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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)!

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[*]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 Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
2025 Xinjiang High School Mathematics Competition Q11
sqing   2
N 6 minutes ago by sqing
Source: China
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(1+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 16 $$
2 replies
1 viewing
sqing
Saturday at 4:30 PM
sqing
6 minutes ago
Number of functions satisfying sum inequality
CyclicISLscelesTrapezoid   19
N 11 minutes ago by john0512
Source: ISL 2022 C5
Let $m,n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_1,X_2,\ldots,X_m$ be pairwise distinct non-empty, not necessary disjoint subset of $X$. A function $f \colon X \to \{1,2,\ldots,n+1\}$ is called nice if there exists an index $k$ such that \[\sum_{x \in X_k} f(x)>\sum_{x \in X_i} f(x) \quad \text{for all } i \ne k.\]Prove that the number of nice functions is at least $n^n$.
19 replies
1 viewing
CyclicISLscelesTrapezoid
Jul 9, 2023
john0512
11 minutes ago
Inequality em981
oldbeginner   21
N 15 minutes ago by sqing
Source: Own
Let $a, b, c>0, a+b+c=3$. Prove that
\[\sqrt{a+\frac{9}{b+2c}}+\sqrt{b+\frac{9}{c+2a}}+\sqrt{c+\frac{9}{a+2b}}+\frac{2(ab+bc+ca)}{9}\ge\frac{20}{3}\]
21 replies
oldbeginner
Sep 22, 2016
sqing
15 minutes ago
Find the minimum
sqing   29
N 17 minutes ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
29 replies
sqing
Sep 4, 2018
sqing
17 minutes ago
Interesting inequality
sqing   4
N 19 minutes ago by sqing
Source: Own
Let $a,b\geq 0, 2a+2b+ab=5.$ Prove that
$$a+b^3+a^3b+\frac{101}{8}ab\leq\frac{125}{8}$$
4 replies
1 viewing
sqing
2 hours ago
sqing
19 minutes ago
Interesting inequality
sqing   3
N 20 minutes ago by sqing
Source: Own
Let $ (a+b)^2+(a-b)^2=1. $ Prove that
$$0\geq (a+b-1)(a-b+1)\geq -\frac{3}{2}-\sqrt 2$$$$ -\frac{9}{2}+2\sqrt 2\geq (a+b-2)(a-b+2)\geq -\frac{9}{2}-2\sqrt 2$$
3 replies
sqing
an hour ago
sqing
20 minutes ago
Simple triangle geometry [a fixed point]
darij grinberg   50
N 21 minutes ago by ezpotd
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
50 replies
darij grinberg
May 18, 2004
ezpotd
21 minutes ago
Inspired by RMO 2006
sqing   6
N 21 minutes ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
6 replies
sqing
Saturday at 3:24 PM
sqing
21 minutes ago
IMO 2009, Problem 2
orl   143
N 26 minutes ago by ezpotd
Source: IMO 2009, Problem 2
Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$

Proposed by Sergei Berlov, Russia
143 replies
orl
Jul 15, 2009
ezpotd
26 minutes ago
A sharp one with 3 var (2)
mihaig   0
an hour ago
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$a+b+c+\sqrt{abc}\geq4.$$
0 replies
mihaig
an hour ago
0 replies
f(1)f(2)...f(n) has at most n prime factors
MarkBcc168   39
N an hour ago by cursed_tangent1434
Source: 2020 Cyberspace Mathematical Competition P2
Let $f(x) = 3x^2 + 1$. Prove that for any given positive integer $n$, the product
$$f(1)\cdot f(2)\cdot\dots\cdot f(n)$$has at most $n$ distinct prime divisors.

Proposed by Géza Kós
39 replies
MarkBcc168
Jul 15, 2020
cursed_tangent1434
an hour ago
Inspired by 2025 Beijing
sqing   11
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
sqing
Saturday at 4:56 PM
sqing
2 hours ago
A functional equation
super1978   1
N 2 hours ago by CheerfulZebra68
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(x)(y+f(y)))=xf(y)+f(xy) $$for all $x,y \in \mathbb R$
1 reply
super1978
2 hours ago
CheerfulZebra68
2 hours ago
Prove that IMO is isosceles
YLG_123   4
N 4 hours ago by Blackbeam999
Source: 2024 Brazil Ibero TST P2
Let \( ABC \) be an acute-angled scalene triangle with circumcenter \( O \). Denote by \( M \), \( N \), and \( P \) the midpoints of sides \( BC \), \( CA \), and \( AB \), respectively. Let \( \omega \) be the circle passing through \( A \) and tangent to \( OM \) at \( O \). The circle \( \omega \) intersects \( AB \) and \( AC \) at points \( E \) and \( F \), respectively (where \( E \) and \( F \) are distinct from \( A \)). Let \( I \) be the midpoint of segment \( EF \), and let \( K \) be the intersection of lines \( EF \) and \( NP \). Prove that \( AO = 2IK \) and that triangle \( IMO \) is isosceles.
4 replies
YLG_123
Oct 12, 2024
Blackbeam999
4 hours ago
Functional xf(x+f(y))=(y-x)f(f(x)) for all reals x,y
cretanman   58
N Apr 27, 2025 by GreekIdiot
Source: BMO 2023 Problem 1
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
Proposed by Nikola Velov, Macedonia
58 replies
cretanman
May 10, 2023
GreekIdiot
Apr 27, 2025
Functional xf(x+f(y))=(y-x)f(f(x)) for all reals x,y
G H J
G H BBookmark kLocked kLocked NReply
Source: BMO 2023 Problem 1
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SomeonesPenguin
129 posts
#50 • 1 Y
Y by zzSpartan
Cute FE. :) $P(x,y):xf(x+f(y))=(y-x)f(f(x))$

Case 1. $f(f(x)) = 0\ \forall x\in \mathbb R$. We have that $xf(x+f(y)) = 0$. If $f$ is constant, then $f \equiv 0$. If not, there exists $a\neq b \in \text{Im}f$. Take $y$ such that $f(y) = a$ and vary $x$ such that $x+a$ covers $\mathbb R$ and for the case where $x=0$ just swap $a$ with $b$ and take $x=a-b \neq 0$. Hence we get $f \equiv 0$.

Case 2. there exists $a$ such that $f(f(a)) \neq 0$. Suppose that there exists $y_{1}\neq y_{2}$ with $f(y_{1})=f(y_{2})$. From $P(a,y_{1})$ and $P(a,y_{2})$ we get a contradiction. So $f$ is injective.

$P(0,1) \Rightarrow f(f(0)) = 0$. And from $P(x,x)$ with $x\neq 0$ we get that $f(x+f(x)) = 0$ so from injectivity we have $x+f(x)=f(0)$. Notice that $0$ also satisfies this relation so we get that $f(x)=c-x$ which does satisfy the FE.
This post has been edited 1 time. Last edited by SomeonesPenguin, Aug 5, 2024, 3:23 PM
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onyqz
195 posts
#51
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Let $P(x,y)$ denote the given assertion.
Now note that:
$P(x,x)$ gives $xf(x+f(x))=0$
$P(0,x)$ gives $f(f(0))=0$
Moreover, $P(f(0),x)$ and $P(0,x)$ give $f(0)f(f(0)+f(x))=(x-f(0))f(0)$, therefore we check the two cases $f(0)=0$ and $f(0)\neq0$.

Case 1: $f(0)=0$
$P(x,0)$ gives $f(x)=-f(f(x))$
$P(-f(x),x)$ and $P(x,0)$ give $0=(x+f(x))(f(-f(x))$
Case 1.1: $x+f(x)=0$, hence $f(x)=-x, \forall x\in\mathbb{R}$
Case 1.2: $f(-f(x))=0$. $P(x,0)$ and $P(x,-f(x))$ give $xf(x)=f(x)^2+xf(x)$ so $f(x)\equiv 0$.

Case 2: $f(0)\neq 0$ and $f(f(0)+f(x))=x-f(0)$ (which in fact also shows surjectivity).
Let $f(0)=a$, then from $f(f(0))=0$ we know $f(a)=0$.

Claim: $f$ is injective.
Proof: Suppose there exist $b, c \in\mathbb{R}$ s.t. $f(b)=f(c)$. Now look at $P(a,b): a(b-a)=af(a+f(b))=af(a+f(c))=a(c-a)$, therefore $b=c$ and $f$ is injective.

Looking back at $P(x,x)$ and using injectivity finally shows $x+f(x)=a$ or $f(x)=a-x$.
We conclude that $\boxed{f(x)=c-x}$ for some constant $c$ and $\boxed{f(x)\equiv 0}$ are the only solutions. $\square$
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Flint_Steel
38 posts
#52
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Lunch break!
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Warideeb
60 posts
#53
Y by
Let $P(x,y)$ be the assertion of
$xf(x+f(y))=(y-x)f(f(x))$
\(Our\) \(claim\) \(is\) $f \equiv 0$ or $f(x)=u-x$ for some constant $u$
Now $f \equiv 0$ is obvious
We assume $f \not\equiv 0$
Now $P(0,y)$ gives us
$y f(f(0))=0$ so $f(f(0))=0$ Let $f(0)$ \(be\) $u$ then
$f(u)=0$ Now
$P(x,x)$ gives us
$xf(x+f(x))=0$ then $f(x+f(x))=0$
Now as $f \not\equiv 0$ there exists some $c \ne 0$ such $f(f(c)) \ne 0$ Now
$P(c,y)$ gives us
$cf(c+f(y))=(y-c)f(f(c))$ proving its bijectivity
Now then $f(x+f(x))=0=f(u)$
Then $x+f(x)=u$ then $f(x)=u-x$ for some constant $u$
This post has been edited 6 times. Last edited by Warideeb, Aug 15, 2024, 1:28 PM
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kinnikuma
9 posts
#54
Y by
Let $f$ be a solution to this equation.

Letting $x = y$ gives $xf(x + f(x)) = 0$.
Letting $x = 0$ and $y \neq 0$ gives $f(f(0)) = 0$.
Let's then take $y$ such that $f(y) = 0$ : we get $xf(x) = (y-x)f(f(x))$.

We are going to distinguish two cases :

- if there exists $x$ such that $f(f(x)) \neq 0$, then $y = \frac{xf(x)}{f(f(x))} + x$. So if there are two different $y$ such that $f(y) = 0$ then from this equation these two $y$ should be equal, absurd. Therefore there exists at most one $y$ such that $f(y) = 0$ : we have already established that $f(f(0)) = 0$, so this $y$ must be $f(0)$. For $x \neq 0$, we have $f(x+f(x)) = 0$ from the $x=y$ equation. So $x+f(x) = f(0) \Longrightarrow f(x) = -x + f(0)$. Combining with $f(0) = -0 + f(0)$, we can affirm that $f$ is of the form $- x + k$ where $k$ is a constant.

- if not, this means that $f(f(x)) = 0$ for all $x$. Therefore, $xf(x) = 0$. Clearly for $x \neq 0$, we necessarily have $f(x) = 0$. By the way, with $y = 0$, we have $xf(x+f(0)) = 0$ since $f(f(x)) = 0$. Letting $x = -f(0)$ we have $-f(0)^2 = 0$ so $f(0) = 0$. In summary, $f \equiv 0$.

To conclude there are two potential solutions : $f(x) = 0$ and $f(x) = -x + k$ where $k$ is a constant. We easily check that there are indeed solutions.
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Acclab
33 posts
#55
Y by
$P(x,x)$ and $P(0,x)$ gives that $x+f(x)$ is always annihilated by $f$, then $P(x,x+f(x))$ yields that
$$ xf(x)=f(x)f(f(x)) $$giving either $f(x)=0$ or $f(f(x))=x$ for each $x$. If there is $x \neq 0$ such that the latter hold, $f$ is injective, giving $f(x) \equiv c-x$, otherwise we have the constant $0$ function everywhere except $0$ and a quick check reveals at $0$ too, both of which works.
This post has been edited 1 time. Last edited by Acclab, Oct 7, 2024, 3:12 PM
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mathematical717
34 posts
#56
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Let the problem statement be denoted by $P(x,y)$, we have that \[P(x,x): xf(x+f(x))=0 \implies f(x+f(x))=0, \ \ \ \forall x \in \mathbb{R*}\]wherein $\mathbb{R*}$ is reals without $0$. Now we have \[P(-f(y),y): -f(y)f(0)=(y+f(y))f(f(-f(y)))\]where letting $y \ne 0$ we would have, $f(y)f(0)=0$ forcing either $f(0)=0$ otherwise $f(y) \equiv 0 \ \ \forall y \in \mathbb{R*}$ Now if we have $f \equiv 0$ that acts as a solution, assuming there are more we must have some $f(f(x)) \neq 0$ otherwise we have, \[xf(x+f(y))=0 \ \forall (x,y) \in \mathbb{R} \implies xf(x+f(f(0)))=0 \implies xf(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R*}\]however for some $y \ne 0$, we must have that $0=f(f(y))=f(0)$ which indicates identically $0$ a direct contradiction to our assumption. Now we have established our claim so let $f(y)=f(y'), \ y \ne y'$ then we compare $P(x,y)$ and $P(x,y')$ to obtain that $f$ is injective. However since \[P(0,1): f(f(0))=0 \ \land \ f(x+f(x))=0 \ \ \forall x \in \mathbb{R*}\]so we must have $f(x)+x=f(0)$ for some constant $f(0)$ and thus $f(x)=f(0)-x \ \forall x \in \mathbb{R*}$ however we must also have that, $f(0)=f(0)-0$ and so we have $f \equiv f(0)-x \ \ \forall x$.
Thus our solutions are $\boxed{f \equiv 0; f(x)=f(0)-x \ \ \forall x}$ which are trivial to check that they work.
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cosdealfa
27 posts
#59 • 1 Y
Y by pb_ana
Denote by $P(x, y)$ the given assertion.
For an $y \neq 0$ $P(0, y): f(f(0)) = 0$
For an $x \neq 0$ $P(x, x): f(x + f(x))=0$

Case 1: $f$ is injective
$\Rightarrow x + f(x) = f(0)$ so $f(x) = f(0) - x, \forall x \in \mathbb{R} $

Case 2: $f$ is not injective
Then we can find $a, b \in \mathbb{R}$ such that $a \neq b$ and $f(a) = f(b)$

Then for any $x$ such that $x \neq a, x \neq b, x \neq 0$ we have:
$P(x, a) : xf(x + f(a)) = (a-x)f(f(x)) $
$P(x, b): xf(x + f(b)) = (b-x)f(f(x)) $

Assume LHS $\neq 0$ then RHS $\neq 0$. It follows that $a-x = b-x$ which is false, by the way we picked $x$.
So LHS $= 0$ therefore $f(x + f(a)) = 0$. $\forall x$ such that $x \neq a, x \neq b, x \neq 0$.
We are only left to check if $f(a + f(a)), f(b+ f(b)), f(f(a))$ are $0$.
By $P(a,a)$ and $P(b,b)$ we get $af(a + f(a)) = bf(b + f(b))= 0$ If $a, b \neq 0$ we are done. So assume one of them is $0$. If $a = 0$:
$\Rightarrow f(b + f(0)) = 0$ and by the first observation $f(f(0)) =0$. We also have $f(x + f(0)) = 0$ for all $x \neq 0, x\neq b$. So basically, $f \equiv 0$. $\blacksquare$
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iStud
268 posts
#60
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nice problem for P1 level of BMO!

Clearly $\boxed{f(x)=0\quad\forall x\in\mathbb{R}}$ is a solution, so we'll now look for all nonzero solutions. Let $P(x,y)$ be the assertion of the functional equation given.

Take $x=y$, so $xf(x+f(x))=0$. For $x\ne 0$, the equation yields $f(x+f(x))=0\dots(1)$. Now taking $P(0,x)$ easily gives us that $xf(f(0))=0$, which is $f(f(0))=0\dots(2)$ for $x\ne 0$. Suppose there exists $a,b\in\mathbb{R}$ so that $f(a)=f(b)$. Hence, by comparing $P(1,a)$ and $P(1,b)$ yields $a-1=b-1$ $\Longleftrightarrow$ $a=b$ $\Longleftrightarrow$ $f$ is injective. So by applying injectivity at $(1)$ and $(2)$ will implies $x+f(x)=f(0)$ $\Longleftrightarrow$ $f(x)=f(0)-x$. Now by letting $f(0)=c$ for some $c\in\mathbb{R}$, we easily get $\boxed{f(x)=c-x\quad\forall x\in\mathbb{R}}$, which is indeed a solution. Therefore, we are done. $\blacksquare$
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Davud29_09
20 posts
#61
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P(1,0) implies that f(f(0))=0.if f(f(x)) isn't constant then f is injective .And P(x,x) implies f(x+f(x))=0=f(f(0)) and we get f(x)=c-x from injectivity.If f(f(x)) is constant f(f(x))=0 and y>>f(0) implies f(x)=0 Answers: f(x)=0 and f(x)=c-x.
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TestX01
341 posts
#63
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Suppose $f(f(x))=0$ holds for all $x$. Then if $x\neq 0$, we have $f(x+f(y))=0$. Fix $y=y_0$. Then $f(f(y_0))=0$, and $f(f(y_0)+x)=0$ for $x\neq 0$ but $x+f(y_0)$ along with $f(y_0)$ produce entire set of reals. Hence $f$ is $0$. We check easily that this works.

Now, subbing $x=0$ gives $0=yf(f(0))$. Pick $y$ nonzero, then $f(f(0))=0$. Hence, assume that $f(f(x_0))\neq 0$ for some $x_0\neq 0$. Substituting that for $x$, we have
\[f(c+f(y))=c'(y-x_0)\]for constants $c',c,x_0$ all non-zero. $f$ is injective as there is an isolated $f(y)$ on the LHS, noting $c'\neq 0$. $f$ is also surjective because shifting and scaling the reals still gives the reals on the RHS. Hence, there is an unique $k$ such $f(k)=0$. Letting $x=y$ initially, we have $xf(x+f(x))=0$, so when $x\neq 0$, then $f(x+f(x))=0=f(k)$, injectiveness yields $f(x)=k-x$. Of course, when $x=0$, then $f(0)=k=k-0$ too. One checks this works easily.
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awesomeming327.
1735 posts
#64
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The answers are $f(x)=c-x$ and $f(x)=0$. These clearly work, so we will now prove that they are the only functions. Let $P(x,y)$ denote the assertion.

Claim 1: $f(x+f(x))=0$.
If $x\neq 0$ then use $P(x,x)$ to get $xf(x+f(x))=0$. If $x=0$ then use $P(0,1)$ to get that $f(f(0))=0$, which finishes the claim.

If $f(f(x))=0$ for all $x$ then $xf(x+f(y))=0$ for all $x$. Then $P(x,f(x))$ gives $xf(x)=0$ for all $x$. If $x\neq 0$ then $f(x)=0$. If $x=0$ then take $P(-f(0),0)$ to get $-f(0)^2=0$ which implies $f(0)=0$.

If $f(f(x))\neq 0$ for even a single value of $x$ then if $f(y_1)=f(y_2)$, then comparing $P(x,y_1)$ and $P(x,y_2)$ immediately gives $y_1=y_2$, so $f$ is injective. Then, by Claim $1$, $x+f(x)$ is a constant. We are done.
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GreekIdiot
258 posts
#65
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$f$ is injective only for values $t$ such that $f(f(t)) \neq 0$
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jasperE3
11384 posts
#66
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cretanman wrote:
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
Proposed by Nikola Velov, Macedonia

Let $P(x,y)$ be the assertion $xf(x+f(y))=(y-x)f(f(x))$ and $S=\{x\in\mathbb R\mid f(x)=0\}$.

Case 1: $|S|=0$
$P(0,1)\Rightarrow f(f(0))=0$, contradiction.

Case 2: $|S|=1$
Let $k\in S$. Note that $f(x+f(x))=0$, since:
$P(x,x)\Rightarrow f(x+f(x))=0$ for all $x\ne0$
$P(0,1)\Rightarrow f(f(0))=0\Rightarrow f(x+f(x))=0$ for $x=0$
Then $x+f(x)=k$ for all $x$, or $\boxed{f(x)=k-x}$ which satisfies the equation for any $k\in\mathbb R$.

Case 3: $|S|\ge2$
Let $a,b\in S$ with $a\ne b$. Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$.
$P(a,b)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,0)\Rightarrow jf(j)=-jf(f(j))\Rightarrow f(f(j))\ne0$
$P(j,a)\Rightarrow jf(j)=(a-j)f(f(j))$
$P(j,b)\Rightarrow jf(j)=(b-j)f(f(j))$
Comparing, we have $a=b$, contradiction. So no further solutions.
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GreekIdiot
258 posts
#67
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pointwise avoided with this one
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