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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:
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[*]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]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Suggestion Form
jwelsh   0
May 6, 2021
Hello!

Given the number of suggestions we’ve been receiving, we’re transitioning to a suggestion form. If you have a suggestion for the AoPS website, please submit the Google Form:
Suggestion Form

To keep all new suggestions together, any new suggestion threads posted will be deleted.

Please remember that if you find a bug outside of FTW! (after refreshing to make sure it’s not a glitch), make sure you’re following the How to write a bug report instructions and using the proper format to report the bug.

Please check the FTW! thread for bugs and post any new ones in the For the Win! and Other Games Support Forum.
0 replies
jwelsh
May 6, 2021
0 replies
k i Read me first / How to write a bug report
slester   3
N May 4, 2019 by LauraZed
Greetings, AoPS users!

If you're reading this post, that means you've come across some kind of bug, error, or misbehavior, which nobody likes! To help us developers solve the problem as quickly as possible, we need enough information to understand what happened. Following these guidelines will help us squash those bugs more effectively.

Before submitting a bug report, please confirm the issue exists in other browsers or other computers if you have access to them.

For a list of many common questions and issues, please see our user created FAQ, Community FAQ, or For the Win! FAQ.

What is a bug?
A bug is a misbehavior that is reproducible. If a refresh makes it go away 100% of the time, then it isn't a bug, but rather a glitch. That's when your browser has some strange file cached, or for some reason doesn't render the page like it should. Please don't report glitches, since we generally cannot fix them. A glitch that happens more than a few times, though, could be an intermittent bug.

If something is wrong in the wiki, you can change it! The AoPS Wiki is user-editable, and it may be defaced from time to time. You can revert these changes yourself, but if you notice a particular user defacing the wiki, please let an admin know.

The subject
The subject line should explain as clearly as possible what went wrong.

Bad: Forum doesn't work
Good: Switching between threads quickly shows blank page.

The report
Use this format to report bugs. Be as specific as possible. If you don't know the answer exactly, give us as much information as you know. Attaching a screenshot is helpful if you can take one.

Summary of the problem:
Page URL:
Steps to reproduce:
1.
2.
3.
...
Expected behavior:
Frequency:
Operating system(s):
Browser(s), including version:
Additional information:


If your computer or tablet is school issued, please indicate this under Additional information.

Example
Summary of the problem: When I click back and forth between two threads in the site support section, the content of the threads no longer show up. (See attached screenshot.)
Page URL: http://artofproblemsolving.com/community/c10_site_support
Steps to reproduce:
1. Go to the Site Support forum.
2. Click on any thread.
3. Click quickly on a different thread.
Expected behavior: To see the second thread.
Frequency: Every time
Operating system: Mac OS X
Browser: Chrome and Firefox
Additional information: Only happens in the Site Support forum. My tablet is school issued, but I have the problem at both school and home.

How to take a screenshot
Mac OS X: If you type ⌘+Shift+4, you'll get a "crosshairs" that lets you take a custom screenshot size. Just click and drag to select the area you want to take a picture of. If you type ⌘+Shift+4+space, you can take a screenshot of a specific window. All screenshots will show up on your desktop.

Windows: Hit the Windows logo key+PrtScn, and a screenshot of your entire screen. Alternatively, you can hit Alt+PrtScn to take a screenshot of the currently selected window. All screenshots are saved to the Pictures → Screenshots folder.

Advanced
If you're a bit more comfortable with how browsers work, you can also show us what happens in the JavaScript console.

In Chrome, type CTRL+Shift+J (Windows, Linux) or ⌘+Option+J (Mac).
In Firefox, type CTRL+Shift+K (Windows, Linux) or ⌘+Option+K (Mac).
In Internet Explorer, it's the F12 key.
In Safari, first enable the Develop menu: Preferences → Advanced, click "Show Develop menu in menu bar." Then either go to Develop → Show Error console or type Option+⌘+C.

It'll look something like this:
IMAGE
3 replies
slester
Apr 9, 2015
LauraZed
May 4, 2019
k i Community Safety
dcouchman   0
Jan 18, 2018
If you find content on the AoPS Community that makes you concerned for a user's health or safety, please alert AoPS Administrators using the report button (Z) or by emailing sheriff@aops.com . You should provide a description of the content and a link in your message. If it's an emergency, call 911 or whatever the local emergency services are in your country.

Please also use those steps to alert us if bullying behavior is being directed at you or another user. Content that is "unlawful, harmful, threatening, abusive, harassing, tortuous, defamatory, vulgar, obscene, libelous, invasive of another's privacy, hateful, or racially, ethnically or otherwise objectionable" (AoPS Terms of Service 5.d) or that otherwise bullies people is not tolerated on AoPS, and accounts that post such content may be terminated or suspended.
0 replies
dcouchman
Jan 18, 2018
0 replies
PE is bisector of BPC
goldeneagle   44
N 6 minutes ago by cursed_tangent1434
Source: Iran TST 2012 -first day- problem 2
Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$.

Proposed by Mr.Etesami
44 replies
1 viewing
goldeneagle
Apr 23, 2012
cursed_tangent1434
6 minutes ago
find question
mathematical-forest   9
N 12 minutes ago by JARP091
Are there any contest questions that seem simple but are actually difficult? :-D
9 replies
mathematical-forest
May 29, 2025
JARP091
12 minutes ago
Interesting inequality
sqing   1
N 24 minutes ago by Zok_G8D
Source: Own
Let $  a, b,c>0,b+c\geq 3a$. Prove that
$$ \sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{1}{\sqrt 2}$$$$ \frac{3}{2}\sqrt{\frac{a}{b+c-a}}-\frac{ 2a^2-b^2-c^2}{(a+b)(a+c)}\geq \frac{2}{5}+\frac{3}{2\sqrt 2}$$
1 reply
sqing
Yesterday at 2:49 AM
Zok_G8D
24 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   6
N 26 minutes ago by Adywastaken
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
6 replies
Maths_VC
May 27, 2025
Adywastaken
26 minutes ago
k cannot make new forum
CuteBaby   10
N Sep 23, 2024 by bpan2021
i cant make new forum
10 replies
CuteBaby
Sep 21, 2024
bpan2021
Sep 23, 2024
Problems From The Book Forum
kingu   2
N Jun 12, 2024 by kingu
I already know that this is a common question, but how does one gain accesses to the PFTB forum's if possible? If it is not, then does this mean that I can create the relevant forum?
I'm thankful for any response.
2 replies
kingu
Jun 12, 2024
kingu
Jun 12, 2024
k Forum question
Kawhi2   3
N May 29, 2024 by KangarooPrecise
How does Aops rank user created forums? Like, do they rank it by posts, or threads?
3 replies
Kawhi2
May 29, 2024
KangarooPrecise
May 29, 2024
k Closing a Dialogue Box Causes Inability to Create Forum
bpan2021   2
N Dec 19, 2023 by jlacosta
Summary of the problem: When making a forum, you must set your forum to be public or private. When you click on the "CREATE" button, you are asked if your forum will be public or private. When you click on the "X" button on the dialogue, you are unable to create the forum anymore, because the "CREATE" button turns unresponsive.
Page URL: https://artofproblemsolving.com/community/category-admin/forum.
Steps to reproduce:
1. Click on the link.
2. Insert random information into the boxes. ...
3. Click the "CREATE" button.
4. Undo that action by pressing the upper "X" above the dialogue box. ...
5. Observe what happens when you try to create the forum afterward.
Expected behavior: The forum will proceed to create when the button is clicked.
Frequency: 6/6 = 100%.
Operating system(s): Windows 11, ChromeOS.
Browser(s), including version: Google Chrome 120.
Additional information: Refreshing allowed me to create a forum again, only if the steps three and four were not followed.
2 replies
bpan2021
Dec 18, 2023
jlacosta
Dec 19, 2023
k How do you make a forum
Blue_banana4   2
N Nov 6, 2023 by Blue_banana4
how do you make a forum
I would like to create a forum that allows people to be free with each other, post about things they think are important, and just generally support each other. I am wondering how one creates a forum.
2 replies
Blue_banana4
Nov 5, 2023
Blue_banana4
Nov 6, 2023
k How to scroll to bottom of topic
Crosstan81   2
N Apr 27, 2023 by Amkan2022
Is there a way to automatically go to the bottom of a topic since it's kind of annoying to have to manually scroll all the way down?
2 replies
Crosstan81
Apr 27, 2023
Amkan2022
Apr 27, 2023
k How to draw these grey lines?
PhilippineMonkey   4
N Dec 11, 2022 by s12d34
Could someone help me to draw such grey line?
4 replies
PhilippineMonkey
Dec 11, 2022
s12d34
Dec 11, 2022
k Private Forum Emptied
Mbean16   4
N Nov 20, 2022 by Sotowa
I made a private forum where I kept my solutions to writing problems from AoPS classes along with feedback and scores, mainly to notice progress and save some of my favorite solutions. I've used the forum for about 6 months, and today I noticed that all of the topics were deleted. Is there a reason for this? Is there anything I can do to reverse it?
4 replies
Mbean16
Nov 20, 2022
Sotowa
Nov 20, 2022
k New Forum?
ZoBro23   3
N Oct 31, 2022 by ZoBro23
Hi,
Is there any way I can create a new AoPS Forum? There seem to be many forums, like the Cubers Forum and the Coding Forum, so can I create a new forum? Or do I need to be an administrator to do that?

Thanks!
3 replies
ZoBro23
Oct 31, 2022
ZoBro23
Oct 31, 2022
k Re-Admin?
megahertz13   2
N Oct 19, 2022 by jlacosta
Hi AoPS Admins, in my forum
2 replies
megahertz13
Oct 19, 2022
jlacosta
Oct 19, 2022
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
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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 $P(-x,x)$.
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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
23515 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)
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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
23515 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.
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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
Y by
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
Y by
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
5611 posts
#39
Y by
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
Y by
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
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 $x,y$.
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
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#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
1557 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
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#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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