

1. From Number Theory and Special Relativity to Game Theory
2. Applications to Economics

Subjects Covered:

Number Theory - Group Theory - RSA Encryption - Game Theory - Estimating Pi - Complex Numbers - Quaternions - Topology of Surfaces - Introduction to Differential Geometry - Collective Decision Making - Survey of Calculus - Applications to Economics - Statistics and the Central Limit Theorem - Special Relativity"

website(?): https://www.teenlife.com/l/summer/masterscholar-north-carolina-math-camp/
17 replies
1 viewing
Ruegerbyrd
Yesterday at 3:15 AM
fake123
Today at 4:51 AM
Segment Product
worthawholebean   25
N Today at 4:37 AM by deduck
Source: AIME 2009II Problem 13
Let $ A$ and $ B$ be the endpoints of a semicircular arc of radius $ 2$. The arc is divided into seven congruent arcs by six equally spaced points $ C_1,C_2,\ldots,C_6$. All chords of the form $ \overline{AC_i}$ or $ \overline{BC_i}$ are drawn. Let $ n$ be the product of the lengths of these twelve chords. Find the remainder when $ n$ is divided by $ 1000$.
25 replies
worthawholebean
Apr 2, 2009
deduck
Today at 4:37 AM
[April 2025] MEX MAGAZINE
Possible   1
N Today at 4:16 AM by Possible
Hello AoPS!

MathEXplained MAGAZINE aims to be a monthly newsletter that publishes articles in a newspaper-style format about various math related subjects in a way which can appeal to a broader audience. Topics may range from trending news stories within the math community to in-depth dissections of niche and interesting mathematical topics for more advanced readers.

Our Goal

While many of us on this platform might feel as if math is something super familiar to us, the truth is that this isn't the case outside of our tightly-knit community. Math doesn't get much coverage in mainstream media due to it being wrongfully seen by the general population either as a nerd's hobby or extremely foreign to the average person. Our mission as an organization is to promote mathematics through recognizing top mathletes, covering interesting math topics, and painting mathematics as something which can be fun and easy to get into.

April 2025 Edition

Celebrate the coming of Spring with MEX MAGAZINE's new publication! Check out this file to solve some Spring related math problems! Make sure to also read about our other spreads which cover fibonacci in nature and fractals in nature!

Website

If you are interested in reading our past issues, you can check out our website at this link. Over there, you can find links to our discord community where problem discussions can happen.

Staff

Our program is currently looking for new staff members! Please fill out this form to be considered.
1 reply
Possible
Today at 4:15 AM
Possible
Today at 4:16 AM
Trying to outreach for my math competition!
joeylllll   0
Today at 3:56 AM
Hi! I'm Joey, and right now I'm hosting a free community type math competition in my state (FL) and a few others across the country with my organization ExcelAcademe. This month will be our third official contest. We do a middle school as well as high school test, each is 60 minutes, and they have a MCQ and short answer section at about mid AMC10/12 to low/mid AIME difficulty. We have past tests up on our website at https://excelacademe.com/rankings/ if anybody's interested.

We're looking for people who are interested in starting up chapters in their own areas and helping expand our competition, or anyone who wants to help out with test-writing. We have been concentrating on hosting the competitions at local libraries and schools, and the top winners each month get a prize from one of our sponsors (Wolfram, Jane Street, (and hopefully more soon)). Anyone who's interested in hosting or test writing please let me know :)
0 replies
joeylllll
Today at 3:56 AM
0 replies
MathILy 2025 Decisions Thread
mysterynotfound   20
N Today at 3:09 AM by mysterynotfound
Discuss your decisions here!
also share any relevant details about your decisions if you want
20 replies
mysterynotfound
Apr 21, 2025
mysterynotfound
Today at 3:09 AM
Integer Coefficient Polynomial with order
MNJ2357   9
N Apr 5, 2025 by v_Enhance
Source: 2019 Korea Winter Program Practice Test 1 Problem 3
Find all polynomials $P(x)$ with integer coefficients such that for all positive number $n$ and prime $p$ satisfying $p\nmid nP(n)$, we have $ord_p(n)\ge ord_p(P(n))$.
9 replies
MNJ2357
Jan 12, 2019
v_Enhance
Apr 5, 2025
Integer Coefficient Polynomial with order
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Source: 2019 Korea Winter Program Practice Test 1 Problem 3
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MNJ2357
644 posts
#1 • 2 Y
Y by Superguy, Adventure10
Find all polynomials $P(x)$ with integer coefficients such that for all positive number $n$ and prime $p$ satisfying $p\nmid nP(n)$, we have $ord_p(n)\ge ord_p(P(n))$.
This post has been edited 1 time. Last edited by MNJ2357, Jan 12, 2019, 10:17 PM
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stroller
894 posts
#2 • 2 Y
Y by Adventure10, Mango247
Wrong, thanks to post below for pointing out :stretcher:
This post has been edited 3 times. Last edited by stroller, Jan 16, 2019, 2:00 AM
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fattypiggy123
615 posts
#3 • 2 Y
Y by stroller, Adventure10
stroller wrote:

The given condition translates to $p | n- \omega_k \implies p | P(n)^k -1 $

Shouldn't this be $p \mid P(n)^m - 1$ for some $m \leq k$?
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stroller
894 posts
#4 • 1 Y
Y by Adventure10
fattypiggy123 wrote:
stroller wrote:

The given condition translates to $p | n- \omega_k \implies p | P(n)^k -1 $

Shouldn't this be $p \mid P(n)^m - 1$ for some $m \leq k$?

Sorry my mistake; I misread the problem as $\text{ord}_p(P(n))|\text{ord}_p(n)$
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stroller
894 posts
#5 • 2 Y
Y by Adventure10, Mango247
My fix for the error pointed out in post #3.

What I was most concerned about being incorrect
This post has been edited 1 time. Last edited by stroller, Jan 16, 2019, 1:55 AM
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fattypiggy123
615 posts
#6 • 4 Y
Y by Kayak, stroller, Adventure10, Mango247
Yes your solution has a few points of contention, for instance $\omega_k$ starts off life as an element of $\mathbb{F}_p$ but somehow becomes an actual complex root of unity halfway through. All of this can be fixed/made rigorous by some algebraic number theory. I will post more on it when I am free.
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fattypiggy123
615 posts
#7 • 7 Y
Y by Loppukilpailija, stroller, MNJ2357, Gaussian_cyber, Superguy, Adventure10, Mango247
The main idea to stroller's solution is to attempt to lift the local conditions, $P(u_p) \equiv v_p \pmod p$ where $u_p,v_p$ are integers satisfying $u_p^k , v_p^m \equiv 1 \pmod p$, into a global condition $p \mid P(x) - y$ for some numbers $x,y$ to deduce that $P(x) - y = 0$ since it cannot have infinitely many distinct prime factors. For instance if $k = m = 1$, then we can simply choose $x = y = 1$. The problem comes when $k$ and $m$ are larger than $2$ and one cannot find any integer $x$ such that $x^k \equiv 1 \pmod p$ for infinitely many primes $p$ where $x \not = 1$. As indicated in stroller's solution, we should bring in the complex primitive roots of unity $\omega_k$ and $\omega_m$. For simplicity, I will just assume $m = k$.

Now there is a natural place (but not the only one!) to do arithmetic with $\omega_k$. The field $\mathbb{Q}(\omega_k)$ is a number field and so its ring of integers is a Dedekind domain. When working beyond $\mathbb{Z}$, it is known that we cannot hope for unique factorization of integers to continue to hold. What one can achieve however is unique factorization into ideals, which is a property that Dedekind domains possess (and can also be taken as its defining property) and so makes Dedekind domains a nice place to do arithmetic in. In the case of $\mathbb{Q}(\omega_k)$, its ring of integers is simply $\mathbb{Z}[\omega_k]$, which is just sums of $\omega_k^i$ with $\mathbb{Z}$-coefficients.

For a prime $p$, let $\mathbb{F}_q$ be the smallest field extension of $\mathbb{F}_p$ where you have a primitive $k^{th}$ root of unity. This always exists if say $p \nmid k$. Then just like how in the usual integers $\mathbb{Z}$, we have the "reduction mod p" maps from $\mathbb{Z}$ to $\mathbb{F}_p$, there will exist reduction maps from $\mathbb{Z}[\omega_k]$ to $\mathbb{F}_q$ for each such $q$ where $\omega_k$ is sent to one of the primitive $k^{th}$ roots of unity in $\mathbb{F}_q$. When $p \nmid k$, the ideal $(p)$ in $\mathbb{Z}[\omega_k]$ factors as a product of distinct prime ideals $\mathfrak{p}_1 \cdots \mathfrak{p}_g$. These reduction maps then arise from reducing modulo $\mathfrak{p}_i$ instead of just mod $p$.

For example, if we work over the Gaussain integers $\mathbb{Z}[i]$ instead then we are looking at field extensions $\mathbb{F}_q$ where $x^2 = -1$ has a solution. For $p = 3$, there are no solutions within $\mathbb{F}_3$ itself and so we have to extend to $\mathbb{F}_9$, the field with $9$ elements. Correspondingly, the element/ideal $(3)$ is prime in $\mathbb{Z}[i]$ and so the map $\mathbb{Z}[i] \to \mathbb{F}_9$ arises from $a + bi \mapsto a \pmod 3 + b \pmod 3 i$. You can check that there are then $9$ elements just like $\mathbb{F}_9$. Howeverfor $p = 5$, within $\mathbb{F}_5$ there is already a solution to $x^2 = -1$ and so $\mathbb{F}_q = \mathbb{F}_5$. Our map $\mathbb{Z}[i] \to \mathbb{F}_5$ can't be just reducing mod $5$ then as that will give us $25$ elements instead. Instead one should factor $(5) = (2+i)(2-i)$ and then reduce mod $(2+i)$ or $(2-i)$. Since $x^2 + 1 = 0$ has two solutions in $\mathbb{F}_5$, there are two possible elements one can map $i$ to and each mod corresponds to one of them.

As a result, we can lift the local conditions to get for each $p \equiv 1 \pmod k$, there is one prime ideal factor $\mathfrak{p}$ of the ideal $(p)$ that divides $P(\omega_k) - \omega_k^m$ which implies that it must be $0$ by unique factorization of ideals.
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MarkBcc168
1595 posts
#8 • 3 Y
Y by stroller, Gaussian_cyber, Adventure10
Here is alternative way to make stroller's solution rigorous, using elementary method. We will prove that $P(\omega_k)^{k!}=1$.

The key idea is to transfer from $\mathbb{Q}(\omega_k)$ to $\mathbb{Z}_p$ properly. However, the difficulty is $\omega_k$ is defined through roots of polynomials, which means we don't know the order of root. We will circumvent this issue by using symmetric polynomials instead.

Fix a positive integer $n$ and let $m=\varphi(n)$. Let $\omega_1, \omega_2, ...,\omega_m$ be primitive $n$-th root unity. By Newton's theorem on polynomial on symmetric polynomial, polynomial
$$Q(X) = (X-P(\omega_1))(X-P(\omega_2))...(X-P(\omega_m))$$has integer coefficients.

Fix a prime $p\equiv 1\pmod n$ and working on $\mathbb{F}_p[X]$. Let $a_1,a_2,...,a_m$ be all residues $\pmod p$ having order $n$. We claim that
Claim 1 : Let
$$Q^*(X) = (X-P(a_1))(X-P(a_2))...(X-P(a_m))$$then $Q^*(X) - Q(X)$ has all coefficients divisible by $p$.

Proof : Let $e_1, e_2,...,e_m$ be Elementary symmetric polynomials of $m$ variables. Then in $\mathbb{F}_p[X]$,
$$(X-a_1)(X-a_2)...(X-a_m) = \varPhi_n(X) = (X-\omega_1)(X-\omega_2)...(X-\omega_m)$$Thus comparing coefficients give $e_i(a_1,a_2,...,a_m) \equiv e_i(\omega_1,\omega_2,...,\omega_m) \pmod{p}$ for any $i=1,2,...,m$. (Recall that both numbers are integer!). Therefore, by Newton's theorem on symmetric polynomial,
$$e_i(P(a_1),P(a_2),...,P(a_m)) \equiv e_i(P(\omega_1),P(\omega_2),...,P(\omega_m)) \pmod p$$implying the result.
Claim 2 : $Q^*(X)$ divides $X^{n!}-1$ (in $\mathbb{F}_p[X]$).

Proof : Since $\mathrm{ord}_p(a_1) = n$, we have $\mathrm{ord}_p(P(a_1))\leqslant n$ or $P(a_1)^{n!}\equiv 1\pmod p$. Hence each root of $Q^*(X)$ is root of $X^{n!}-1$, done.
Now we are ready to shift the local conditions into global conditions. Since $Q(X)$ must congruent to $Q*(X)$ in $\mathbb{F}_p[X]$, polynomial $Q(X)$ must congruent to some divisor of $X^{n!}-1$ in $\mathbb{F}_p[X]$.

By Pigeonhole's principle there exists infinitely many prime $q\equiv 1\pmod n$ which $Q(X)$ is congruent to same divisor $R(X)$ of $X^{n!}-1$ in $\mathbb{F}_q[X]$. This force all coefficients of $Q(X)-R(X)$ to be divisible by infinitely many prime $q$. Hence $Q(X) = R(X)\mid X^{n!}-1$, implying $P(\omega_i)^{n!} = 1$ as desired.
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pieater314159
202 posts
#9 • 2 Y
Y by Adventure10, Mango247
Here's another elementary solution:

Fix $n$, and let $p$ vary across all primes equivalent to $1\bmod n$. Now, for all $x$ for which $x^n\equiv 1\bmod p$, we have

$$P(x)^{n!}\in\{0,1\}\bmod p\implies P(x)^{n!+1}-P(x).$$
(we get $1$ if we can apply our condition and otherwise we get $0$). Let $Q(x)=P(x)^{n!+1}-P(x)$, and write

$$Q(x)=R(x)+S(x)(x^n-1)$$
where $\deg(R)<n$ (we do this in $\mathbb{Z}[x]$). Now, for all $p\equiv 1\bmod n$, we have that $Q(x)\equiv 0\bmod p$ if $x^n=\equiv 1\bmod p$, which means that $R(x)\equiv 0\bmod p$. As $R$ has $n$ roots in $\mathbb{F}_p$ and it is of degree $\leq n-1$, we have that $R$ must be equivalent to the zero polynomial $\bmod p$, and thus all of the coefficients of $R$ are divisible by $p$. Now, as this holds for infinitely many $p$, we see that $R(x)$ is identically $0$, or that

$$x^n-1\big|P(x)^{n!+1}-P(x).$$
This implies that, if $x=\zeta_n$ is a primitive $n$th root of unity,

$$P(\zeta_n)=0\mathrm{\ or\ }P(\zeta_n)^{n!}=1\implies |P(\zeta_n)|=1.$$
One of these two criteria must hold for infinitely many $n$. If it is the first, then $P$ has infinitely many roots and is identically $0$. If it is the second, then, letting $d$ be the degree of $P$,

$$P(z)z^dP\left(\frac{1}{z}\right)=z^d$$
holds for infinitely many $z$; as this is a polynomial equation these must be identical as polynomials. We claim the only solutions to this are of the form $\pm x^d$. Assume we have a counterexample $P$ with minimal degree. As $P(x)/x$ would be a counterexample if $P(0)$ were equal to $0$, we conclude that $P(0)\neq 0$. However, as $z\to\infty$ ($z$ real), the left side is of order $P(0)z^d(cz^d)$ ($c$ is the leading coefficient) while the right side is $z^d$, allowing us to conclude (as $cP(0)\neq 0$), that $d=0$ and $P$ is constant, which implies that if $P\equiv c$ we have $c^2=1$, which is exactly our solution set. Now, if $P(x)=x^d$ we have a valid solution and if $P(x)=-x^d$ we have a contradiction at $n=1$ and $p=3$, so the only solutions are $P(x)=0$ and $P(x)=x^d$, finishing the proof.
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v_Enhance
6877 posts
#10 • 1 Y
Y by ihategeo_1969
Solution from Twitch Solves ISL:

The answer is $P(n) = n^{d}$ only for $d \ge 0$, which clearly works.
For the other direction, we assume $P$ is nonconstant, and let $\ell$ be a large prime, say $\ell > 100 (\deg P + 100)^2$.

Claim: For this prime $\ell$, we have a divisibility of ${\mathbb Z}[X]$-polynomials \[ \Phi(X) \coloneqq X^{\ell-1} + X^{\ell-2} + \dots + 1 \mid P(X)^\ell - 1. \]Proof. Because $\Phi$ is irreducible and has large degree, it is coprime to both $P(n)$ and $P(n) \pm1$. Then there exists a constant $C$ such that \[ \gcd(\Phi(n), P(n)(P(n)^2-1)) \le C \]for all $n$, by Bezout.
Consider large primes $p > C$ such that
  • $p \not\equiv 1 \pmod q$ for any prime $3 \le q < \ell$,
  • $p \equiv 1 \pmod \ell$.
There are infinitely many such primes by Dirichlet. For each such $p$, we can find $n$ such that $p \mid \Phi(n)$, simply by taking $g$ to be a primitive modulo $p$, and choosing $n = g^{\frac{p-1}{\ell}}$.
For that $n$ we have $p \mid n^\ell-1$, so the order of $n$ is at most $\ell$. Consequently, $p$ divides \[ P(n) \cdot (P(n)-1) \cdot (P(n)^2-1) \cdot \dots \cdot (P(n)^{\ell}-1). \]Now $p$ has to divide these factors. It doesn't divide $P(n)$ as $p > C$. And if it divided $P(n)^k-1$ for some $k < \ell$, then the order of $P(n)$ modulo $p$ would be a divisor of $k$. But it should also divide $\ell-1$, and because of constraints on $p$ this would be force the order to be at most $2$. From $p > C$, that's impossible too. Hence $p$ divides $P(n)^\ell-1$, the last factor.
In other words, we have shown there are arbitrarily large primes such that $p$ divides $\Phi(n)$ and $P(n)^\ell-1$ for some $\ell$. Hence (via the same Bezout argument) it follows $\Phi(X)$ is not coprime to $P(X)^\ell-1$ and hence divides it. $\blacksquare$
Now let $\zeta$ be a primitive $\ell$th root of unity. Then $P(\zeta)$ is an $\ell$th root of unity as well, so \[ P(\zeta) - \zeta^d = 0. \]for some integer $r$, say $0 \le d \le \ell-1$. However, $\Phi(X) \coloneqq X^{\ell-1} + X^{\ell-2} + \dots + 1$ is the minimal polynomial of $\zeta$, and $\deg P \ll \ell$. Reading the coefficients of our nonconstant $P$, this could only happen if $P(X) = X^d$ exactly, as desired.

Remark: The last step of the argument really uses the fact we have $P(X)^\ell-1$. It would not work if we instead had $P(X)^k-1$ for some $k < \ell$, because then the $\operatorname{lcm}(k,\ell)$th cyclotomic polynomial may not be so well-behaved. That's why in the proof of the claim we had to some modular condition on $p$ (with $p \not\equiv 1 \pmod q$ for all $q$) to rule out the possibility that $p$ divided any of the other factors. If one tries the argument at first with just generic large $p$ dividing an element in the range of $\Phi$, one would instead get $\Phi(X) \mid P(X)^k-1$ for some $k$ depending on $\ell$, leading to the issue above.
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