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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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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.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

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MATHCOUNTS/AMC 8 Basics
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Visit the pages linked for full schedule details for each of these programs!


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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
AIME score for college apps
Happyllamaalways   56
N 5 hours ago 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
+1 w
Happyllamaalways
Mar 13, 2025
Countmath1
5 hours ago
MIT Beaverworks Summer Institute
PowerOfPi_09   0
5 hours ago
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
5 hours ago
0 replies
Convolution of order f(n)
trumpeter   70
N 5 hours ago by HamstPan38825
Source: 2019 USAMO Problem 1
Let $\mathbb{N}$ be the set of positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the equation \[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]for all positive integers $n$. Given this information, determine all possible values of $f(1000)$.

Proposed by Evan Chen
70 replies
trumpeter
Apr 17, 2019
HamstPan38825
5 hours ago
k HOT TAKE: MIT SHOULD NOT RELEASE THEIR DECISIONS ON PI DAY
alcumusftwgrind   8
N Today 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
1 viewing
alcumusftwgrind
Today at 2:11 AM
maxamc
Today at 10:13 AM
rows are DERANGED and a SOCOURGE to usajmo .
GrantStar   26
N Today 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
Today at 6:00 AM
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Today 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
Today at 5:38 AM
average FE
KevinYang2.71   74
N Today 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
Today at 4:55 AM
Rip Red/Blue, Long live Amber/Bronze
AwesomeYRY   50
N Today at 3:35 AM by MathLuis
Source: USAMO 2022/1, JMO 2022/2
Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.
50 replies
AwesomeYRY
Mar 24, 2022
MathLuis
Today at 3:35 AM
Erecting Rectangles
franchester   101
N Today at 3:12 AM by Ilikeminecraft
Source: 2021 USAMO Problem 1/2021 USAJMO Problem 2
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that \[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.
101 replies
franchester
Apr 15, 2021
Ilikeminecraft
Today at 3:12 AM
2025 AMC 8 Problem
Kexinshi   8
N Today at 3:00 AM by CJB19
Source: 2025 AMC 8 Problem #15
Kei draws a $6$-by-$6$ grid. He colors $13$ of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let $m$ and $M$ equal equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Had fun doing this one!
8 replies
Kexinshi
Jan 31, 2025
CJB19
Today at 3:00 AM
How to get better at AMC 10
Dream9   2
N Today at 2:01 AM by hashbrown2009
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).
2 replies
Dream9
Today at 1:17 AM
hashbrown2009
Today at 2:01 AM
have you done DCX-Russian?
GoodMorning   80
N Today at 1:23 AM by bjump
Source: 2023 USAJMO Problem 3
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.

Proposed by Holden Mui
80 replies
GoodMorning
Mar 23, 2023
bjump
Today at 1:23 AM
Stanford Math Tournament (SMT) Online 2025
stanford-math-tournament   5
N Today at 12:28 AM by stanford-math-tournament
[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]
5 replies
stanford-math-tournament
Mar 9, 2025
stanford-math-tournament
Today at 12:28 AM
AIME Math History
hashbrown2009   82
N Yesterday at 11:35 PM by stjwyl
Idk why but I wanted to see how good ppl are
Post all your AIME scores ever (if you qualified for USA(J)MO, you may put that score, too)

(Note: Please do not post fake scores. I legit want to see how good ppl are and see how good I am)
I'll start:

5th grade: AIME : 2 lol
6th grade: AIME : 5
7th grade: AIME : 8
8th grade : AIME : 13 USAJMO: 18
9th grade (rn): AIME: 11 (sold)
82 replies
hashbrown2009
Feb 20, 2025
stjwyl
Yesterday at 11:35 PM
average FE
G H J
Source: USAJMO 2024/5
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KevinYang2.71
391 posts
#65 • 1 Y
Y by deduck
We claim the only such functions are $\boxed{f(x)\equiv -x^2}$, $\boxed{f(x)\equiv 0}$, and $\boxed{f(x)\equiv x^2}$. It is easy to check that these work.

Let $P(x,y)$ be the given assertion. $P(0,0)$ gives us $f(f(0))=0$. We have
\begin{align*}
P(1,0)&\implies f(1)=f(f(1))+f(0)\\
P(1,1)&\implies f(0)+2f(1)=f(f(1))+f(1)
\end{align*}so $f(0)=0$. Now
\begin{align*}
P(x,0)&\implies f(x^2)=f(f(x))\\
P(x,x^2)&\implies 2x^2f(x)=f(f(x))+f(x^2)
\end{align*}so $f(f(x))=f(x^2)=x^2f(x)$. It follows that $f(x)=f(-x)$.

Claim 1. If $f(a)=f(b)\neq 0$ then $a=\pm b$.

Proof. Let $a,b\in\mathbb{R}$ satisfy $f(a)=f(b)\neq 0$. Then
\[
P(a,b^2)\implies f(a^2-b^2)+2b^2f(a)=a^2f(a)+b^2f(b)\implies f(a^2-b^2)=f(a)(a^2-b^2)
\]so
\[
f(a)(a^2-b^2)=f(a^2-b^2)=f(b^2-a^2)=f(a)(b^2-a^2).
\]It follows that $a^2=b^2$, as desired. Thus $f(x)\in\{-x^2,0,x^2\}$. $\square$

Claim 2. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=a^2$ and $f(b)=-b^2$.

Proof. Assume the contrary. Then
\begin{align*}
P(a,b^2)&\implies f(a^2-b^2)=a^4-b^4-2a^2b^2\\
P(b,a^2)&\implies f(b^2-a^2)=a^4-b^4+2a^2b^2
\end{align*}so $ab=0$, a contradiction. $\square$

Claim 3. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=a^2$ and $f(b)=0$.

Proof. Assume the contrary. Then
\begin{align*}
P(a,b^2)&\implies f(a^2-b^2)=a^4-2a^2b^2\\
P(b,a^2)&\implies f(a^2-b^2)=a^4
\end{align*}so $ab=0$, a contradiction. $\square$

Claim 4. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=-a^2$ and $f(b)=0$.

Proof. Assume the contrary. Then
\begin{align*}
P(a,b^2)&\implies f(a^2-b^2)=-a^4+2a^2b^2\\
P(b,a^2)&\implies f(a^2-b^2)=-a^4
\end{align*}so $ab=0$, a contradiction. $\square$

The conclusion follows. $\square$
This post has been edited 1 time. Last edited by KevinYang2.71, May 8, 2024, 7:00 AM
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KevinChen_Yay
185 posts
#66
Y by
Awesomeness_in_a_bun wrote:
they gave 0 for just including the functions

Yea I wrote the correct answer along with full proof, only error being assuming polynomial and got a 0 rips
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OptimalFEian
10 posts
#67
Y by
It is obvious that the functions $f(x) = x^2 \thinspace \forall x \in \mathbb{R}$ and $f(x)\equiv 0$ satisfy the given equation. Now, we will prove that these are the only solutions. Let $P(x, y)$ denote the given assertion.
\begin{align*}
		P(0, 0)&: f(f(0)) = 0 \\
		P(f(0), f(0)^2/2)&: f(f(0)^2/2) = f(0) + f(f(0)^2/2) \implies f(0) = 0 \\
		P(x, 0)&: f(x^2) = f(f(x)) \\
		P(0, x)&: f(-x) = f(x) \\
		P(x, x^2/2)&: f(f(x)) = x^2 f(x).
	\end{align*}Then, we can rewrite $P(x, y)$ as
\[f(x^2 - y) = (x^2 - 2y) f(x) + f(y).\]Replacing $y$ by $-y^2$ gives
\[f(x^2 + y^2) = x^2 f(x) + y^2 f(y) + 2y^2 f(x).\]By symmetry, we get $y^2 f(x) = x^2 f(y)$. Setting $y \to 1$, we have $f(x) = cx^2$ for some constant $c$. Since $f(f(x)) = x^2 f(x)$, it follows that $c = 0$ or $1$.
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ihatemath123
3426 posts
#70
Y by
The solutions are $f(x) = x^2$, $f(x) = -x^2$ and $f(x) = 0$.

Claim: We have $f(0) = 0$.
Proof: The claim follows if we subtract $P(1, \tfrac{1}{2} )$ from $P(1,1)$.

Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$, so our FE is equivalent to
\[f(x^2 - y) = (x^2 - 2y) f(x) + f(y).\]Let $Q(x,y)$ denote the above assertion.

Claim: If there exists a nonzero root of $f$, then $f(x) = 0$ for all $x$.
Proof: Let this nonzero root be $r$. Subtracting $Q(0,x)$ from $Q(r,x)$ gives us
\[f(r^2 - x) = f(-x),\]so $f$ is periodic with period $r^2$. Now, subtracting $Q(x, r^2)$ from $Q(x,0)$ gives us $r^2 f(x) = 0$ for all $x$, so $f$ is identically $0$.

From hereon, assume $0$ is the only root of $f$. Taking $P(x, \tfrac{x^2}{2} )$ where $x \neq 0$ gives us $x^2 f(x) = f(f(x))$, implying "quasi-injectivity": that if $f(a) = f(b)$, then $|a|  = |b|$. Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$, so $f(x) = \pm x^2$ for each $x$.

Claim: Either $f(x) = x^2$ for all $x$ or $f(x) = -x^2$ for all $x$.
Proof: Suppose otherwise. Notice that if $f$ is a solution to our rewritten FE, then $-f$ is also a solution. So, assume WLOG that $f(1) = 1$. Now, let $x$ and $y$ be nonzero reals with such that $x^2 - y = 1$:
  • If, FTSOC, $f(x) = x^2$ and $f(y) = -y^2$, taking $Q(x,y)$ gives us $y = 0$, which is a contradiction.
  • If, FTSOC, $f(x) = -x^2$ and $f(y) = y^2$, taking $Q(x,y)$ gives us
    \[x^2(x^2 - 2y) = 0 \implies y =  x^2 - y = 1 \implies x = \sqrt{2}.\]But now, taking $Q(1, \sqrt{2})$ gives us $\left| (1 - \sqrt{2})^2 \right| = \left| -1 - 2 \sqrt{2} \right|$, contradiction.
  • If, FTSOC, $f(x) = -x^2$ and $f(y) = -y^2$, taking $Q(x,y)$ gives us $2(x^2 - y)^2 = 0$, contradiction.

It follows that $f(x) = x^2$ or $f(x) = -x^2$.
This post has been edited 1 time. Last edited by ihatemath123, Aug 23, 2024, 3:30 AM
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CyclicISLscelesTrapezoid
371 posts
#71 • 4 Y
Y by GrantStar, OronSH, megarnie, centslordm
The solutions are $f(x)=0$, $f(x) \equiv x$, and $f(x) \equiv -x^2$, which work. Let $P(x,y)$ denote the assertion in the problem statement.

$P(0,0)$ gives $f(f(0))=0$.
$P(f(0),\tfrac{f(0)^2}{2})$ gives $f(0)=0$.
$P(0,x)$ gives $f(x)=f(-x)$.
$P(x,0)$ gives $f(f(x))=f(x^2)$.

$P(x,y^2)$ gives
\[f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)=f(x^2)+f(y^2).\]Swapping $x$ and $y$ gives
\[f(x^2-y^2)+2y^2f(x)=f(y^2-x^2)+2x^2f(y)=f(x^2-y^2)+2x^2f(y),\]so $y^2f(x)=x^2f(y)$. Plugging in $y=1$ gives $f(x)=f(1)x^2$.

Since $f(f(x))=f(x^2)$, we have
\[f(1)(f(1)x^2)^2=f(1)x^4 \implies f(1)^3=f(1),\]so $f(1) \in \{0,1,-1\}$, as desired. $\blacksquare$
This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Sep 1, 2024, 9:15 PM
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fearsum_fyz
48 posts
#72
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We claim the only answers are $\boxed{f(x) = 0}$, $\boxed{f(x) = x^2}$, and $\boxed{f(x) = - x^2}$. It is easy to verify that these work. It remains to show that they the only solutions.

Claim 1: $x^2 f(x) = f(f(x))$
Proof.
$\underline{P}(x, 0) \implies f(x^2) = f(f(x)) + f(0)$
$\underline{P}(x, x^2) \implies f(0) + 2x^2f(x) = f(f(x)) + f(x^2)$
Adding, we get $2x^2 f(x) = 2f(f(x)) \implies \boxed{x^2 f(x) = f(f(x))}$

Claim 2: $f(0) = 0$
Proof.
By Claim 1, we have $f(f(1)) = f(1)$. Therefore:
$\underline{P}(1, 1) \implies f(0) + \cancel{2f(1)} = \cancel{2f(1)}$
$\implies \boxed{f(0) = 0}$.

Claim 3: $f$ is even.
Proof.
Using Claim 2:
$\underline{P}(0, y) \implies \boxed{f(-y) = f(y)}$

Claim 4: If $f \not\equiv 0$, then $f(x) = f(y) \implies x = \pm y$.
Proof.
By Claim 1, $f(x) = f(y) \implies f(f(x)) = f(f(y)) \implies \frac{f(f(x))}{f(x)} = \frac{f(f(y))}{f(y)} \implies x^2 = y^2 \implies x = \pm y$.

It is easy to see that Claim 4 then finishes the problem: If $f \not\equiv 0$, then we have
$f(x^2) = f(f(x)) \implies \boxed{f(x) = \pm x^2}$.

It remains to take care of the pointwise trap, which is not hard to do. However, for the sake of compactness, I have excluded the details from my post.
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eg4334
597 posts
#73 • 1 Y
Y by Marcus_Zhang
I claim the only solutions are $\boxed{f(x) \equiv 0, x^2, -x^2}$ which can easily be verified to work. Let $P(x, y)$ denote the given assertion.

$P(x, \frac{x^2}{2})$ gives $f(f(x)) = x^2f(x)$.

$P(-x, y)$ and $P(x, y)$ and equating gives $x^2f(x)-2yf(x)=x^2f(-x)-2yf(-x)$. $y=0$ so $x^2f(x)=x^2f(-x)$ or $f(x)=f(-x)$ for nonzero $x$ and thus for all $x$. Therefore $f$ is even. Call this statement $1$.

Let $x=0$ immediately gives us $2yf(0)=0$ so $f(0)=0$, coupled with the fact that $f(f(0))=0$. Now let $y=0$, so $f(x^2)=f(f(x))$, and we have the chain: $$f(x^2)=f(f(x))=x^2f(x)$$which is of tremendous use. Let $x=f(x)$ in the latter inequality to get $$f(f(f(x)))=f(x)^2f(f(x))$$$$f(x)^2 f(x^2) = x^4 f(x^2)$$so $f(x) = 0, x^2, -x^2$ as mentioned before for an individual $x$.

We now tackle the pointwise trap. I provide a sketch below.

If $f(a)=a^2$ and $f(b)=0$ for $a, b \neq 0$, then consider $P(b, a)$ and $P(b, -a)$. This readily shows $f(b^2-a) = f(b^2+a)=a^2$. This contradicts $a, b \neq 0$.

If $f(a)=-a^2$ and $f(b)=0$ for $a, b \neq 0$ then consider $P(b, a)$ and $P(b, -a)$ yet again. This shows that $(b^2-a)^2 = (b^2+a)^2$ again which contradicts $a, b \neq 0$.

If $f(a)=a^2$ and $f(b)=-b^2$ then $P(\sqrt{a}, b)$ and $P(\sqrt{b}, a)$ give us another contradiction.
This post has been edited 1 time. Last edited by eg4334, Mar 1, 2025, 1:09 AM
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cappucher
88 posts
#74
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The only solutions that satisfy the given are $f(x) \equiv x^2, -x^2, 0$.

Claim 1: $f(0) = 0$

Proof: Let $P(x, y)$ denote the given equation. From $P(x, 0)$, we obtain

\[f(x^2) = f(f(x)) + f(0)\]
Let $c = f(0)$. From $P(0, 0)$, we obtain

\[c = f(c) + c \implies f(c) = 0 \implies f(f(c)) = c\]
From $P(c, 0)$, we obtain

\[f(c^2) = f(f(c)) + c \implies f(c^2) = 2c\]
However, from $P(c, c^2)$, we know that

\[f(0) + 2c^2f(c) = f(f(c)) + f(c^2)\]\[c = c + f(c^2) \implies f(c^2) = 0\]\[f(c^2) = 2c = 0 \implies c = 0\]
as desired.

Claim 2: $f(f(x)) = f(x^2) = x^2f(x)$.

Proof: From $P(x, 0)$, we have

\[f(x^2) = f(f(x)) + f(0) \implies f(x^2) = f(f(x))\]
From $P(x, x^2)$, we have

\[f(x^2 - x^2) + 2x^2f(x) = f(f(x)) + f(x^2)\]\[2x^2f(x) = 2f(x^2) \implies x^2f(x) = f(x^2)\]
as desired.

Claim 3: $f(x)$ is even.

Proof: From $P(0, y)$, we have

\[P(0 - y) + 2yf(0) = f(f(0)) + f(y)\]\[f(-y) = f(y)\]
as desired.

Using claims $2$ and $3$, we will prove that $f(x) \equiv x^2, -x^2, 0$. After we finish doing so, we will show these are the only solutions by avoiding the pointwise trap.

Note that

\[f(f(f(x))) = f(f(x^2)) = x^4f(x^2) = x^6f(x)\]
and that

\[f(f(f(x))) = (f(x))^2f(f(x)) = x^2(f(x))^3\]
We can then equate the two to obtain

\[x^6f(x) = x^2\left(f(x)\right)^3\]\[x^2f(x)\left(f(x)^2 - x^4\right) = 0\]\[x^2f(x)\left(f(x) + x^2\right)\left(f(x) - x^2\right) = 0\]
The only way for this equation to be satisfied for all $x$ is if $f(x) \equiv -x^2, x^2, 0$.

Avoiding the pointwise trap

There are three cases to tackle: $f(x) \equiv 0, x^2$, $f(x) \equiv 0, -x^2$, and $f(x) \equiv -x^2, x^2$. Because these take a while to type out, I will edit this solution later to include them. The basic idea is to have non-zero variables $a$ and $b$ such that $f(a) = 0$ and $f(b) = b^2$ (for case 1; same logic applies for cases 2 and 3) and show that this leads to a contradiction with the given. Claim 3 ($f(x)$ is even) will prove useful for this.
This post has been edited 2 times. Last edited by cappucher, Dec 23, 2024, 8:19 AM
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Yiyj1
1174 posts
#75 • 1 Y
Y by megarnie
OronSH wrote:
$f(x^2-y)+2yf(x)=f(f(x))+f(y)$

We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$

Denote the assertion as $P(x,y).$
Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$
Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$
From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even.
Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$

Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$

Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$

Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$

Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$

Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work.

Oron orz
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D4N13LCarpenter
13 posts
#76 • 1 Y
Y by Vahe_Arsenyan
Let $P(x, y)$ denote the given assertion. We claim that the only solutions are $f\equiv0$, $f(x)=x^2$ and, $f(x)=-x^2$ which clearly work.

Claim 1.4 We have $f(x^2)=f(f(x))$
Proof $P(x,0)$ yields $f(x^2)=f(f(x))+f(0)$, so it suffices to prove that $f(0)=0$. Observe that by $P(1,1)$, we get $$f(0)+2f(1)=f(f(1))+f(1)$$but after cancelling out terms and substituting $f(f(1))$ for $f(1)-f(0)$, we get $f(0)+f(1)=f(1)-f(0)$ which directly implies the result.
Notice how by using $f(0)=0$ we can now easily prove that $f$ is even, as $P(0,y)$ gives $f(-y)=f(y)$. The key claim is as follows.
Claim 1.5 $f(x)=f(y)$ implies $\vert x\vert=\vert y\vert$ or $f(x)=f(y)=0$
Proof $P(x, y^2)$ states $$f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$$and $P(y, x^2)$ states $$f(y^2-x^2)+2x^2f(y)=f(f(x))+f(x^2)$$but using both that $f(x^2)=f(f(x))$ and that $f$ is even, this is equivalent to $y^2f(x)=x^2f(y)=x^2f(x)$ so either $f(x)=0$ or $x^2=y^2\implies \vert x\vert=\vert y\vert y$
Claim 1.6 If $f(c)=0$ for some positive real $c$, than $f\equiv0$
Proof $P(c,x)$ gives $$f(c^2-y)=f(f(c))+f(y)$$but note that $f(f(c))=f(0)=0$ so we in fact have $f(c^2-y)=f(y)$. However, by Claim 1.5 either $f(c^2-y)=f(y)=0$ or $\vert c^2-y\vert=\vert y\vert$, the latter being only true when $y=\frac{c^2}{2}$. Thus, we have $f(x)=0$ for all $x\neq \frac{c^2}{2}$. But one can now easily find $\frac{c^2}{2}$ taking any $x, y$ satisfying $x^2-y=\frac{c^2}{2}$, as then $P(x,y)$ gives $f(\frac{c^2}{2})=0$.
From now on, suppose that $f(c)\neq0$ for any $c$ greater than $0$. Recall that $f(x^2)=f(f(x))$ so we can apply Claim 1.5 to get $\vert f(x)^2\vert=\vert x^4\vert$, which yields $f(x)=\pm x^2$.

We now aim to prove that $f$ can't change of sign, thus leaving us with only the initially stated solutions. Assume for the sake of contradiction that $f$ changes in sign. By Pigeonhole principle, one of these must repeat an infinite number of times, WLOG say it is $f(x)=x^2$, as the other is completely symmetric. Now choose a sufficiently large $a$ such that $f(a)=a^2$ and a small $b\neq 0$ in comparison which satisfies $f(b)=-b^2$. Now notice that $P(a,b)$ gives $$f(a^2-b)+2a^2b=a^4-b^2$$where we used $f(f(x))=x^2f(x)$, which is a consequence of $P(x^2, y^2)$. We can rewrite this as $f(a^2-b)=a^4-b^2-2a^2b$ but we also know $f(a^2-b)=\pm (a^2-b)^2$. So we can now distinguish two cases.

Case 1. $f(a^2-b)=(a^2-b)^2$
Combining this with $f(a^2-b)=a^4-b^2-2a^2b$ we get $a^4-b^2-2a^2b=a^4-2a^2b+b^2$ so $-b^2=b^2\implies b=0$. But we supposed $b\neq 0$, so this case leads to a contradiction.

Case 2. $f(a^2-b)=-(a^2-b)^2$
Notice how $a^4-b^2-2a^2b$ is positive for sufficiently large $a$, however $-(a^2-b)^2$ is always negative, so we can immediately discard this case.
This post has been edited 2 times. Last edited by D4N13LCarpenter, Jan 10, 2025, 7:02 PM
Reason: I accidentally wrote $f(b)=b^2$ but it's clearly $f(b)=-b^2$
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anandswaroop191
14 posts
#77
Y by
Plug in $y = 0$ to get $f(f(x)) = f(x^2) - f(0)$, substitute in the original equation to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y) - f(0)$*.

Plug in $x = y = 1$ to get $f(0) + 2f(1) = f(1) + f(1) - f(0) \implies f(0) = 0$. Substitute into * to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y)$**.

Plug in $x = 0$ to get $f(y) = f(-y)$.

Substitute $y \mapsto -y$ to get $f(x^2 + y) - 2yf(x) = f(x^2) + f(-y) = f(x^2) + f(y)$. Subtract ** from this to get $f(x^2 + y) - f(x^2 - y) = 4yf(x)$. Substitute $y \mapsto y^2$ to get $f(x^2 + y^2) - f(x^2 - y^2) = 4y^2f(x)$.

Note that $f(x^2 - y^2) = f(y^2 - x^2)$ (because $f$ is even), so $4y^2f(x) = f(x^2 + y^2) - f(x^2 - y^2) = f(y^2 + x^2) - f(y^2 - x^2) = 4x^2f(y)$.

Simplifying, we get $y^2f(x) = x^2f(y)$. Plug in $y = 1$ to get $f(x) = x^2f(1)$. Therefore, $f(x) = cx^2$ for some constant $c$.

Going back to the original equation, we have $f(x^2 - y) + 2yf(x) = cx^4 - 2cx^2y + cy^2 + 2x^2y = cx^4 + cy^2$, and $f(f(x)) + f(y) = c^3x^4 + cy^2$. Equating these, we have $(c^3 - c)x^4 = 0$ for all real $x$; solving $c^3 - c = 0$, we get $c \in \{-1, 0, 1\}$.

The desired functions are $f(x) = -x^2$, $f(x) = 0$, and $f(x) = x^2$. $\square$
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mathwiz_1207
91 posts
#78
Y by
Avoids pointwise I think. We claim the only solutions are $f \equiv \boxed{0, x^2, -x^2}$. It can be easily verified that these work. We will now prove that these are the only solutions.

Claim: $f(f(x)) = x^2f(x) = f(x^2)$
Proof. $x = x, y = \frac{x^2}{2}$ yields
\[f(\frac{x^2}{2})+ x^2f(x) = f(f(x)) + f(\frac{x^2}{2}) \implies f(f(x)) = x^2f(x)\]Now, $x = x, y = x$ gives
\[f(x^2 - x) + 2xf(x) = x^2f(x) + f(x)\]\[f(x^2 - x) = (x - 1)^2f(x)\]Plugging in $x = 1$ gives $f(0) = 0$. Now, $x = x, y = 0$ gives us
\[f(x^2) = f(f(x)) = x^2f(x)\]so the claim is proven.

We now prove the second key claim, which almost finishes the problem:

Claim 2: $f$ is even.
Proof. By claim $1$, we have that
\[f(x^2) = x^2f(x)\]and that
\[f((-x)^2) = x^2f(-x)\]This implies that $x^2f(x) = x^2f( -x)$. If $x \neq 0$, we get $f(x) = f(-x)$. Otherwise, if $x = 0$, it is trivial that $f(0) = f(-0)$. Thus, $f$ is even.

Now for the finish. Plugging in $x = x, y = t^2$ yields
\[f(x^2 - t^2) + 2t^2f(x) = f(f(x)) + f(t^2) = f(f(x)) + f(f(t))\]Swapping $x$ and $t$ gives
\[f(t^2 - x^2) + 2x^2f(t) = f(f(t)) + f(x^2) = f(f(x)) + f(f(t))\]However, since $f$ is even, the first term of both equations are equal, so we may subtract to get
\[2x^2f(t) = 2t^2f(x)\]Assume $x, t$ are nonzero, then this means that $\frac{f(x)}{x^2}$ is equal to a constant for all nonzero values of $x$. This means $f(x) = cx^2$, and upon substituting, we find $c = 0, 1, -1$. Clearly, $x = 0$ also satisfies this, so we find $f \equiv 0, x^2, -x^2$ and we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Jan 30, 2025, 6:59 AM
Reason: skill issue
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bjump
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#79 • 2 Y
Y by KenWuMath, megarnie
Setting $x=y=0$ gives $f(f(0))=0$. Setting $x=1$, and $y=0$ gives $f(1)=f(f(1))+f(0)$. Setting $x=y=1$ gives $f(0)+f(1)=f(f(1))$. Subtracting the previous equations we have made gives $f(0)=0$. Therefore setting $y=0$ gives $f(x^2)=f(f(x))$. Setting $x=0$ now gives $f(y)=f(-y)$. Now I claim that $f(x)=g(x)$ is a valid solution to the functional equation iff $f(x)=-g(x)$ is a valid solution. This follows because :$$-f(x^2-y)-2yf(x)=-f(-f(x)) -f(y) = -f(f(x))-f(y) \iff f(x^2-y)+2yf(x) = f(f(x)) + f(y).$$Therefore we may assume that $f(1) \ge 0$. Notice that setting $x=1$ and $y=x^2$ implies that $f(x^2-1)+2x^2f(1)=f(1)$, and setting $x=x$, and $y=1$ gives $f(x^2-1)+2f(x)=f(x^2)+f(1)$ subtracting equations gives $f(x)=f(1)x^2$. We also have from previosly derived properties $f(1)=f(f(1))= f(1)^3$. Therefore $f(x)=-x^2, 0 , x^2$.
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Ilikeminecraft
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#80
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I claim that $f(x) = 0, x^2, -x^2$

Let $P(x,y)$ be our assertion. We start by plugging in a lot of values.

$P(x, 0)$ implies that $f(x^2) = f(f(x)) + f(0),$ and $P(x, x^2)$ implies that $f(0) + 2x^2f(x) = f(f(x)) + f(x^2) = f(f(x)) + f(f(x)) + f(0),$ and thus $x^2f(x) = f(f(x)), (1)$. $P(0, 0)$ implies that $f(f(0)) = 0.$

\begin{lemma}
$f(0) = 0.$
\end{lemma}
\begin{proof}
Let $f(0) = k.$ Thus, we also have that $f(k) = 0.$

Now, we take $P(0, k),$ which tells us that $f(-k) = -2k^2.$ $P(-k, 0)$ gives us that $f(x^2) = f(f(-k)) + k = k^2f(-k) + k = -2k^4 + k$, but $P(k, 0)$ gives us that $f(x^2) = f(f(k)) + f(0) = 2k.$ Thus, $(2k^3 + 1)k = 0.$

If $k \neq 0,$ we have that $k = -2^{-\frac13}.$ Thus, $f(-k) = -2^{\frac13} = \frac1k.$ We also know that $k^2f(-k) = f(f(-k)) = f(\frac1k),$ and hence $f(\frac1k) = k.$ Thus, $f(f(\frac1k)) = 0,$ but then we also have that $f(f(\frac1k)) = \frac1{k^2}f(\frac1k) = \frac 1k$ by $(1),$ and hence $\frac1k = 0,$ which is a contradiction. Thus, $f(0) = 0.$
\end{proof}

$P(0, x)$ tells us that $f$ is even. Now, $P(x, y^2)$ tells us that $f(x^2 - y^2) + 2y^2f(x) = f(f(x)) + f(y^2) = x^2f(x) + y^2f(y).$ However, by swapping the variables, we get that $f(y^2 - x^2) + 2x^2f(y) = y^2f(y) + x^2f(x).$ By subtracting these two, we get that $2x^2f(y) + x^2f(x) + y^2f(y) = 2y^2f(x) + y^2f(y) + x^2f(x).$ Thus, $\frac{f(x)}{x^2} = \frac{f(y)}{y^2}.$

Let this constant be $c.$ By plugging it in our equation, we see that $c = 1, 0, -1.$ Thus, $f(x) = 0, -x^2, x^2$
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joshualiu315
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#81
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i believe this dodges pointwise trap


The answer is $f(x) = \boxed{0, \pm x^2}$.

Let the given assertion be $P(x,y)$. Note that $P(x,0)$ gives

\[f(x^2) = f(f(x))+f(0), \quad \bigstar\]
and $P(x,\tfrac{x^2}{2})$ gives $x^2f(x) = f(f(x))$. We plug this into the original equation to get a new assertion:

\[Q(x,y): \ f(x^2-y) + 2yf(x) = x^2f(x)+f(y).\]
Plugging in $Q(-x,y)$ gives

\[f(x^2-y) + 2yf(-x) = x^2f(-x) + f(y).\]
Comparing this to $Q(x,y)$ gives

\[(x^2-2y)(f(x)-f(-x)) = 0,\]
so for $y \neq \tfrac{x^2}{2}$, we have $f(x) = f(-x)$ for all $x$; in other words, $f$ is even. Then, $Q(0,y)$ gives

\[2yf(x) = 0,\]
which means $f(0) = 0$. Plugging this into $\bigstar$ yields $f(x^2) = f(f(x)) = x^2f(x)$. Comparing $Q(x,y^2)$ and $Q(y,x^2)$ gives

\[f(x^2-y^2) + 2y^2f(x) = x^2f(x) + f(y^2) = f(x^2)+f(y^2),\]\[f(y^2-x^2) + 2x^2f(y) = y^2f(y)+f(x^2) = f(y^2)+f(x^2).\]
Since $f(x^2-y^2) = f(y^2-x^2)$, we have $2y^2f(x) = 2x^2f(y)$, or that $\tfrac{f(x)}{x^2} = \tfrac{f(y)}{y^2}$. Setting this equal to a constant yields $f(x) = cx^2$ for some value of $c$ and all values of $x$. Then,

\[cx^4 = f(x^2) = f(f(x)) = c^3x^4 \implies c \in \{-1,0,1\}.\]
This gives us our desired solution set.
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