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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
H not needed
dchenmathcounts   44
N 6 minutes ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
+1 w
dchenmathcounts
May 23, 2020
Ilikeminecraft
6 minutes ago
IZHO 2017 Functional equations
user01   51
N 27 minutes ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
27 minutes ago
chat gpt
fuv870   2
N 29 minutes ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
41 minutes ago
fuv870
29 minutes ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N 31 minutes ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
31 minutes ago
Segment has Length Equal to Circumradius
djmathman   72
N an hour ago by Zhaom
Source: 2014 USAMO Problem 5
Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.
72 replies
djmathman
Apr 30, 2014
Zhaom
an hour ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   22
N an hour ago by RainbowSquirrel53B
Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
[list]
[*] Medals for the top teams
[*] Shirts, pins, stickers and certificates for all participants
[*] Additional awards provided by our wonderful sponsors!
[/list]

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
22 replies
MustangMathTournament
Mar 8, 2025
RainbowSquirrel53B
an hour ago
2025 ROSS Program
scls140511   11
N 2 hours ago by fuzimiao2013
Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
11 replies
scls140511
Today at 2:36 AM
fuzimiao2013
2 hours ago
d_k-eja Vu
ihatemath123   46
N 3 hours ago by Ilikeminecraft
Source: 2024 USAMO Problem 1
Find all integers $n \geq 3$ such that the following property holds: if we list the divisors of $n!$ in increasing order as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
Proposed by Luke Robitaille.
46 replies
ihatemath123
Mar 20, 2024
Ilikeminecraft
3 hours ago
average FE
KevinYang2.71   75
N 4 hours ago by Marcus_Zhang
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
75 replies
KevinYang2.71
Mar 21, 2024
Marcus_Zhang
4 hours ago
apparently circles have two intersections :'(
itised   76
N 4 hours ago by Ilikeminecraft
Source: 2020 USOJMO Problem 2
Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\ell$ be a variable line that is tangent to $\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\ell$.

Proposed by Titu Andreescu and Waldemar Pompe
76 replies
itised
Jun 21, 2020
Ilikeminecraft
4 hours ago
Too Bad I'm Lactose Intolerant
hwl0304   216
N 5 hours ago by AshAuktober
Source: 2018 USAMO Problem 1/USAJMO Problem 2
Let \(a,b,c\) be positive real numbers such that \(a+b+c=4\sqrt[3]{abc}\). Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]
216 replies
hwl0304
Apr 18, 2018
AshAuktober
5 hours ago
Easy Combinatorics
JetFire008   1
N 5 hours ago by Marcus_Zhang
Source: AMC 12 2001
How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
1 reply
JetFire008
6 hours ago
Marcus_Zhang
5 hours ago
Did this get posted yet
pog   25
N Today at 2:03 PM by santhoshn
Source: 2024 AMC 8 #1
What is the ones digit of \[222{,}222-22{,}222-2{,}222-222-22-2?\]
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }6\qquad\textbf{(E) }8$
25 replies
pog
Oct 11, 2024
santhoshn
Today at 2:03 PM
Stanford Math Tournament (SMT) Online 2025
stanford-math-tournament   6
N Today at 3:49 AM by Vkmsd
[center]Register for Stanford Math Tournament (SMT) Online 2025[/center]


[center] :surf: Stanford Math Tournament (SMT) Online is happening on April 13, 2025! :surf:[/center]

[center]IMAGE[/center]

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online

When? The contest will take place April 13, 2025. The pre-contest puzzle hunt will take place on April 12, 2025 (optional, but highly encouraged!).

What? The competition features a Power, Team, Guts, General, and Subject (choose two of Algebra, Calculus, Discrete, Geometry) rounds.

Who? You!!!!! Students in high school or below, from anywhere in the world. Register in a team of 6-8 or as an individual.

Where? Online - compete from anywhere!

Check out our Instagram: https://www.instagram.com/stanfordmathtournament/

Register and learn more here:
https://www.stanfordmathtournament.com/competitions/smt-2025-online


[center]IMAGE[/center]


[center] :surf: :surf: :surf: :surf: :surf: [/center]
6 replies
stanford-math-tournament
Mar 9, 2025
Vkmsd
Today at 3:49 AM
Another NT FE
nukelauncher   59
N 4 hours ago by AshAuktober
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 $a+b>2019$.
59 replies
nukelauncher
Sep 22, 2020
AshAuktober
4 hours ago
Another NT FE
G H J
Source: ISL 2019 N4
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nukelauncher
354 posts
#1 • 8 Y
Y by TheLiker, itslumi, Kanep, centslordm, megarnie, ImSh95, TheHU-1729, deplasmanyollari
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 $a+b>2019$.
Z K Y
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nukelauncher
354 posts
#2 • 3 Y
Y by Mathematicsislovely, centslordm, ImSh95
Solution
Z K Y
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anyone__42
92 posts
#3 • 11 Y
Y by idkwhatthisis, Kanep, Aryan-23, centslordm, TheMath_boy, uvuvwevwevwe, megarnie, sabkx, Vladimir_Djurica, ImSh95, TensorGuy666
can be done easily using dirichlet's theorem
Z K Y
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dangerousliri
924 posts
#4 • 4 Y
Y by A-Thought-Of-God, centslordm, sabkx, ImSh95
Here it is a solution that I did find when I did get the shortlist. Somehow for me it took me faster to solve this then N1 which was on the test. Also it should be $a+b>C$ and not $a+b>2019.$

Since $gcd(f(1),1)=1$ from Dirichlet's theorem we have there exist infinitely prime numbers of the form $1+nf(1)$. Taking all such $n>C$ then for $a=1$ and $b=n$ we have,
$$1+f(n)|1+nf(1)\Rightarrow f(n)=nf(1).$$
Now taking $a$ fixed and $b=n>C$ large enough we have,
$$a+nf(1)|a+f(n)|a^2+nf(a)|f(1)a^2+f(1)nf(a)\Rightarrow a+nf(1)|f(1)a^2+f(1)nf(a)-(a+nf(1))f(a)$$$$\Rightarrow a+nf(1)|a(f(1)a-f(a))\Rightarrow f(a)=f(1)a=ka.$$
Hence $f(a)=ka$ for all positive integers $a$, and clearly for any positive integer $k$ hold.
This post has been edited 1 time. Last edited by dangerousliri, Sep 23, 2020, 12:10 AM
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blacksheep2003
1081 posts
#5 • 3 Y
Y by centslordm, sabkx, ImSh95
Solution

Oh wow, I thought my solution was pretty unique to use Dirichlet's Theorem, but it seems other people had the same idea lol oops :P
This post has been edited 4 times. Last edited by blacksheep2003, Sep 23, 2020, 12:49 AM
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MarkBcc168
1593 posts
#6 • 12 Y
Y by mijail, Imayormaynotknowcalculus, k12byda5h, A-Thought-Of-God, Aryan-23, Kanep, centslordm, tigerzhang, GeoKing, ImSh95, Stuffybear, iamnotgentle
Comment: Extremely standard NT functional divisibility. This trick used to solve this problem have been appeared many many times.

As in every functional divisibility problems, the main point is to prove that $f$ is linear on some infinite set. We prove the following claim.

Claim: There exists infinite sequence $C<x_1<x_2<\hdots$ such that $f(x_i)=kx_i$ for some constant $k$.

Proof: The idea is to force the right hand side to be prime. By Dirichlet, there exists infinitely many $x>C$ which $1+xf(1)=p$ is prime. However,
$$1+f(x)\mid xf(1)+1\implies f(x)+1\in\{1,p\}\implies f(x)=xf(1)$$for infinitely many $x$. Hence we are done. $\blacksquare$

Once we have this we are almost done. Fix $n\in\mathbb{Z}^+$. Since
\begin{align*}
&n+kx_i\mid n^2+x_if(n) \\ 
\implies &n+kx_i\mid k(n^2+x_if(n)) - f(n)(n+kx_i) \\
\implies &n+kx_i\mid kn^2-nf(n)
\end{align*}and $x_i$ is arbitrarily large. We get $f(n)=kn$ for any $n$.
This post has been edited 1 time. Last edited by MarkBcc168, Sep 23, 2020, 1:43 AM
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FAA2533
65 posts
#7 • 3 Y
Y by Aryan-23, centslordm, ImSh95
nukelauncher wrote:
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 $a+b>2019$.
It somehow seemed very easy to me than N1.

Solution
This post has been edited 6 times. Last edited by FAA2533, Jul 18, 2022, 2:35 PM
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ubermensch
820 posts
#9 • 4 Y
Y by ATGY, Illuzion, centslordm, ImSh95
Let $P(a,b)$ denote the assertion that $a+f(b) \mid a^2+bf(a)$.

Naturally, a $P(1,b)$ substitution gives us $1+f(b) \mid 1+bf(1) \implies bf(1) \geq f(b)$. We'll now denote $f(1)$ by $k$, as we shall be using it quite extensively.

$P(a,1): a+k \mid a^2+f(a) \implies a+k \mid a^2+f(a)-a(a+k) \implies a+k \mid f(a)-ak \implies a+k \mid f(a)-ak +k(a+k)$.

Hence we get $a+k \mid f(a)+k^2$. Coupled with our first bound, this gives us that $f(a)=at+tk-k^2$ for large enough $a$.

Here, although it is quite instinctual to substitute $a$ as a prime, we can in fact avoid that altogether and deal just with our regular naturals- although it will get quite ugly, but at the very least it will remain very mechanical.

Our aim, as it always is in such problems, is to manipulate the divisibility so as to be able to make the left hand side arbitrarily large without altering the expression on the right, hence forcing the right hand side to be become zero. A simple $(a,b)$ substitution in the original expression with a little modification would give us what we're looking for.

Firstly-
$$a+f(b) \mid a^2+bf(a) \implies a+f(b) \mid a^2+bf(a)-a(a+f(b))$$$$\implies a+f(b) \mid bf(a)-af(b) +f(b)(a+f(b)) \implies a+f(b) \mid bf(a)+f(b)^2$$
Thus we have $a+f(b) \mid f(b)^2 + bf(a)$. We now substitute $f(a)=at+(k^2-kt)$ to get-

$$a+f(b) \mid b(at+k^2-kt)+f(b)^2$$$$\implies a+f(b) \mid abt+bk^2-bkt+f(b)^2-bt(a+f(b)) \implies a+f(b) \mid b(k^2-kt)+f(b)^2-btf(b)$$Observe now that we can make our $a$ arbitrarily large for a constant $b$, and noting that $k=f(1)$ has a fixed value along with $t \leq k$ forces $f(b)^2-btf(b)+b(k^2-kt)=0$, as planned.

Writing $f(b)=rb+rk-k^2$ gives us $(rb+rk-k^2)^2-bt(rb+rk-k^2)+b(k^2-kt)=0$. For large $b$, since our $k$ and $t$ are small (compared to our now large $b$), we can simply equate the coefficients of $b^2$ and $b$ and the constant term zero individually. Hence, $r^2-rt=0 \implies r=t$ and equating the constant term to zero gives us $r=k$.

Thus, for arbitrarily large $n$, we have $f(n)=kn$. The finish is now quite simple-
$$P(n,b): n+f(b) \mid n^2+knb-n(n+f(b))-(kb-f(b))(n+f(b))$$$$\implies n+f(b) \mid (kb-f(b))f(b)$$for large $n$, hence forcing $f(b)=kb$ for all $b$.

Clearly $f(n)=kn$ satisfies the original condition, and hence we're done.

Quite a standard one, and rather mechanical, especially if you've done something like this before (POP). Nevertheless, quite an instructive exercise on functional divisibilities...
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dchenmathcounts
2443 posts
#10 • 3 Y
Y by centslordm, Marie36, ImSh95
guess this has been said before, but here goes

We claim the answer is $f(x)=kx$ for $k\in \mathbb{Z}_{>0}.$

Fix $a=1.$ Note that $1+bf(1)$ is a prime infinitely often by Dirichlet (is this how you spell it?). So let $b$ be some absurdly large number such that $a+b>2019$ and $1+bf(1)$ is prime; then $1+f(b)\mid 1+bf(1).$ Since $1+bf(1)$ is prime we either have $1+f(b)=1$ or $1+f(b)=1+bf(1).$ Clearly the latter holds, so there are infinitely many $b$ such that $f(b)=bf(1).$

Now take some mega ultra large $b$ such that $f(b)=bf(1),$ and note that $a+f(b)=a+bf(1)\mid a^2+bf(a)$ and $a+bf(1)\mid a^2+abf(1).$ Subtracting the latter from the former gives $a+bf(1)\mid b(f(a)-af(1)).$ Note that $\gcd(a+bf(1),b)=\gcd(a,b),$ so if $a$ is prime this implies that $a+bf(1)\mid f(a)-af(1).$ Since $b$ is, as I quote, "ultra mega large," $|a+bf(1)|>|f(a)-af(1)|,$ so for all prime $a$ we have $f(a)=af(1).$

Now take some ultra mega large prime $b,$ which we have just shown exists, and note that
\[a+bf(1)\mid b(f(a)-af(1)), \text{ which implies that}\]\[a+bf(1)\mid f(a)-af(1),\]and since $b$ is ultra mega large, we have shown that $f(a)=af(1)$ in general.
This post has been edited 1 time. Last edited by dchenmathcounts, Sep 23, 2020, 5:14 PM
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ayan_mathematics_king
1527 posts
#11 • 6 Y
Y by DapperPeppermint, Mathematicsislovely, amar_04, A-Thought-Of-God, centslordm, ImSh95
Solution by my friend RustyFox who is banned now.So I am posting his solution.
nukelauncher wrote:
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 $a+b>C$.

Solution : Let $P(x, y)$ be the assertion. $P(1, k) \implies kf(1)\geq f(k)$.

This is going to be useful. Let $p \mid x$ where $p$ is a prime.

Let $k = Cf(x)$

$P(kp-f(x), x \implies kp \mid (kp-f(x))^2 + f(kp-f(x))x$ which means that $p \mid f(x)$. So, $p \mid f(p)$ for all primes $p$.

Let us say that $f(p)= g(p)p$ for primes $p$. Observe that $0 < g(p) \leq f(1)$. If $g(p)\neq g(q)$ for two distinct primes $p, q$ which are much much greater than $C$ (say it's much greater than $(C+2019f(1))!\uparrow \uparrow (C+2019f(1))!)$, then $P(p, q) \implies p + qg(q) \mid p^2 + pqg(p)$ or in other words $p + qg(q) \mid pq(g(p)-g(q))$. Now since $gcd(p, q) = gcd(p, g(q)) = gcd(p, qg(q)) = 1$, we havs $p + qg(q) \nmid pq$. Now, $p + qg(q) \mid g(p)-g(q)$ but the dividend is less than $f(1)$ and due to our value of $p, q$, $p + qg(q) >> f(1)$ which means that for all sufficiently large primes we have $g(p) = g(q) = c$. Now for any prime $p$ such that $f(p) = pc$ and any other prime $q$ for which we aren't sure if $f(q)$ equals $qc$ or not and obviously here $q < p$,we take $P(q, p)$ to get that $q + pc \mid p(f(q)-qc)$ and since $pc  + q > pc \geq p$, we havs $q + pc \mid f(q)-qc$ and so see that $f(q) - qc \leq qf(1)-qc = q(f(1)-c)$ and so choosing $p > q(f(1)-c)$ would suffice and so for all primes $p$, we have $f(p)=pc$. Now, let $p >>a$. Then $P(a, p) \implies a + pc \mid pf(a)-pca$ but $gcd(p, a)=1$ and so $a+pc \mid f(a)-ac$. Choosing large enough $p$ yields that $\boxed{f(a)=ac}$ for all naturals $a$
This post has been edited 3 times. Last edited by ayan_mathematics_king, Sep 24, 2020, 7:51 AM
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ProblemSolver2048
104 posts
#15 • 2 Y
Y by centslordm, ImSh95
Since gcd(f(1,),1)=1 from Dirichlet's theorem we have there infinitely prime numbers of the form 1+nf(1). Taking all such n>C then for all a=1 and b=n we have $$1+f(n)|1+nf(1)\Rightarrow f(n)=nf(1).$$Now taking a fixed and b=n> C large enough we have,
$$a+nf(1)|a+f(n)|a^2+nf(a)|f(1)a^2+f(1)nf(a)\Rightarrow a+nf(1)|f(1)a^2+f(1)nf(a)-(a+nf(1))f(a)$$$$\Rightarrow a+nf(1)|a(f(1)a-f(a))\Rightarrow f(a)=f(1)a=ka.$$hence f(a)=ka for all positibe integers , clearly for any positive integer k hold.
Acknowledgements to dangerousliri's in post 4. I didn't make it.
This post has been edited 1 time. Last edited by ProblemSolver2048, Dec 14, 2020, 7:39 PM
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Mathematicsislovely
245 posts
#16 • 5 Y
Y by amar_04, centslordm, ImSh95, sabkx, ylmath123
Let $P(a,b)$ denote the assertion of the question.

Claim:For all prime $p$ we have $p|f(p)$.
proof.Fix a prime $p$.$k$ be an integer such that $(k+1)p-f(p)>2020^{2020}$.Then $P(kp-f(p),p)\implies p|kp-f(p)+f(p)|(kp-f(p))^2+pf(stuff)\implies p|f(p)$. $\blacksquare$.

Then $f(p_n)=c_np_n$ where $p_n$ is the $n^{th}$ prime.But as $P(1,b)\implies f(b)\le f(1)b$ $\forall 
 b\ge 2020$. So $c_n$ is bounded sequence.Hence,there exists an integer $r$ such that it is equal to infinitely many $c_i$.So there exists an increasing sequence $(q_n)_{n=1}^{\infty}$ of prime integers such that $f(q_s)=rq_s$ $\forall s\in \mathbb N$.

Now, for a fix $a$ take a big $q_i$ such that $gcd(a,q_i)=1$.Then,
$P(a,q_i)\implies\\
 a+rq_i|a^2+q_if(a)\\
\implies a+rq_i|a^2+q_if(a)-a^2-arq_i\\
\implies a+rq_i|q_i(f(a)-ar)\\
\implies a+rq_i|(f(a)-ar)$.
Now for sufficiently large $q_i$ we get $f(a)=ar$.Hence,$\boxed{f(x)=rx}$ is the only solution which indeed satisfies the functional equation. $\blacksquare$.
This post has been edited 4 times. Last edited by Mathematicsislovely, Sep 26, 2020, 11:00 PM
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Ilove_mathematics
188 posts
#17 • 3 Y
Y by centslordm, odkaI, ImSh95
Let $P(a,b)$ denote the divisibility $(a+f(b)) \mid (a^2+bf(a)).  ($For simplicity say $C = 2019)$

Let $n$ denote $f(1).$ We are going to show that $f(k) = kn,$ for all $k \in \mathbb{Z}_{>0}.$ Let $\mathbb{A} = \{C+1, C+2, ...\}.$

Lemma 1: Define $g:\mathbb{A} \rightarrow \mathbb{Q}$ as $g(a) = \frac{an-f(a)}{a+n},$ then $g$ is limited.

Proof: Let $a$ be a positive integer greater than $C.$ Then $P(a,1)$ gives $a+n \mid a^2 + f(a) \implies (a+n).q = a^2 + f(a)$ with $q \in \mathbb{N}.$ On the other hand $P(1, a) \implies 1+f(a) \mid 1+an \implies f(a) \leq an,$ therefore $(a+n).a =$ $a^2 + an \geq a^2 + f(a), $ suppose for the sake of contradiction that $f(a) < an \implies q < a \implies q = a-k$ so $(a+n)(a-k) = a^2 + f(a) > a^2,$ since $(a+n)(a-n)  < a^2$ we have $k < n \implies k$ is bounded as $a$ varies along the natural numbers greater than $C,$ yielding $k = \frac{an-f(a)}{a+n} = g(a)$ is limited.$\blacksquare$

Lemma 2: There are infinitely many positive integers $s$ such that $f(s) = sn.$

Proof: Let $S = \{a_1, a_2, ...\}$ be the set of positive integers such that $\frac{a_in-f(a_i)}{a_i+n} = t$ a constant, with $C < a_1 < a_2 < ...,$ there exists such $t,$ because the set $\mathbb{A}$ is unbounded. So $\frac{a_in-f(a_i)}{a_i+n} = t \implies$ $f(a_i) = a_in-t(a_i+n),$ for all $i.$
Now fix $j$ such that $a_j(n-t+1) \ne nt.$ Plugging $a = a_i, b = a_j$ gives: $$(a_i + f(a_j)) \mid (a_i^2 + a_jf(a_i)) \implies$$$$(a_i + a_jn - t(a_j+n)) \mid [(a_i^2 + a_ja_in - a_jt(a_i+n)) - a_i(a_i+a_jn-t(a_j+n))] \implies$$$$(a_i+a_jn-t(a_j+n)) \mid (a_i-a_j)nt.$$Let $d_i$ be $gcd(a_i + a_jn-t(a_j+n), a_i - a_j),$ (of course $i \ne j$) if $d_i > max(a_j(n-t+1), nt) \implies$ $a_i + a_jn-t(a_j+n) \equiv a_j + a_j(n-t) - nt \pmod {d_i} \implies d_i \mid a_j(n-t+1)-nt,$ as supposed we must have $a_j(n-t+1) = nt,$ contradiction. Therefore $d_i$ is bounded. Observe $(a_i + a_jn-t(a_j+n))|d_int,$ $\forall i \in \mathbb{N}, i \ne j,$ setting $i \rightarrow \infty \implies a_i \rightarrow \infty$ thus, we eventually have: $a_i + a_jn -t(a_j+n) > d_int \implies d_int = 0 \implies t = 0 \implies f(a_i) = a_in,$ for all $i,$ sufficiently large. $\blacksquare$


Fix $k$ a positive integer.
Plug $a = k, b = a_i \implies k+a_in \mid k^2 + a_if(k) \implies (k+a_in) \mid (k^2+a_if(k) - k(k+a_in)),$ yielding $(k+a_in) \mid a_i(f(k)-kn)$ so $(k+a_in).q_i = a_i.|f(k)-kn|, q_i \in \mathbb{Z}_{\geq 0} \implies q_i = \frac{a_i.|f(k)-kn|}{k+a_in}$ call $t = |f(k)-kn|$ a constant. If $q_i > t \implies (k+a_in).t < a_i.t$ contradiction. We conclude $q_i$ is bounded. On the other hand if $i \ne j$ and $q_i = q_j \implies \frac{a_it}{k+a_in} = \frac{a_jt}{k+a_jn}$ $\implies a_itk + a_ia_jnt = a_jtk + a_ja_int \implies (a_i-a_j)kt = 0$ then $t$ would be $0.$ Since $q_i, q_j$ are bounded, both positive and we can choose $i, j \in \mathbb{N}$ such that $q_i = q_j, f(k)-kn = 0$ for all $k \in \mathbb{N},$ as $(a+kb).a = a^2 + akb \implies f(a) = an$ for all positive integers $n.$
This post has been edited 1 time. Last edited by Ilove_mathematics, Sep 27, 2020, 4:22 AM
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solver1104
510 posts
#18 • 2 Y
Y by centslordm, ImSh95
Remark: Standard problem, seems pretty easy in difficulty unless I messed up. Also, I cite the so called Weak - Dirichlet theorem many times over this proof. Although no proof of Dirichlet Theorem that I know of is elementary, Weak Dirichlet can easily be proven by cyclotomic polynomials - see Evan Chen's Orders handout for more details.

random stuff

We claim that the answer is all functions $f(x) \equiv cx$ for any positive integer $c$. It's easy to see that all such functions work, because $a + bc | a^2 + abc = a(a+bc)$. Now we show that these are the only functions that work.

Consider the more general problem, replacing $2019$ with any constant $C$. Denote the given assertion$P(a,b)$.

Now we claim that $f(b) = bf(1)$ holds for infinitely many $b$. Indeed, $P(1,b)$ gives $1 + f(b) | 1 + bf(1)$, and now by Weak Dirichlet, there exists infinitely many $b$ such that $bf(1) + 1$ is prime, and since $1 + f(b) > 1$, we have $f(b) = bf(1)$, as desired.

Now we show that this actually implies $f(a) = af(1)$ for all $a$. Indeed, by Weak Dirichlet again, there exists infinitely many primes that are $1 \pmod{f(a) \cdot f(1)}$ for any $a$. Now take a prime of this form, $p$, and let $p$ be arbitrarily large. Note that $p$ is relatively prime to $f(a)$, and it is also of the form $cf(1) + 1$ for some $c$, hence by the first claim, $f(p) = pf(1)$. Then $P(p,a)$ gives $p + f(a) | p^2 + apf(1)$. Now $\gcd(p, p+f(a)) = \gcd(p,f(a)) = 1$ by the assumption that $p$ is $1 \pmod{f(a)}$, so the extra factor of $p$ on the $RHS$ can be taken out. Now we have $p + f(a) | p + af(1)$ for all $a$, and since $p$ is arbitrarily large, so that $p >>> f(a), af(1)$, so the two sides are actually equal, and we are done.
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rcorreaa
238 posts
#19 • 2 Y
Y by centslordm, ImSh95
Pretty standard NT FE problem... :maybe:

From the problem, $1+f(n)|1+nf(1)$ for all $n \in \mathbb{N}$. From Dirichlet's Theorem, since $gcd(1,f(1))=1$, there are infinetly prime numbers in the AP $\{1+nf(1)\}_{n \geq 1}$.

Thus, since $1+f(n)>1, 1+f(n)=1+nf(1)$ when $1+nf(1)$ is prime $\implies f(n)=nf(1)$ in this case.

Hence, there exists a sequence $\{a_n\}_{n=1}^{\infty}$ such that $f(a_n)=a_nf(1)$ for all $n \in \mathbb{N}$. Then, from the problem,
$$a+a_nf(1)|a^2+a_nf(a) \implies a+a_nf(1)|a^2f(1)+a_nf(a)f(1),af(a)+f(a)f(1)a_n$$$$\implies a+a_nf(1)|a^2f(1)-af(a)$$for all $n \in \mathbb{N}$

Taking $n$ large enough, $a^2f(1)-af(a)$ must be $0$.

Therefore, $f(a)=af(1)$ for all $a \in \mathbb{N}$, which clearly works.

$\blacksquare$
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