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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/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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0 replies
jlacosta
Mar 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
<QBC =<PCB if BM = CN, <PMC = <MAB, <QNB = < NAC
parmenides51   1
N 6 minutes ago by dotscom26
Source: 2005 Estonia IMO Training Test p2
On the side BC of triangle $ABC$, the points $M$ and $N$ are taken such that the point $M$ lies between the points $B$ and $N$, and $| BM | = | CN |$. On segments $AN$ and $AM$, points $P$ and $Q$ are taken so that $\angle PMC = \angle  MAB$ and $\angle QNB = \angle NAC$. Prove that $\angle QBC = \angle PCB$.
1 reply
parmenides51
Sep 24, 2020
dotscom26
6 minutes ago
Bosnia and Herzegovina EGMO TST 2017 Problem 2
gobathegreat   2
N 18 minutes ago by anvarbek0813
Source: Bosnia and Herzegovina EGMO Team Selection Test 2017
It is given triangle $ABC$ and points $P$ and $Q$ on sides $AB$ and $AC$, respectively, such that $PQ\mid\mid BC$. Let $X$ and $Y$ be intersection points of lines $BQ$ and $CP$ with circumcircle $k$ of triangle $APQ$, and $D$ and $E$ intersection points of lines $AX$ and $AY$ with side $BC$. If $2\cdot DE=BC$, prove that circle $k$ contains intersection point of angle bisector of $\angle BAC$ with $BC$
2 replies
gobathegreat
Sep 19, 2018
anvarbek0813
18 minutes ago
Another NT FE
nukelauncher   58
N 29 minutes ago by andrewthenerd
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
58 replies
nukelauncher
Sep 22, 2020
andrewthenerd
29 minutes ago
Easiest Functional Equation
NCbutAN   7
N 31 minutes ago by InftyByond
Source: Random book
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
7 replies
NCbutAN
Mar 2, 2025
InftyByond
31 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   27
N 5 hours ago by sadas123
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AIME level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
27 replies
TennesseeMathTournament
Mar 9, 2025
sadas123
5 hours ago
OTIS Mock AIME 2025 airs Dec 19th
v_Enhance   38
N Today at 1:32 AM by smileapple
Source: https://web.evanchen.cc/mockaime.html
Satisfactory. Keep cooking.
IMAGE

Problems are posted at https://web.evanchen.cc/mockaime.html#current now!

Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:

[list]
[*] The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
[*] The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
[/list]
Like last time, all 30 problems are set by current and past OTIS students.

Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
[list]
[*] Free, obviously. Anyone can participate.
[*]Both tests will release sometime Dec 19th. You can do either/both.
[*]If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
[*]Solutions, statistics, and maybe some high scores will be published shortly after that.
[/list]
Feel free to post questions, hype comments, etc. in this thread.
38 replies
v_Enhance
Dec 6, 2024
smileapple
Today at 1:32 AM
How to get better at AMC 10
Dream9   6
N Today at 1:31 AM by sadas123
I'm nearly in high school now but only average like 75 on AMC 10 sadly. I want to get better so I'm doing like the first 11 questions of previous AMC 10's almost every day because I also did previous years for AMC 8. Is there any specific way to get better scores and understand more difficult problems past AMC 8? I have almost no trouble with AMC 8 problem given enough time (like 23-24 right with enough time).
6 replies
Dream9
Yesterday at 1:17 AM
sadas123
Today at 1:31 AM
An FE. Who woulda thunk it?
nikenissan   112
N Today at 1:04 AM by Marcus_Zhang
Source: 2021 USAJMO Problem 1
Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for positive integers $a$ and $b,$ \[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]
112 replies
nikenissan
Apr 15, 2021
Marcus_Zhang
Today at 1:04 AM
AIME score for college apps
Happyllamaalways   56
N Yesterday at 4:45 PM by Countmath1
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
56 replies
Happyllamaalways
Mar 13, 2025
Countmath1
Yesterday at 4:45 PM
MIT Beaverworks Summer Institute
PowerOfPi_09   0
Yesterday at 4:30 PM
Hi! I was wondering if anyone here has completed this program, and if so, which track did you choose? Do rising juniors have a chance, or is it mainly rising seniors that they accept? Also, how long did it take you to complete the prerequisites?
Thanks!
0 replies
PowerOfPi_09
Yesterday at 4:30 PM
0 replies
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Yesterday at 10:13 AM by maxamc
rant lol

Imagine a poor senior waiting for their MIT decisions just to have their hopes CRUSHED on 3/14 and they can't even celebrate pi day...

and even worse, this year's pi day is special because this year is a very special number...

8 replies
alcumusftwgrind
Yesterday at 2:11 AM
maxamc
Yesterday at 10:13 AM
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Yesterday at 6:00 AM by joshualiu315
Source: USAJMO 2024/4
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?

Proposed by Alec Sun
26 replies
GrantStar
Mar 21, 2024
joshualiu315
Yesterday at 6:00 AM
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Yesterday at 5:38 AM by joshualiu315
Source: 2024 USAMO Problem 5, JMO Problem 6
Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE = EC$. Let $M$ be the midpoint of $BC$.

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

Proposed by Anton Trygub
73 replies
ihatemath123
Mar 21, 2024
joshualiu315
Yesterday at 5:38 AM
average FE
KevinYang2.71   74
N Yesterday at 4:55 AM by joshualiu315
Source: USAJMO 2024/5
Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy
\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]for all $x,y\in\mathbb{R}$.

Proposed by Carl Schildkraut
74 replies
KevinYang2.71
Mar 21, 2024
joshualiu315
Yesterday at 4:55 AM
Eisenstein Irreducibility Criterion revisited
orl   34
N Jul 28, 2024 by almagest3001
Source: IMO 1993, Day 1, Problem 1
Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$
34 replies
orl
Nov 19, 2005
almagest3001
Jul 28, 2024
Eisenstein Irreducibility Criterion revisited
G H J
Source: IMO 1993, Day 1, Problem 1
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orl
3647 posts
#1 • 6 Y
Y by Davi-8191, Inconsistent, Adventure10, Rounak_iitr, Amir Hossein, and 1 other user
Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$
Z K Y
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joshua_mex
48 posts
#2 • 2 Y
Y by Adventure10, Mango247
well i think is an easy one...lets say that $g(x)=x^g+a_{g-1}x^{g-1}+...+a_{0}$ and $q(x)=x^q+b_{q-1}x^{q-1}+...+b_{0}$ then $a_{0}b_{0}=3$ so WLOG $b_{0}$ is multiple of 3 so lets prove by induction that $b_{n}$ is multiple of 3
first lets see that $b_{0}$ is and we know that the coefficient of $x^1$ in $p(x)$ is 0 so $b_{0}a_{1}+b_{1}a_{0}=0$ or $b_{0}a_{1}=-b_{1}a_{0]}$ and $a_{0}$ is not multiple of 3 so $b_{1}$ must be and that way you can get the result.

that is looking for the coefficient 0 of $x^{n}$ in $p(x)$ and having that $b_{i}=3k$ for $i=0,1,...,n-1$ all that is sufficient if the coefficient is 0 in $x^{q}$ and doesnt occur if $q=n-1$ if that is true then $g(x)=x+g$ with g not multiple of 3. so then $g^n+5g^{n-1}=g^{n-1}(g+5)=-3$ for some integrer g, but then $g^{n-1}$ divides 3, so $g=1$ or $g=+/-3,n=2$ and both are wrong :!: :D

is it right? :roll: :lol: ?
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Joao Pedro Santos
152 posts
#3 • 3 Y
Y by Illuzion, Adventure10, and 1 other user
We want to prove that $ f(x)$ is irreducible over $ \mathbb{Z}$. Since $ f(x)$ largest degree's term is $ 1$, its factors' largest degree's terms are $ 1$. By the Extended Eisenstein Criterion, $ f(x)$ has an irreducible factor over $ \mathbb{Z}$ with degree larger that $ n-2$. So it has an irreducible factor over $ \mathbb{Z}$ with degree $ n-1$ or $ n$. Let's suppose $ f(x)$ has an irreducible factor $ \mathbb{Z}$ with degree $ n-1$. So it also has an irreducible factor with degree $ 1$, and since its largest degree's term is $ 1$, it has a zero over $ \mathbb{Z}$, which means $ f(x)$ also has a zero over $ \mathbb{Z}$. But $ f(x)$ is always odd, therefore it can't be $ 0$. Contradiction! So $ f(x)$ has a irreducible factor over $ \mathbb{Z}$ with degree $ n$, which is $ f(x)$, QED.
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darij grinberg
6555 posts
#4 • 2 Y
Y by Adventure10, Mango247
Just edited my post at http://www.mathlinks.ro/viewtopic.php?t=37593 to include this problem as well. Just take $ p = 3$ (which is prime) and $ q = 1$ (which is squarefree and not divisible by $ p$). And consider the case $ n=2$ separately.

darij
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Pascal96
124 posts
#5 • 2 Y
Y by Adventure10, Mango247
It is a direct consequence of extended Eisenstein's criterion and rational root theorem
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mmmath
14 posts
#6 • 2 Y
Y by popdit, Adventure10
you can use peron criterion it's obvious with peron
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Pascal96
124 posts
#7 • 1 Y
Y by Adventure10
What is Peron criterion?
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darij grinberg
6555 posts
#8 • 1 Y
Y by Adventure10
He wants to say Perron. It's Theorem A in http://rms.unibuc.ro/bulletin/pdf/53-3/perron.pdf . Applying it here is pretty much overkill, though.
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filetmignon821
292 posts
#9 • 8 Y
Y by Supercali, Adventure10, Mango247, aidan0626, and 4 other users
You could use perron's, but saying 5>4 is a little boring, don't you think?

Another way to do this problem is to reduce it mod 3. We have $x^n+5x^{n-1}+3 \equiv x^n+2x^{n-1}$ (mod 3). Simplifying gives $x^{n-1}(x-2)$. Suppose f(x)=g(x)h(x), for some $g, h \in \mathbb{Z}[x]$ nonconstant. Since x and x-2 are irreducible in $\mathbb{F}_3[x]$, we have (WLOG) $g(x)=x^k+3g_1(x)$ and $h(x)=x^{n-k-1}(n-2)+3h_1(x)$, for some integer polynomials $g_1$ and $h_1$. First we consider the case where k=0. This means that g is nonconstant, a contradiction. If k=n-1, then h(x) is a linear polynomial, so f must have an integer root. We can easily check that this is false. Finally, if 1<k<n-1, multiplying g and h gives $x^{n-1}(x-2)+3g_1(x)x^{n-k-1}(x-2)+3h_1(x)x^k+9g_1(x)h_1(x)$. Setting this equal to our original polynomial and plugging in x=0 gives $9g_1(0)h_1(0)=3$. Since $g_1, h_1 \in \mathbb{Z}[x]$, this is clearly a contradiction. Thus, our polynomial is irreducible in $\mathbb{Z}[x]$.

I think this problem could also be classified as number theory. Irreducible polynomials is one of the biggest overlaps of NT and algebra.
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arkanark
255 posts
#10 • 2 Y
Y by Adventure10, Mango247
EDIT: Ignore this.
This post has been edited 1 time. Last edited by arkanark, Jul 3, 2013, 1:57 PM
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mavropnevma
15142 posts
#11 • 6 Y
Y by arkanark, coolcheetah157, FlakeLCR, quangminhltv99, Adventure10, Mango247
So, first, from $f(a_i) = a_i^n + 5a_i^{n-1} + 3 = 0$, you deduce $a_i^n + 5a_i^{n-1} = -3$, which is correct, only that you write $f(a_i) = a_i^n + 5a_i^{n-1} = -3$. Nice piece of algebra! All we have is $f(a_i) -3 = a_i^n + 5a_i^{n-1} = -3$. And now we will have $g(a_i)h(a_i) -3 = -3$, i.e. $g(a_i)h(a_i) = 0$; obviously, since $g,h$ are the factors of $f$.

Therefore $(g(a_i), h(a_i)) = (\pm 1, \mp 3)$ is clearly wrong, since at least one of them must be $0$. Moreover, in your mistaken way, you miss the possibility $(g(a_i), h(a_i)) = (\pm 3, \mp 1)$.

Nevermind, indeed under these wrong conclusions we would have $p(a_i) = g(a_i)+h(a_i) \in \{-2,2\}$, which you then write $p(a_i) =\pm 2$. True, $\deg p \leq n-1$. But you don't know that the $a_i$'s, $1\leq i \leq n$ are distinct, and even if they were, this does not contradict that Fundamental Theorem (rather the simple theorem that a polynomial cannot have more roots than its degree), since $\pm 2$ is not just one value; for example for the polynomial $p(x) = x^2-3$ we have $p(\pm 1) = -2$ and $p(\pm\sqrt{5}) = 2$. So even if your argumentation up to this point were correct, you still have no contradiction. When only knowing $\deg p \leq n-1$, you need $p(x) = \pm 2$ for at least $2n-1$ distinct arguments, in order to get a contradiction.
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neelanjan
382 posts
#14 • 2 Y
Y by Adventure10, Mango247
Suppose
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x + 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x + 1)$. We show that all the a's are divisible by 3 and using that we will establish a contradiction .[Here , we will show using only + ve sign of 3 , the other case is
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x - 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x - 1)$ , in which constant terms contains only - ve terms and the casee follows only the + ve case . So , we will show here only the first case] .
r and s must be greater than 1.Because
for if r = 1, then 3 is a root . Now we will make two cases :
case 1:
n is even, then it follows that 0 = $3^n +5\cdot$$3^{n- 1} + 3 = 3^{n-1}( 3 +5)+ 3$, which is false since 3+5 = 8.
case 2 :
if n is odd we would have 0 = $3^{n-1}(3 + 5) + 3$, which is false since 3 + 5 = 8.
If s = 1, then 1 is a root and we will argue by contradiction in the same way .
So r$\leq$n - 2, and hence the coefficients of x, $x^2, ... , x^r$ are all zero. Since the coefficient of x is zero, we have: $a_1+3b_1 = 0$, so $a_1$ is divisible by 3.
Now , we will use the induction hypothesis .
We can
now proceed by induction.
Assume $a_1, ... , a_t$ are all divisible by 3. Then
consider the coefficient of $x_{t+1}$. If s-1 $\geq $t+1, then $a_{t+1}$ = linear combination of $a_1,... , a_t + 3b_{t+1}$. If s-1 $\leq$ t+1, then $a_{t+1}$ = linear combination of some or all of $a_1,... , a_t.$
Either way, $a_{t+1}$ is divisible by 3.
Now we will consider the coefficients of x, $x^2$,
... , $x^{r-1}$ which gives us that all the a's are multiples of 3. Now consider the coefficientof $x^r$ which is also zero.
Now we will get , that is a sum of of terms which are multiples of 3 .
Then it does not becomes 0.
So , follows a contradiction !
So , the factorization is not possible .( proved ).
-NEELANJAN MONDAL.
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AlgebraFC
512 posts
#16 • 3 Y
Y by Hedy, Adventure10, Mango247
Solution
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ubermensch
820 posts
#17 • 2 Y
Y by Adventure10, Mango247
Extended Eisenstein $->$ RRT $->$ $\mod 2$ $->$ Done.
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ubermensch
820 posts
#18 • 2 Y
Y by Adventure10, Mango247
My point is, would your solution even count if you directly apply Eisenstein on this problem, because it completely trivialises the problem... (apparently any use of a theorem is prohibited if it trivialises the problem, and this theorem certainly does, right?)
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Idio-logy
206 posts
#19 • 1 Y
Y by Adventure10
It is a direct result of Perron's criterion. Alternatively, suppose $x^n+5x^{n-1}+3 = f(x)g(x)$. Reduce it modulo 3 to get $f(x)g(x) \equiv x^{n-1}(x+5)$, so we can assume that $f(x) = x^{p}(x+5) + 3F(x)$, $g(x) = x^{q} + 3G(x)$. Multiply them to get $x^qF(x)+x^p(x+5)G(x)+3F(x)G(x) = 1$. Plug in $x=0$, then $3F(0)G(0)=1$, clearly impossible.
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pi_is_3.141
145 posts
#20 • 1 Y
Y by Mango247
Click to reveal hidden text
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vvluo
1574 posts
#21
Y by
pi_is_3.141 wrote:
Click to reveal hidden text

oh lol someone is taking AMSP Alg 3...
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mathiscool12
80 posts
#22
Y by
Solution 1
Solution 2
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fukano_2
492 posts
#24
Y by
$                    $
This post has been edited 3 times. Last edited by fukano_2, Oct 27, 2021, 10:50 AM
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CALCMAN
249 posts
#25 • 1 Y
Y by Frestho
fukano_2 wrote:
one liner
-_- Remark

I think this is wrong? $\overline{f(x)}$ only needs to be irreducible $\pmod 5$, not completely irreducible (for example, if I took $f(x)=x^2+2x+1$ and had $x^2+2x+1\equiv x^2+1\pmod 2$, $x^2+1$ being irreducible doesn't imply $x^2+2x+1$ is irreducible, since $x^2+1$ could still be reducible $\pmod 2$).
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R-sk
429 posts
#26
Y by
This can be done by using perrons criterian
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jasperE3
11091 posts
#27
Y by
By pure coincidence, I had been looking at irreducibility criterion when I encountered this problem. Perron's kills it.

Since $5>1+3$, $f$ is irreducible over $\mathbb Z[x]$. $\square$
This post has been edited 1 time. Last edited by jasperE3, Apr 2, 2021, 4:00 AM
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OlympusHero
17017 posts
#28
Y by
What on earth?!

Solution
This post has been edited 1 time. Last edited by OlympusHero, Aug 10, 2021, 7:17 PM
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RedFlame2112
1444 posts
#29 • 1 Y
Y by centslordm
orl wrote:
Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$

1 liner by Perron's

By IG's inequality, $5$ is a greater number than $1+3=4$. Hence, we are done since Perron's finishes over $\mathbb{Z}\left[x\right]$ $\square$

Good thing i still have my irreducibility handouts :gleam:

it's also not too hard to prove perron's irreducibility criterion in olympiads, assume the assertion and suppose that $f(x)=g(x)h(x)$. This means that $f$ has only 1 root of modulus greater than 1, which means either $g$ or $h$ has roots not outside the unit circle. You'll reach a contradiction that $|g(0)| < 1$ yet $g(0)$ is, by definition, a nonnegative integer.
This post has been edited 3 times. Last edited by RedFlame2112, Aug 10, 2021, 8:36 PM
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bryanguo
1031 posts
#30 • 2 Y
Y by channing421, centslordm
Everyone says "Perron's" but there's no explanation of what Perron's Criterion is, so I'll put the general statement here.
Quote:
Let $f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ be a polynomial with integer coefficients such that $a_0\neq 0$ and \[|a_{n-1}|>1+|a_{n-2}+\cdots+|a_1|+|a_0|.\]Then $f(X)$ is irreducible in $\mathbb{Z}[X].$
We come back to the original problem.
Quote:
Prove that the polynomial $X^n + 5X^{n-1} + 3$ is irreducible in $\mathbb{Z}[X].$

Direct application of Perron's gives $5>1+3,$ so the polynomial is irreducible as required. Below is an alternate solution through Eisenstein's Criterion, of which here is the statement:
Quote:
Let $f(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ be a polynomial with integer coefficients. Then $f(X)$ is irreducible in $\mathbb{Z}[X]$ if there exists a prime $p$ such that
(i) $p \mid a_0, p \mid a_1, \cdots, p \mid a_{n-1},$ and
(i) $p \nmid a_n, p^2 \nmid a_0.$

It is easy to verify by the rational root theorem that no linear factors divide $X^n+5X^{n-1}+3.$ Assume on the contrary that $P(X)=X^n+5X^{n-1}+3=f(X)g(X).$ Taking mod three, we get \[f(X)g(X)\equiv X^{n-1}(X-1)\pmod 3.\]Let $f(X)\equiv X^a\pmod 3,g(X)\pmod X^b(x-1)\pmod 3.$ Now, $a,b\neq 0$ since $f(X),g(X)$ both have degree at least two. Thus, we have $3\mid f(0)$ and $3\mid g(0)$ so $9\mid P(0)$ which is clearly false because $P(0)=3.$
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djmathman
7934 posts
#31
Y by
Wow, I have finally done an irreducibility problem. Let's not use Perron's Criterion.

Write $x^n + 5x^{n-1} + 3 = g(x) h(x)$ for some polynomials $g\in \mathbb Z[x]$ and $h\in \mathbb Z[x]$. Plugging in $x = 0$ yields $g(0)h(0) = 3$, so without loss of generality $3\nmid h(0)$. However, taking $f$ modulo $3$ yields
\[
x^n - x^{n-1} \equiv \bar g(x) \bar h(x)\pmod 3.
\]Thus, $x$ cannot divide $\bar h(x)$.

There are two cases to consider. In the first case, $g(x) = x^{n-1} + 3p(x)$ and $h(x) = x - 1 + 3q(x)$ for some polynomials $p\in \mathbb Z[x]$ and $q\in\mathbb Z[x]$. Since $\deg g \geq n - 1$ and $\deg h \geq 1$, these inequalities are actually equalities. That is, $h$ is linear and $f$ has a rational root. But the only rational roots $f$ can possibly have are $-1$ and $-3$ (RRT combined with the fact that $f$ has no positive roots), and both of these fail.

In the second case, $g(x) = x^n - x^{n-1} + 3p(x)$ and $h(x) = 1 + 3q(x)$ for some polynomials $p\in \mathbb Z[x]$ and $q\in\mathbb Z[x]$. Comparing degrees in the same way implies that $h$ is a constant. In turn, $f$ is irreducible.
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ZETA_in_olympiad
2211 posts
#32
Y by
orl wrote:
Let $n > 1$ be an integer and let $f(x) = x^n + 5 \cdot x^{n-1} + 3.$ Prove that there do not exist polynomials $g(x),h(x),$ each having integer coefficients and degree at least one, such that $f(x) = g(x) \cdot h(x).$

Also PEN Q7
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lifeismathematics
1188 posts
#33 • 1 Y
Y by Rounak_iitr
For storage:

FTSOC assume the contrary and set $f(x)=h(x)\cdot g(x)$ where $h(x),g(x) \in \mathbb{Z}[x]$. plugging in $x=0$ yields $h(0)g(0)=3$ and plugging $x=-5$ we get $h(-5)g(-5)=3$.

now by RRT, we can clearly observe that there is no integer root of $f(x)$ hence $h(x)$ and $g(x)$ are not linear.

set $h(x)=(x-z_{1})(x-z_{2})\cdots (x-z_{r})$ where $\text{deg(h(x))}=r$ and $z_{i}$'s are root of $h(x)$ for $1\leqslant i \leqslant r$

now WLOG take $h(0)=\pm 1$ ( since $h(x) \in \mathbb{Z}[x]$)

we get $\left|\prod_{i=1}^{r} z_{i}\right|=1$

and $\left|\prod_{i=1}^{r}(z_{i}+5)\right|=|(z_{1}+5)(z_{2}+5)\cdots (z_{r}+5)|$

now notice for $z_{l} \in \{z_{1},z_{2},\cdots z_{r}\}$ we get $z_{l}^{n-1}(z_{l}+5)=-3$ hence we get:

$\left|\prod_{i=1}^{r}(z_{i}+5)\right|=\left|\prod_{i=1}^{r}\frac{-3}{z_{i}^{n-1}}\right|=3^{r}$ which gives that $|h(-5)|=3^{r}$

but also since $|h(-5)||3$ ( as $h(-5)g(-5)=3$ and $h(x)\in \mathbb{Z}[x]$) we have that $3^{r}|3$ for $r>1$ which gives a contradiction , and hence the result follows $\blacksquare$
This post has been edited 6 times. Last edited by lifeismathematics, Mar 16, 2023, 8:58 AM
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Haris1
63 posts
#34
Y by
Using Einstein's Criterion we take p=3, in this case 3 divides a0 , 3 dosent divide an and 3^2 dosent divide 3 therefore by Einstein's criterion it implies that f(x) is irreducible.
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djmathman
7934 posts
#35
Y by
@above: that doesn't work. Eisenstein's Criterion requires $p = 3$ to divide every coefficient except the leading one. Because $3$ does not divide $5$, you can't naively use Eisenstein here.
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Haris1
63 posts
#36
Y by
Yes my mistake
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almagest3001
6 posts
#37
Y by
It is enough to show that $f$ is irreducible in $\mathbb{Q}$ since it is primitive. Observe that $p(x) = 5x^{n-1} + 3$ is irreducible in $\mathbb{Q}$ by Eisenstein. Then $x^n+p \in \mathbb{Q}[x][x]$, where $p$ is taken as the constant coefficient, is irreducible in $\mathbb{Q}[x]$ by the general Eisenstein criterion, hence $f$ is irreducible in $\mathbb{Q}$.
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djmathman
7934 posts
#38
Y by
That doesn't work. Unless I'm misunderstanding your argument, by that logic, if $p(x)$ is any irreducible polynomial of degree less than $k$, then $x^k + p(x)$ is also irreducible. This is clearly false, take $p(x) = 3x^2 + 3x + 1$ and $k = 3$.

(Also, what is $\mathbb Q[x][x]$? If you mean "the ring of polynomials in $x$ whose coefficients are polynomials in $x$", then this really is the same thing as $\mathbb Q[x]$. In particular, given a ring $K$, the polynomial ring $K[X]$ can be thought of taking the ring $K$ and adding a new variable $X$ which commutes with elements of $K$ but otherwise has no additional properties. This means something like $\mathbb Q[x][x]$ is inherently problematic, because the first $x$ and the second $x$ interact with each other nontrivially.)
This post has been edited 3 times. Last edited by djmathman, Jul 28, 2024, 5:52 AM
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almagest3001
6 posts
#39
Y by
You're right. I misremembered the generalized Eisenstein criterion.
This post has been edited 2 times. Last edited by almagest3001, Jul 28, 2024, 6:53 PM
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