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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
May 1, 2025
0 replies
k i Peer-to-Peer Programs Forum
jwelsh   157
N Dec 11, 2023 by cw357
Many of our AoPS Community members share their knowledge with their peers in a variety of ways, ranging from creating mock contests to creating real contests to writing handouts to hosting sessions as part of our partnership with schoolhouse.world.

To facilitate students in these efforts, we have created a new Peer-to-Peer Programs forum. With the creation of this forum, we are starting a new process for those of you who want to advertise your efforts. These advertisements and ensuing discussions have been cluttering up some of the forums that were meant for other purposes, so we’re gathering these topics in one place. This also allows students to find new peer-to-peer learning opportunities without having to poke around all the other forums.

To announce your program, or to invite others to work with you on it, here’s what to do:

1) Post a new topic in the Peer-to-Peer Programs forum. This will be the discussion thread for your program.

2) Post a single brief post in this thread that links the discussion thread of your program in the Peer-to-Peer Programs forum.

Please note that we’ll move or delete any future advertisement posts that are outside the Peer-to-Peer Programs forum, as well as any posts in this topic that are not brief announcements of new opportunities. In particular, this topic should not be used to discuss specific programs; those discussions should occur in topics in the Peer-to-Peer Programs forum.

Your post in this thread should have what you're sharing (class, session, tutoring, handout, math or coding game/other program) and a link to the thread in the Peer-to-Peer Programs forum, which should have more information (like where to find what you're sharing).
157 replies
jwelsh
Mar 15, 2021
cw357
Dec 11, 2023
k i C&P posting recs by mods
v_Enhance   0
Jun 12, 2020
The purpose of this post is to lay out a few suggestions about what kind of posts work well for the C&P forum. Except in a few cases these are mostly meant to be "suggestions based on historical trends" rather than firm hard rules; we may eventually replace this with an actual list of firm rules but that requires admin approval :) That said, if you post something in the "discouraged" category, you should not be totally surprised if it gets locked; they are discouraged exactly because past experience shows they tend to go badly.
-----------------------------
1. Program discussion: Allowed
If you have questions about specific camps or programs (e.g. which classes are good at X camp?), these questions fit well here. Many camps/programs have specific sub-forums too but we understand a lot of them are not active.
-----------------------------
2. Results discussion: Allowed
You can make threads about e.g. how you did on contests (including AMC), though on AMC day when there is a lot of discussion. Moderators and administrators may do a lot of thread-merging / forum-wrangling to keep things in one place.
-----------------------------
3. Reposting solutions or questions to past AMC/AIME/USAMO problems: Allowed
This forum contains a post for nearly every problem from AMC8, AMC10, AMC12, AIME, USAJMO, USAMO (and these links give you an index of all these posts). It is always permitted to post a full solution to any problem in its own thread (linked above), regardless of how old the problem is, and even if this solution is similar to one that has already been posted. We encourage this type of posting because it is helpful for the user to explain their solution in full to an audience, and for future users who want to see multiple approaches to a problem or even just the frequency distribution of common approaches. We do ask for some explanation; if you just post "the answer is (B); ez" then you are not adding anything useful.

You are also encouraged to post questions about a specific problem in the specific thread for that problem, or about previous user's solutions. It's almost always better to use the existing thread than to start a new one, to keep all the discussion in one place easily searchable for future visitors.
-----------------------------
4. Advice posts: Allowed, but read below first
You can use this forum to ask for advice about how to prepare for math competitions in general. But you should be aware that this question has been asked many many times. Before making a post, you are encouraged to look at the following:
[list]
[*] Stop looking for the right training: A generic post about advice that keeps getting stickied :)
[*] There is an enormous list of links on the Wiki of books / problems / etc for all levels.
[/list]
When you do post, we really encourage you to be as specific as possible in your question. Tell us about your background, what you've tried already, etc.

Actually, the absolute best way to get a helpful response is to take a few examples of problems that you tried to solve but couldn't, and explain what you tried on them / why you couldn't solve them. Here is a great example of a specific question.
-----------------------------
5. Publicity: use P2P forum instead
See https://artofproblemsolving.com/community/c5h2489297_peertopeer_programs_forum.
Some exceptions have been allowed in the past, but these require approval from administrators. (I am not totally sure what the criteria is. I am not an administrator.)
-----------------------------
6. Mock contests: use Mock Contests forum instead
Mock contests should be posted in the dedicated forum instead:
https://artofproblemsolving.com/community/c594864_aops_mock_contests
-----------------------------
7. AMC procedural questions: suggest to contact the AMC HQ instead
If you have a question like "how do I submit a change of venue form for the AIME" or "why is my name not on the qualifiers list even though I have a 300 index", you would be better off calling or emailing the AMC program to ask, they are the ones who can help you :)
-----------------------------
8. Discussion of random math problems: suggest to use MSM/HSM/HSO instead
If you are discussing a specific math problem that isn't from the AMC/AIME/USAMO, it's better to post these in Middle School Math, High School Math, High School Olympiads instead.
-----------------------------
9. Politics: suggest to use Round Table instead
There are important conversations to be had about things like gender diversity in math contests, etc., for sure. However, from experience we think that C&P is historically not a good place to have these conversations, as they go off the rails very quickly. We encourage you to use the Round Table instead, where it is much more clear that all posts need to be serious.
-----------------------------
10. MAA complaints: discouraged
We don't want to pretend that the MAA is perfect or that we agree with everything they do. However, we chose to discourage this sort of behavior because in practice most of the comments we see are not useful and some are frankly offensive.
[list] [*] If you just want to blow off steam, do it on your blog instead.
[*] When you have criticism, it should be reasoned, well-thought and constructive. What we mean by this is, for example, when the AOIME was announced, there was great outrage about potential cheating. Well, do you really think that this is something the organizers didn't think about too? Simply posting that "people will cheat and steal my USAMOO qualification, the MAA are idiots!" is not helpful as it is not bringing any new information to the table.
[*] Even if you do have reasoned, well-thought, constructive criticism, we think it is actually better to email it the MAA instead, rather than post it here. Experience shows that even polite, well-meaning suggestions posted in C&P are often derailed by less mature users who insist on complaining about everything.
[/list]
-----------------------------
11. Memes and joke posts: discouraged
It's fine to make jokes or lighthearted posts every so often. But it should be done with discretion. Ideally, jokes should be done within a longer post that has other content. For example, in my response to one user's question about olympiad combinatorics, I used a silly picture of Sogiita Gunha, but it was done within a context of a much longer post where it was meant to actually make a point.

On the other hand, there are many threads which consist largely of posts whose only content is an attached meme with the word "MAA" in it. When done in excess like this, the jokes reflect poorly on the community, so we explicitly discourage them.
-----------------------------
12. Questions that no one can answer: discouraged
Examples of this: "will MIT ask for AOIME scores?", "what will the AIME 2021 cutoffs be (asked in 2020)", etc. Basically, if you ask a question on this forum, it's better if the question is something that a user can plausibly answer :)
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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.
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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.
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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
solve in positive integers: 3 \cdot 2^x +4 =n^2
parmenides51   4
N 8 minutes ago by AylyGayypow009
Source: Greece JBMO TST 2019 p2
Find all pairs of positive integers $(x,n) $ that are solutions of the equation $3 \cdot 2^x +4 =n^2$.
4 replies
parmenides51
Apr 29, 2019
AylyGayypow009
8 minutes ago
A strong inequality problem
hn111009   3
N 8 minutes ago by ZeltaQN2008
Source: Somewhere
Let $a,b,c$ be the positive number satisfied $a^2+b^2+c^2=3.$ Find the minimum of $$P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3abc}{2(ab+bc+ca)}.$$
3 replies
hn111009
Yesterday at 2:02 AM
ZeltaQN2008
8 minutes ago
ISI UGB 2025 P7
SomeonecoolLovesMaths   10
N 10 minutes ago by Mathworld314
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
10 replies
SomeonecoolLovesMaths
Yesterday at 11:28 AM
Mathworld314
10 minutes ago
2018 JBMO TST- Macedonia, problem 4
Lukaluce   3
N 12 minutes ago by Erto2011_
Source: 2018 JBMO TST- Macedonia
Determine all pairs $(p, q)$, $p, q \in \mathbb {N}$, such that

$(p + 1)^{p - 1} + (p - 1)^{p + 1} = q^q$.
3 replies
Lukaluce
May 28, 2019
Erto2011_
12 minutes ago
Beams inside a cube
AOPS12142015   32
N 5 hours ago by cursed_tangent1434
Source: USOMO 2020 Problem 2, USOJMO 2020 Problem 3
An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
[list=]
[*]The two $1 \times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are $3 \cdot {2020}^2$ possible positions for a beam.)
[*]No two beams have intersecting interiors.
[*]The interiors of each of the four $1 \times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.
[/list]
What is the smallest positive number of beams that can be placed to satisfy these conditions?

Proposed by Alex Zhai
32 replies
AOPS12142015
Jun 21, 2020
cursed_tangent1434
5 hours ago
Summer internships/research opportunists in STEM
o99999   8
N 6 hours ago by Craftybutterfly
Hi, I am a current high school student and was looking for internships and research opportunities in STEM. Do you guys know any summer programs that do research such as RSI, but for high school freshmen that are open?
Thanks.
8 replies
o99999
Apr 22, 2020
Craftybutterfly
6 hours ago
9 Will I make JMO?
EaZ_Shadow   20
N Today at 5:38 AM by Craftybutterfly
will I be able to make it... will the cutoffs will be pre-2024
20 replies
EaZ_Shadow
Feb 7, 2025
Craftybutterfly
Today at 5:38 AM
Sequence of integers
tenniskidperson3   32
N Today at 4:33 AM by N3bula
Source: 2012 USAMO problem #3
Determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1, a_2, a_3, \ldots$ of nonzero integers such that the equality \[a_k+2a_{2k}+\ldots+na_{nk}=0\]holds for every positive integer $k$.
32 replies
tenniskidperson3
Apr 24, 2012
N3bula
Today at 4:33 AM
Isogonal Conjugates: 2011 USAMO #5
tenniskidperson3   77
N Today at 12:35 AM by N3bula
Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that
\[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\] Prove that $\overline{Q_1Q_2}\parallel\overline{AB}$ if and only if $\overline{Q_1Q_2}\parallel\overline{CD}$.
77 replies
tenniskidperson3
Apr 28, 2011
N3bula
Today at 12:35 AM
camp/class recommendations for incoming freshman
walterboro   7
N Today at 12:33 AM by Ruegerbyrd
hi guys, i'm about to be an incoming freshman, does anyone have recommendations for classes to take next year and camps this summer? i am sure that i can aime qual but not jmo qual yet. ty
7 replies
walterboro
Saturday at 6:45 PM
Ruegerbyrd
Today at 12:33 AM
Vertices of a pentagon invariant: 2011 USAMO #2
tenniskidperson3   54
N Yesterday at 11:21 PM by N3bula
An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.
54 replies
tenniskidperson3
Apr 28, 2011
N3bula
Yesterday at 11:21 PM
HCSSiM results
SurvivingInEnglish   69
N Yesterday at 10:23 PM by MathWizardThatCanBeatYou
Anyone already got results for HCSSiM? Are there any point in sending additional work if I applied on March 19?
69 replies
SurvivingInEnglish
Apr 5, 2024
MathWizardThatCanBeatYou
Yesterday at 10:23 PM
[CASH PRIZES] IndyINTEGIRLS Spring Math Competition
Indy_Integirls   1
N Yesterday at 10:09 PM by Indy_Integirls
[center]IMAGE

Greetings, AoPS! IndyINTEGIRLS will be hosting a virtual math competition on May 25,
2024 from 12 PM to 3 PM EST.
Join other woman-identifying and/or non-binary "STEMinists" in solving problems, socializing, playing games, winning prizes, and more! If you are interested in competing, please register here![/center]

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[center]Important Information[/center]

Eligibility: This competition is open to all woman-identifying and non-binary students in middle and high school. Non-Indiana residents and international students are welcome as well!

Format: There will be a middle school and high school division. In each separate division, there will be an individual round and a team round, where students are grouped into teams of 3-4 and collaboratively solve a set of difficult problems. There will also be a buzzer/countdown/Kahoot-style round, where students from both divisions are grouped together to compete in a MATHCOUNTS-style countdown round! There will be prizes for the top competitors in each division.

Problem Difficulty: Our amazing team of problem writers is working hard to ensure that there will be problems for problem-solvers of all levels! The middle school problems will range from MATHCOUNTS school round to AMC 10 level, while the high school problems will be for more advanced problem-solvers. The team round problems will cover various difficulty levels and are meant to be more difficult, while the countdown/buzzer/Kahoot round questions will be similar to MATHCOUNTS state to MATHCOUNTS Nationals countdown round in difficulty.

Platform: This contest will be held virtually through Zoom. All competitors are required to have their cameras turned on at all times unless they have a reason for otherwise. Proctors and volunteers will be monitoring students at all times to prevent cheating and to create a fair environment for all students.

Prizes: At this moment, prizes are TBD, and more information will be provided and attached to this post as the competition date approaches. Rest assured, IndyINTEGIRLS has historically given out very generous cash prizes, and we intend on maintaining this generosity into our Spring Competition.

Contact & Connect With Us: Follow us on Instagram @indy.integirls, join our Discord, follow us on TikTok @indy.integirls, and email us at indy@integirls.org.

---------
[center]Help Us Out

Please help us in sharing the news of this competition! Our amazing team of officers has worked very hard to provide this educational opportunity to as many students as possible, and we would appreciate it if you could help us spread the word!
1 reply
Indy_Integirls
Yesterday at 2:36 AM
Indy_Integirls
Yesterday at 10:09 PM
Past USAMO Medals
sdpandit   4
N Yesterday at 8:28 PM by sadas123
Does anyone know where to find lists of USAMO medalists from past years? I can find the 2025 list on their website, but they don't seem to keep lists from previous years and I can't find it anywhere else. Thanks!
4 replies
sdpandit
May 8, 2025
sadas123
Yesterday at 8:28 PM
Another FE
Ankoganit   43
N Apr 3, 2025 by jasperE3
Source: India IMO Training Camp 2016, Practice test 2, Problem 2
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$.
43 replies
Ankoganit
Jul 22, 2016
jasperE3
Apr 3, 2025
Another FE
G H J
G H BBookmark kLocked kLocked NReply
Source: India IMO Training Camp 2016, Practice test 2, Problem 2
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Ankoganit
3070 posts
#1 • 4 Y
Y by doxuanlong15052000, Davi-8191, toilaDang, Adventure10
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$.
This post has been edited 1 time. Last edited by Ankoganit, Jul 22, 2016, 9:04 AM
Z K Y
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RedRising
22 posts
#2 • 1 Y
Y by Adventure10
I found that $f(x)=-x$ is the solution, but I don't have time to write it now, it may also be wrong. Anyways, I analysed $P(x,0),P(x,-x)$ and $P(-x,x)$.
Z K Y
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Diamondhead
167 posts
#3 • 1 Y
Y by Adventure10
Wrong solution.

Edit: Yes. Thanks pco.
This post has been edited 1 time. Last edited by Diamondhead, Jul 22, 2016, 2:57 PM
Reason: Wrong
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pco
23511 posts
#4 • 3 Y
Y by Diamondhead, Adventure10, Mango247
Diamondhead wrote:
...
$P\left(\frac{x}{f(k)},k-x\right)$: $f\left(\left(\frac{x}{f(k)}\right)^2+\frac{x}{f(k)}f(k-x)\right)=x$ for all $x\in\mathbb{R}$, so $f$ is a bijective function....
No, you can only conclude from this that $f(x)$ is surjective. This is not enough to prove injectivity (and you need injectivity for your final step)
Z K Y
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darthsid
413 posts
#5 • 2 Y
Y by Adventure10, Mango247
Put $x,y=0$ to get $f(0)=0$
also put $y=0$ to obtain $f(x^{2})=xf(x)$

Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$.

Putting $y=k$ we obtain $f(x^{2})=xf(x)=xf(x+k)$ which means for $x\not= 0$ we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.

Suppose no such $k$ exists.

Then putting $y=-x$ we get $f(x^2+xf(-x))=0$ $\implies x^2 +xf(-x)=0$ which gives $f(-x)=-x$ replacing $-x$ with $x$ we get $f(x)=x$

Hence $f(x)=x$ and $f(x)=0$ are the only solutions.

Edit- Wrong solution. Thanks @Ashutoshmaths for pointing it out.
This post has been edited 1 time. Last edited by darthsid, Jul 23, 2016, 8:49 AM
Reason: wrong sol
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Ashutoshmaths
976 posts
#6 • 2 Y
Y by Adventure10, Mango247
darthsid wrote:
Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$.

we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.
This is wrong.
I don't think you can solve this problem without using surjectivity.
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darthsid
413 posts
#7 • 2 Y
Y by Adventure10, Mango247
Ashutoshmaths wrote:
darthsid wrote:
Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$.

we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.
This is wrong.
I don't think you can solve this problem without using surjectivity.

I don't get it. Why is it wrong? Edit: I get it now.
This post has been edited 1 time. Last edited by darthsid, Jul 23, 2016, 11:45 AM
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fractals
3028 posts
#8 • 8 Y
Y by ThE-dArK-lOrD, darthsid, kapilpavase, Ankoganit, anantmudgal09, opptoinfinity, Adventure10, Mango247
Solution
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WizardMath
2487 posts
#9 • 2 Y
Y by sa2001, Adventure10
Note: This problem had a lot of fake solves at the Camp and was by far the most troll problem there. And btw only 4-5 students solved it correctly in the time limit (excluding fake solves which almost all others did).
This post has been edited 1 time. Last edited by WizardMath, Jul 23, 2016, 1:44 AM
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Reynan
634 posts
#10 • 2 Y
Y by Adventure10, Mango247
darthsid wrote:
Ashutoshmaths wrote:
darthsid wrote:
Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$.

we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution.
This is wrong.
I don't think you can solve this problem without using surjectivity.

I don't get it. Why is it wrong?

$f(x+k)=f(x)$ does not mean $f(x)$ is constant, $f$ can be periodic with period $k$
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elVerde
25 posts
#11 • 6 Y
Y by lebathanh, Excuse, babu2001, Timmy456, Adventure10, D_S
Let $E(x,y)$ be the statement of the functional equation.
From $E(0,0)$ we get $f(0)=0$. Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution of the equation, henceforth suppose that $f\not\equiv 0$.
Let $a$ be such that $f(a)\neq 0$. Then $E(x, a-x)$ implies $f(z)=xf(a)$, where $z=x^2+xf(a-x)$. Since $f(a)\neq 0$, it follows that $f$ is surjective.

Suppose that $f(n)=f(m)$ for real numbers $n,m\in\mathbb{R}$. From $E(x,n)$ and $E(x,m)$ we get $xf(x+n)=f(x^2+xf(n))=f(x^2+xf(m))=xf(x+m)$, therefore $f(x+n)=f(x+m), \forall x\in\mathbb{R}$. Set $k=m-n$; then with $z=x+n$ the previous equation can be written as $f(z)=f(z+k),\forall z\in\mathbb{R}\,\,(*)$.
Choose $y$ such that $f(y)=-k$, which exists because $f$ is surjective. Now using $E(k,y)$ we have
\begin{align*}
0=f(k^2-k^2)=f(k^2+kf(y))=kf(k+y)\stackrel{*}{=}kf(y)=-k^2
\end{align*}hence $k=0$, so $n=m$. Therefore $f$ is injective.
$E(1,x)$ gives $f(1+f(x))=f(1+x),\forall x\in\mathbb{R}$, which is just $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ because of injectivity.
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acegikmoqsuwy2000
767 posts
#12 • 4 Y
Y by Ankoganit, sa2001, Adventure10, Mango247
Considerably different solution
This post has been edited 1 time. Last edited by acegikmoqsuwy2000, Jan 2, 2017, 7:30 AM
Reason: no confirmation needed
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MonsterS
148 posts
#14 • 2 Y
Y by Adventure10, Mango247
Let $P(x,y)$ be the assertion.
If f is constant. we get $f(x)=0$
$P(0,0) \implies f(0)=0$
Suppose there 's $f(k)=0$ and$ k\not=0$
$P(x,0) \implies f(x^2)=xf(x) $
set $x=k \implies f(k^2)=0 $
$P(x,k^2)$and$P(x,0) \implies f(x^2)=xf(x+k^2)=xf(x)  \implies f(x)=f(x+k^2)$
$P(x,k)$and$P(x,0) \implies f(x^2)=xf(x+k)=xf(x) \implies f(x)=f(x+k)$
$P(k,x) \implies f(k^2+kf(x))=kf(x+k) \implies f(kf(x))=kf(x)$ _(1)
And easy to see $f$ is bijections.
from $(1)$ we get $f(kx)=kx \implies f(x)=x $contradictions if $x=k$
So $f(z)=0$ if only if $z=0$
$P(x,-x) \implies f(x^2+xf(-x))=0 so f(x)=x$
This post has been edited 1 time. Last edited by MonsterS, Mar 9, 2017, 5:31 PM
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TRYTOSOLVE
255 posts
#15 • 2 Y
Y by Adventure10, Mango247
guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$.
$p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant)
Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is$\boxed {f(x)=x}$
This post has been edited 2 times. Last edited by TRYTOSOLVE, Mar 11, 2017, 5:17 PM
Reason: so so
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pco
23511 posts
#16 • 1 Y
Y by Adventure10
TRYTOSOLVE wrote:
guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$.
$p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(x,\omega)$ implies that $f(\omega)=f(\omega+y)$

For me, $p(x,\omega)$ implies $f(x^2)=xf(x+\omega)$ and not $f(\omega)=f(\omega+y)$
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TRYTOSOLVE
255 posts
#17 • 2 Y
Y by Adventure10, Mango247
ups oh yes I will сhange it now
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TRYTOSOLVE
255 posts
#18 • 2 Y
Y by Adventure10, Mango247
it is not easy like I thought.
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TRYTOSOLVE
255 posts
#19 • 2 Y
Y by Adventure10, Mango247
Is my solution true guys?
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TRYTOSOLVE
255 posts
#20 • 2 Y
Y by Adventure10, Mango247
.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$.
$p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant)
Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is$\boxed {f(x)=x}$
This post has been edited 3 times. Last edited by TRYTOSOLVE, Mar 12, 2017, 4:02 AM
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TRYTOSOLVE
255 posts
#21 • 2 Y
Y by Adventure10, Mango247
guys I can not find mistake in my solution.What about you?
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bobthesmartypants
4337 posts
#22 • 1 Y
Y by Adventure10
TRYTOSOLVE wrote:
If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant)
And how might this show that $f(x)=0$ for all $x\in\mathbb{R}$?
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sa2001
281 posts
#23 • 2 Y
Y by Adventure10, Mango247
........................................
This post has been edited 15 times. Last edited by sa2001, Feb 10, 2018, 7:37 AM
Reason: Incorrect solution
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sa2001
281 posts
#24 • 2 Y
Y by Adventure10, Mango247
...............
This post has been edited 1 time. Last edited by sa2001, Feb 10, 2018, 7:37 AM
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Vrangr
1600 posts
#25 • 3 Y
Y by Arhaan, Adventure10, Mango247
Here's a solution in the vain of INMO 2015 P3. (proving injectivity at $0$)
Solution
This post has been edited 3 times. Last edited by Vrangr, Feb 19, 2018, 1:48 PM
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Vrangr
1600 posts
#26 • 3 Y
Y by math-o-enthu, Adventure10, Mango247
@2above and @3above, your solution seems to be true but needlessly complicated in places. Great solution nevertheless.
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abbosjon2002
114 posts
#28 • 2 Y
Y by Adventure10, Mango247
Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$. $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective
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sa2001
281 posts
#29 • 1 Y
Y by Adventure10
abbosjon2002 wrote:
Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$. $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective

I think you mean $f \not\equiv 0$. Then, yes you can.
This post has been edited 1 time. Last edited by sa2001, Feb 10, 2018, 7:24 PM
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ThE-dArK-lOrD
4071 posts
#30 • 2 Y
Y by Adventure10, Mango247
Probably the silliest way to solve this problem:

As usual, let $P(x,y)$ denote $f(x^2+xf(y))=xf(x+y)$ for all $x,y\in \mathbb{R}$.
$P(0,0)$ easily gives us $f(0)=0$ and then $P(x,0)$ gives $f(x^2)=xf(x)$.

Suppose there exists non-zero real number $a$ that $f(a)=0$.
$P(x,a)$ gives us $f(x^2)=xf(x+a)=xf(x)$ for all $x\in \mathbb{R}$. So, $f(x)=f(x+a)$ for all $x\in \mathbb{R} -\{ 0\}$.
Since $f(0)=f(a)=0$, we conclude that $f(x)=f(x+a)$ for all $x\in \mathbb{R}$.
We can easily prove by induction that $f(x)=f(x+na)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}^+$.
$P(x+a,y)$ gives us $f(x^2+2xa+a^2+xf(y)+af(y))=(x+a)f(x+a+y)=(x+a)f(x+y)$ for all $x,y \in \mathbb{R}$.
So, $f\Big( x^2+xf(y) +a(2x+a+f(y))\Big) =xf(x+y)+af(x+y)$ for all $x,y \in \mathbb{R}$.
Plugging in $x$ by $t=\frac{n-a-f(y)}{2}$ where $n$ is a positive integer gives us
$$f(t^2+tf(y)+na)=f(t^2+tf(y))=tf(t+y)=tf(t+y)+af(t+y)\implies f(t+y)=0.$$Hence, we get that for any $a\neq 0$ that $f(a)=0$, $f\Big( \frac{n-a-f(y)}{2}+y\Big) =0$ for all positive integer $n$ and real number $y$.

Suppose there exists non-zero real number $a$ that $f(a)=0$.
We've $f\Big( \frac{1-a-f(1)}{2}+1\Big) =0$ and $f\Big( \frac{2-a-f(1)}{2}+1\Big) =0$.
From the preceding paragraph, we've $f(x)=f(x+\frac{1-a-f(1)}{2}+1 )=f(x+\frac{2-a-f(1)}{2}+1)$ for all $x\in \mathbb{R}$.
So, $f(x)=f(x+\frac{1}{2})$ for all real number $x$. From $f(0)=0$, we can easily prove that $f(m)=0$ for all integer $m$.
Hence, we've $f(x)=f(x+n)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}$.
Using the preceding paragraph again, we've $f\Big( \frac{n-m-f(y)}{2}+y\Big) =0$ for all $n\in \mathbb{Z}^+,m\in \mathbb{Z},y\in \mathbb{R}$.
So, $f(y-\frac{f(y)}{2})=0$ for all real number $y$. This gives $f(n^2+nf(y)-\frac{nf(n+y)}{2})=f(n^2+\frac{nf(y)}{2})=0$ for all $n\in \mathbb{Z}^+$ and $y\in \mathbb{R}$.
So, $f(4+f(y))=f(f(y))=0\implies f(x+f(y))=f(x)$ for all $x,y\in \mathbb{R}$.
And so $f(x+f(y))=f(x)=f(x+f(z))\implies f(f(y)-f(z))=0$ for all $x,y,z\in \mathbb{R}$.

If there exists $k\in \mathbb{R}$ that $f(k)\neq 0$.
For any $r\in \mathbb{R}$, not hard to show that there exists $a,b,c\in \mathbb{R}$ that $b-a=\frac{r}{f(k)}$ and $a+b+c=k$.
Also, there exists $u,v\in \mathbb{R}$ that $f(u)=af(a+b+c)$ and $f(v)=bf(a+b+c)$.
From the preceding paragraph, we get $f\Big( (b-a)f(a+b+c)\Big) =f(r)=0$. This means $f(x)=0$ for all $x\in \mathbb{R}$, contradiction with the existence of $k$.

Hence, we conclude that if there exists non-zero real number $a$ that $f(a)=0$ then $f(x)=0$ for all $x\in \mathbb{R}$.
If there not exists such $a$, $P(x,-x)$ gives $f(x^2+xf(-x))=0\implies x^2+xf(-x)=0$ for all $x\in \mathbb{R}$, this easily gives $f(x)=x$ for all $x\in \mathbb{R}$.
This post has been edited 2 times. Last edited by ThE-dArK-lOrD, Feb 19, 2018, 1:09 PM
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e_plus_pi
756 posts
#31 • 3 Y
Y by HouseofTerror_43, Adventure10, Mango247
Good Problem.
Ankoganit wrote:
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$.

We shall show that $f(x) \equiv x \ \forall x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions to the given functional equation and both of them clearly work.
$  $
Now assume that $f$ is not identically $0$ and hence there exists at least one $\alpha \in \mathbb{R}$ such that $f(\alpha) \neq 0$. Then,
$P(x, \alpha-x) : f(x^2+x \cdot f(\alpha)) = \underbrace{x \cdot f(\alpha)}_{\text{Spans all of} \ \mathbb{R}} \implies f$ is surjective.
$  $
Note that $P(0,0) \implies f(0) = 0$ and $P(x,0) : f(x^2) = x \cdot f(x) \forall  x \ \in \mathbb{R} \dots (\star)$.
Now, suppose there exists $\lambda \in \mathbb{R}$ such that $f(\lambda) = 0 , \lambda \neq 0$. Also by surjectivity of $f$ there must exist $\mu \in \mathbb{R}$ such that $f(\mu) = 1$.
$  $
$P(x , \lambda) : f(x^2) =  \underbrace{x \cdot f(x + \lambda) =f(x)}_{\bigstar}  \implies f$ is periodic.
$P(\lambda , \mu): f(\lambda^2 + \lambda) =  \underbrace{\lambda \cdot f(\lambda + \mu) =\lambda \cdot f(\mu)}_{\text{Periodic}}  = \lambda$.
$  $
But, $f(\lambda^2 + \lambda ) = f(\lambda^2)= \lambda \cdot f(\lambda) = 0 \iff \lambda = 0$. So , $f$ is injective at $0$. Now, the problem easily breaks down as:
$  $
$P(-f(y), y) : f(y) \cdot f(y-f(y))=0 \implies f(y - f(y))=0 \forall y \neq 0 \iff y=f(y) \forall y \neq 0$.
As $f(0) = 0$, we conclude that $\boxed{f(x) \equiv x \forall x \ \in \mathbb{R}}$
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Ali3085
214 posts
#33 • 1 Y
Y by ILOVEMYFAMILY
my short solution: :rotfl:
claim(1):$f $ is surjective
proof:
$P(\frac{x}{f(y)},y-x) \implies f(something)=x$
$\blacksquare$
claim(2): $f$ is injective
proof:
suppose there's $a \neq b : f(a)=f(b)$ note that
$P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$
define $\mathbb{S}=\{a-b | f(a)=f(b)  \}$ then $f(x)=f(x+s)  \forall s \in \mathbb{S}$
now let $f(y_1)=1 ,f(y_2)=2$
$P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$
$\blacksquare$
now just put $P(1,y) \implies f(y)=y$
and we win :D
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Fouad-Almouine
72 posts
#34
Y by
Ankoganit wrote:
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$.

Nice actually and pritty short ;)
Let $P(x,y)$ be the assertion
$$f\left( x^2+xf(y)\right)=xf(x+y)$$Notice that if $f \equiv c$ then $c=0$.
Suppose we are searching for a non-constant solution. In other words, there exist $t \in \mathbb{R}$ with $f(t) \neq 0$, then $P(x,t-x)$ gives
$$ f\left( x^2+xf(t-x)\right)=xf(t) \ \forall  x \in \mathbb{R} $$Thus $f$ is surjective, and $P(0,0)$ gives $f(0)=0$, and by $P(x,0)$ we get
$$f\left( x^2 \right)=xf(x) \quad (1)$$Now for each $z \in \mathbb{R}^*$, there exist $x_0 \in \mathbb{R}^*$ such that $f(x_0)=z \neq 0$, thus $P(-f(x_0),x_0)$ gives
$$0=f(0)=-f(x_0)f\left(x_0-f(x_0)\right)$$because $f(x_0) \neq 0$, it is forced that $f\left(x_0-f(x_0)\right)=0$.
Finally, $P(f(x_0),x_0-f(x_0)$ gives
$$f\left( f(x_0)^2 \right)=f(x_0)^2 $$from $(1)$, we get
$$f(x_0)f\left(f(x_0)\right) = f\left( f(x_0)^2 \right)=f(x_0)^2 $$$$ \Rightarrow f(z) = f\left(f(x_0)\right) = f(x_0) = z \ \forall  z \in \mathbb{R}^* $$Hence $f \equiv x$ for all $x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions. $\blacksquare$
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oVlad
1746 posts
#36 • 1 Y
Y by MS_asdfgzxcvb
Very cool problem! Let $P(x,y)$ be the given assertion. Observe that $P(0,y)$ yields $f(0)=0.$ Thud, $P(x,0)$ implies $f(x^2)=xf(x).$

Note that $f\equiv 0$ is a solution; assume this sin't the case. Then, there exists $c$ such that $f(c)\neq 0$ then by varying $x$ in $P(x,c-x)$ we get that $f$ is surjective.

Claim: If $f(a)=f(b)$ then $f$ is periodic with period $|a-b|.$

Proof: If $f(a)=f(b)$ then by combining $P(x,a)$ and $P(x,b)$ we get \[xf(x+a)=f(x^2+xf(a))=f(x^2+xf(b))=xf(x+b).\]Therefore, for all $x\neq 0,$ we have $f(x+a)=f(x+b).$ But since $f(a)=f(b)$ then $f(x+a)=f(x+b)$ for $x=0$ as well, so $f$ is periodic with period $|a-b|. \ \square$

Case 1: Assume that $f$ is not injective. Then there exist $a\neq b$ such that $f(a)=f(b).$ According to our claim, $f$ is periodic with period $c:=|a-b|\neq 0.$

Now, recall that $f(x^2)=xf(x).$ Combining this with the periodicity of $f$, we can infer that for any integer $n$ we have \begin{align*}\frac{n-c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)&=f\bigg(\frac{n^2+c^2-2nc}{4}\bigg)=f\bigg(\frac{n^2+c^2-2nc}{4}+n\cdot c\bigg)=f\bigg(\frac{n^2+c^2+2nc}{4}\bigg) \\ &=\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}\bigg) =\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}-c\bigg)=\frac{n+c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)\end{align*}Since $c\neq 0$ then $(n-c)/2\neq(n+c)/2$ so we must have $f((n-c)/2)=0.$

But using our claim, since $f(0)=0=f((n-c)/2)$ then $f$ is periodic with period $(n-c)/2.$ In particular, this implies that $f$ is periodic with period $n-c.$ Recall that $n$ could be any integer, so $f$ has periods $1-c$ and $2-c,$ resulting in the fact that $f$ has period $1.$

However, $P(1,y)$ gives us $f(1+f(y))=f(1+y).$ Using the fact that $f$ has period $1,$ we actually have $$f(f(y))=f(1+f(y))=f(1+y)=f(y).$$But since $f$ is surjective, $f(y)$ can take any real value, so we get that $f(x)=x$ for all $x$, contradicting the fact that $f$ is not injective!

Case 2: Assume that $f$ is injective. Well $P(1,y)$ gives us $f(1+f(y))=f(1+y)$ resulting in $f(y)=y$ for all $y.$

Therefore, the only functions are $f\equiv 0$ and $f\equiv\text{id}.$
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trying_to_solve_br
191 posts
#37
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This is one of the best FE's I've seen in a while. Let $P(x,y)$ be the assertion.

$P(0,0)$ implies $f(0)=0$.

First, notice $f$ is surjective by taking $y=k-x$ and $x=L/f(k)$ (as we vary $L$, $f$ goes through all reals).

Then, $y=0$ implies $f(x^2)=xf(x)=-xf(-x)$ and thus $f$ is odd. Now notice that $y=-x$ implies $f(x^2-f(x^2))=0$, and hence if we prove injectivity at 0 we're done, as we'd prove that $f(x^2)=x^2$ and to extend to the negative reals just use it is odd.

To prove injectivity at 0, let $S=\{a_1,...\}$ the set of all real numbers that satisfy $f(a_i)=0$.

But notice that if $a \in S$, then $a^2 \in S$, because of $x=a,y=0$.

Now, to finish, just notice that: $P(x,a_i)$ implies $f(x^2)=xf(x)=xf(x+a_i)$. Thus the function has period $a_i$ ($f(x)=f(x+a_i)$. Now, by surjectivity, take $c$ so that $f(c)=1$, and $P(a_i,c) \implies f(a_i^2+(1.a_i))=f(a_i^2)=a_i.f(a_i+c)=a_i$, because of $f(k+a_i)=f(k)$. Thus $f(a_i^2)=0=a_i$, and we're done. $\blacksquare$
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CT17
1481 posts
#38
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Note that $f(x)\equiv 0$ and $f(x)\equiv x$ are both solutions. From now on, we can assume there exist (not necessarily distinct) $u$ and $v$ such that $f(u)\neq 0$ and $f(v)\neq v$. Let $c = |v| - f(|v|)$.

$P(x,u-x)\implies f$ is surjective.

$P(0,0)\implies f(0) = 0$.

$P(x,0)\implies f(x^2) = xf(x)$. In particular $f$ is odd, so $f(|v|)\neq |v|$.

$P\left(-\sqrt{|v|},\sqrt{|v|}\right)\implies f\left(|v| - \sqrt{|v|}f\left(\sqrt{|v|}\right)\right) = 0\implies f(c) = 0$.

$P(x,c)\implies f(x^2) = xf(x+c)\implies f(x) = f(x + c)$ for $x\neq 0$. But $0 = f(0) = f(c)$, so $f(x) = f(x+c)$ for all $x$.

$P(c,y)\implies f(c^2 + cf(y)) = cf(c+y) = cf(y)$. Since $c\neq 0$ and $f$ is surjective the funtion $cf(y)$ is surjective, so $f(c^2 + x) = x$ for all $x$. In particular $f$ is linear with slope $1$, but plugging back in yields no new solutions.
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megarnie
5607 posts
#39
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Let $P(x,y)$ denote the given assertion.

$P(0,0): f(0)=0$.

$P(x,0): f(x^2)=xf(x)$.

Clearly $\boxed{f\equiv 0}$ is a solution. Henceforth assume there exists a $k$ with $f(k)\ne 0$.

Claim: $f$ is surjective.
Proof: $P(x,k-x): f(x^2+xf(k-x))=xf(k)$. Since $f(k)\ne 0$, $xf(k)$ can take on any value, so $f$ is surjective $\blacksquare$.

Claim: If $f(k)=0$, then $k=0$.
Proof: Suppose there exists a $k\ne 0$ with $f(k)=0$.

$P(x,k): f(x^2)=xf(x+k)$.

So $xf(x)=xf(x+k)\implies f(x)=f(x+k)$.

Let $a$ satisfy $f(a)=-k$.

$P(k,a): f(k^2-k^2)=0=kf(a+k)=kf(a)=-k^2$, a contradiction. $\blacksquare$

$P(x,-x): f(x^2+xf(-x))=0\implies x^2=-xf(-x)$. If $x\ne 0$, then $-f(-x)=x\implies \boxed{f(x)=x}$.
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mathscrazy
113 posts
#40
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We prove that $f(x)=x$ and $f\equiv 0$ are the only solutions.
Let $P(x,y)$ be the assertion in $f\left( x^2+xf(y)\right)=xf(x+y)$.

It can be easily checked that the only constant function is $\boxed{f\equiv0}$, which indeed works and is our first solution.
Let $f$ be non-constant.
Claim : $f$ is injective.
Proof :
Assume possible $f(y_1)=f(y_2)$ and $y_1\neq y_2$.
$P(x,y_1)-P(x,y_2) : f(x+y_1)=f(x+y_2) \forall x$.
Hence, $f$ is periodic with period $|y_2-y_1|$.
Hence, $f$ is bounded above.

But, we can choose very large $x$ and $y$ such that $f(x+y)\neq0$(Note we can select such $y$ because $f\not\equiv 0$) : $f\left( x^2+xf(y)\right)<xf(x+y)$.
Contradiction!
Hence, $f(y_1)=f(y_2) \implies y_1=y_2$.
Hence proved claim!
$P(0,y) : f(0)=0$.
$P(-x,x) : f(x^2-xf(x))=0$.

Hence, $f(x^2-xf(x))=f(0) \overset{\text{injective}}{\implies} x^2-xf(x)=0 \implies f(x)=x \forall x\neq0$.
Combining with $f(0)=0$, we get $\boxed{f(x)=x },\forall x$, which indeed satisfies given equation and is our second solution.

Hence, we are done :D
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ThisNameIsNotAvailable
442 posts
#42
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Ankoganit wrote:
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$.
My solution, without using the injectivity.
Let $P(x,y)$ be the assertion that $f\left( x^2+xf(y)\right)=xf(x+y)$.
$P(0;y) \implies f(0)=0$
$P(x;0) \implies f(x^2)=xf(x)$, which leads to $f$ is odd.
Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution, hence assume that there exist $k$ such that $f(k)\ne 0$.
$P(x;k-x) \implies f(\cdots)=xf(k)$, it follows that $f$ is surjective, so there exists $a \in \mathbb R$ such that $f(a)=0$.

$P(x;a) \implies f(x+a)=f(x) \quad \forall x \in \mathbb R \implies f(x+na)=f(x) \quad \forall n \in \mathbb Z$
$P(x;-x) \implies f(x^2-xf(x))=0 \implies f(a^2)=0 \quad (*)$
$P(x;a^2) \implies f(x+a^2)=f(x) \quad \forall x \in \mathbb R$

Put $x \longrightarrow x+a$ into $f(x^2)=xf(x)$, we have:
$f(x^2+2ax+a^2)=(x+a)f(x+a) \implies f(n^2)=(n+a)f(n)=nf(n) \implies af(n)=0 \quad \forall n \in \mathbb Z$
If $a=0$, from $(*)$ we have $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ is also a solution.

If $f(n)=0 \quad \forall n \in \mathbb Z$:
$P(x;n) \implies f(x+n)=f(x) \quad \forall n \in \mathbb Z$
$P(x+n;y) \implies f(x^2+2nx+n^2+(x+n)f(y))=(x+n)f(x+y+n)$
$\implies  f(x^2+2nx+(x+n)f(y))=(x+n)f(x+y) \quad \forall n \in \mathbb Z$

Because of the surjectivity, for all $x \in \mathbb R, n \in \mathbb Z$, there exists $t$ such that $f(t)=n-x$.
Put $y \longrightarrow t$ into the above equation, we have:
$f(2nx)=(x+n)f(x+t) \quad \forall n \in \mathbb Z$
Put $x \longrightarrow x-n$ into the above equation:
$xf(x+t)=f(2nx) \implies nf(x+t)=0 \implies f(x+t)=0 \quad \forall n \in \mathbb Z$
So all the solutions are $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ or $\boxed{f(x)=x,\forall x\in\mathbb{R}}$.
This post has been edited 1 time. Last edited by ThisNameIsNotAvailable, Apr 10, 2022, 4:38 AM
Reason: Fix a typo
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hood09
117 posts
#43
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let $P(x,y)$ be the natural assertion.
$P(0,0) : f(0)=0$
$p(x,0) : f(x^2)=xf(x)$
if $\exists k \neq 0$ such that $f(k)=0$ then by $P(x,k) : f(x^2)=xf(x+k) \implies f(x)=f(x+k)$ then $f$ is $k-$periodic .
and we have by $P(x+k,-k ) : f((x+k)^2)=(x+k)f(x) \implies f(x^2)=xf(x)+kf(x) \implies k=0$ or $\boxed{\forall x :  f(x)=0}$ is a solution
if $k=0$ then $\forall z\neq 0 : f(z) \neq 0 $ :
$P(-x,x) : f(x^2-xf(x))=0 \implies xf(x)=x^2 \implies f(x)=x$ so we have the second solution $\boxed{\forall x : f(x)=x}$
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MathLuis
1525 posts
#44
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Oly grind back maybe?
Let $P(x,y)$ the assertion of the F.E. clearly if $f$ is constant then $f(x)=0$ works so lets work when $f$ is not constant.
$P(0,x)$
$$f(0)=0$$$P(x,0)$
$$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(f(x),-x)$
$$f(f(x)-x)=0$$$P(x,f(x)-x)$
$$xf(x)=f(x^2)=xf(f(x)) \implies f(x)=f(f(x))$$As $f$ is non-cero there exists $d$ with $f(d) \ne 0$ so by $P(x,d-x)$
$$f(x^2+xf(d-x))=xf(d) \implies f \; \text{surjective}$$Since $f$ is surjective we use that on the last equation by setting $t=f(x)$ to get
$$f(t)=t \; \forall t \in \mathbb R$$Hence $\boxed{f(x)=0,x \; \forall x \in \mathbb R}$ work thus we are done :D
This post has been edited 1 time. Last edited by MathLuis, Jun 4, 2022, 2:39 PM
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VZH
60 posts
#45
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Let $P(x,y)$ be the assertion.
$P(0,0)$: $f(0)=0$
$P(x,0)$: $f(x^2)=xf(x)$

Case 1: $f(x)=0 \iff x=0$
$P(-f(x),x)$: $f(x)f(x-f(x))=0 \implies f(x)=x, \forall x \in\mathbb{R}$, which is a solution.

Case 2: There exists $t\neq0$ such that $f(t)=0$
$P(x,t)$: $f(x^2)=xf(x+t) \implies xf(x+t)=xf(x) \implies f(x)=f(x+t), \forall x \in\mathbb{R}$. This means $f(d)=0 \implies f$ has period $d$.
$f(t^2)=tf(t)=0$, so $f$ has period $t^2$.
Now, $f(x^2+2tx+t^2)=f((x+t)^2)=(x+t)f(x+t)=xf(x)+tf(x)=f(x^2)+tf(x)$. Let $x$ be an integer, then $f(x^2+2tx+t^2)=f(x^2) \implies tf(x)=0 \implies\forall x \in\mathbb{Z}, f(x)=0$, so $f$ has period $1$.
Take a positive integer $k$ that is not a perfect square, $f(k)=\sqrt{k}f(\sqrt{k})=0 \implies f(\sqrt{k})=0$, where $\sqrt{k}$ is irrational.
Thus, $f$ has periods $1$ and $\sqrt{k}$.
We aim to prove $f(x) \equiv 0$. Only need to prove $f(x)=0$ for all $x \in [0,1)$. For fixed irrational number $\alpha$, let $n$ span all integers, $\{n\alpha\}$ is dense in $[0,1)$. Hence $0=f(0)=f(n\sqrt{k}-[n\sqrt{k}])$, and $n\sqrt{k}-[n\sqrt{k}]$ can be any number in $[0,1)$ $\implies f(x)=0, \forall x \in [0,1)$. So $f(x) \equiv 0$, which indeed is a solution.
Conclusion: Solutions are $f(x)=x, \forall x \in\mathbb{R}$ or $f(x) \equiv 0$.

Weird solution I just came up with (any mistakes?).
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ZETA_in_olympiad
2211 posts
#46
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Let $P(x,y)$ be the given assertion.

$P(0,0)$ gives $f(0)=0.$ If constant then $f\equiv 0.$ If not then $P(x,y-x)$ gives $f$ is surjective. Now let $f(u)=f(v).$ Comparing $P(z,u)$ and $P(z,v)$ gives $f(z)=f(z+w)$ where $w=u-v.$

Take $f(m)=0$ and $f(n)=1.$ Comparing $P(w,m)$ and $P(w,n)$ forces $w=0.$ So $f$ is injective and $P(1,x)$ gives $f(x)=x.$ Both work.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, Jun 4, 2022, 4:12 PM
Reason: Typo
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ZETA_in_olympiad
2211 posts
#47
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Ali3085 wrote:
my short solution: :rotfl:
claim(1):$f $ is surjective
proof:
$P(\frac{x}{f(y)},y-x) \implies f(something)=x$
$\blacksquare$
claim(2): $f$ is injective
proof:
suppose there's $a \neq b : f(a)=f(b)$ note that
$P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$
define $\mathbb{S}=\{a-b | f(a)=f(b)  \}$ then $f(x)=f(x+s)  \forall s \in \mathbb{S}$
now let $f(y_1)=1 ,f(y_2)=2$
$P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$
$\blacksquare$
now just put $P(1,y) \implies f(y)=y$
and we win :D

Your solution needs a fix since surjectivity holds $\forall f(x)\neq 0.$
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trinhquockhanh
522 posts
#48
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2016 India IMO Training T2P2 wrote:
Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y),\forall x,y\in \mathbb{R}^{(1)}$$
When $f$ is a constant function, we have $\boxed{f(x)=0,\forall x\in \mathbb{R}}$

Now consider the case when $f$ is not a constant function, then $\exists a \in \mathbb{R}$ such that $f(a)\ne 0$

$(1). P(x,a-x)\Rightarrow f(x^2+xf(a-x))=xf(a),$ subtitute $x\rightarrow \dfrac{x}{f(a)}\Rightarrow f(\text{something})=x\Rightarrow f$ is surjective

$(1).P(0,0)\Rightarrow f(0)=0; (1).P(x,0)\Rightarrow f(x^2)=xf(x)$

Assume that there is $c\ne 0$ such that $f(c)=0$

$(1).P(x,c)\Rightarrow f(x^2)=xf(x+c)\Rightarrow f(x)=f(x+c),\forall x\in \mathbb{R}$ (as we already have $f(x^2)=xf(x),\forall x\in \mathbb{R}$)

$(1).P(c,y)\Rightarrow f(c^2+cf(y))=cf(y+c)=cf(y)\Rightarrow f(xc+c^2)=xc\Rightarrow f(x+c^2)=x$

Subtitute $x\rightarrow c-c^2\Rightarrow c-c^2=0\Rightarrow c=1\Rightarrow f(x+1)=x\Rightarrow f(0)=-1,$ a contradiction.

Hence $f(x)=0\Leftrightarrow x=0; (1).P(x,-x)\Rightarrow f(x^2+xf(-x))=0\Rightarrow x^2=-xf(-x)\Rightarrow \boxed{f(x)=x,\forall x\in \mathbb{R}}$
This post has been edited 9 times. Last edited by trinhquockhanh, Jul 12, 2023, 5:45 AM
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jasperE3
11325 posts
#49
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Let $P(x,y)$ be the assertion $f\left(x^2+xf(y)\right)=xf(x+y)$. Note that $\boxed{f(x)=0}$ is a solution, otherwise there is some $j$ with $f(j)\ne0$.

$P(0,0)\Rightarrow f(0)=0$
$P(-f(x),x)\Rightarrow f(x)f(x-f(x))=0\Rightarrow f(x-f(x))=0$ if $f(x)\ne0$, still obviously true when $f(x)=0$
$P(f(x),0)\Rightarrow f\left(f(x)^2\right)=f(x)f(f(x))$
$P(f(x),x-f(x))\Rightarrow f\left(f(x)^2\right)=f(x)^2$
Comparing the last two of these, we get $f(f(x))=f(x)$ if $f(x)\ne0$, but still true when $f(x)=0$.
$P(x,j-x)\Rightarrow f\left(x^2+xf(j-x)\right)=xf(j)$ and so $f$ is surjective. From $f(f(x))=f(x)$ we get $\boxed{f(x)=x}$ which fits.
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