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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.

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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
Interesting inequalities
sqing   0
6 minutes ago
Source: Own
Let $ a,b>0 $ and $ a+b\leq 1  $ . Prove that
$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-k\left(\frac{a}{b}+\frac{b}{a}\right) \geq 225-2k$$Where $104\geq k\in N^+.$
$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right) \geq 225 $$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-106\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{155}{12}$$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-108\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{26}{3}$$$$\left(\frac{4}{a^2}-1\right)\left(\frac{4}{b^2}-1\right)-110\left(\frac{a}{b}+\frac{b}{a}\right) \geq \frac{17}{4}$$
0 replies
sqing
6 minutes ago
0 replies
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   2
N 8 minutes ago by dangerousliri
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
2 replies
NicoN9
27 minutes ago
dangerousliri
8 minutes ago
Combi Geo
Adywastaken   2
N 32 minutes ago by lakshya2009
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
2 replies
Adywastaken
Yesterday at 3:58 PM
lakshya2009
32 minutes ago
Brilliant guessing game on triples
Assassino9931   1
N an hour ago by Sardor_lil
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
1 reply
Assassino9931
Yesterday at 9:46 AM
Sardor_lil
an hour ago
in n^2-9 has 6 positive divisors than GCD (n-3, n+3)=1
parmenides51   7
N an hour ago by AylyGayypow009
Source: Greece JBMO TST 2016 p3
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
7 replies
parmenides51
Apr 29, 2019
AylyGayypow009
an hour ago
Calculus
youochange   12
N an hour ago by FriendPotato
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
12 replies
youochange
Yesterday at 2:38 PM
FriendPotato
an hour ago
The familiar right angle from the orthocenter
buratinogigle   2
N an hour ago by jainam_luniya
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
2 replies
buratinogigle
4 hours ago
jainam_luniya
an hour ago
a deep thinking topic. either useless or extraordinary , not yet disovered
jainam_luniya   5
N an hour ago by jainam_luniya
Source: 1.99999999999....................................................................1. it this possible or not we can debate
it can be a new discovery in world or NT
5 replies
jainam_luniya
an hour ago
jainam_luniya
an hour ago
Divisibilty...
Sadigly   4
N an hour ago by jainam_luniya
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the sum of their squares divides the square of their product.
4 replies
1 viewing
Sadigly
Yesterday at 9:07 PM
jainam_luniya
an hour ago
ioqm to imo journey
jainam_luniya   2
N an hour ago by jainam_luniya
only imginative ones are alloud .all country and classes or even colleges
2 replies
jainam_luniya
2 hours ago
jainam_luniya
an hour ago
Inequality
Sadigly   5
N an hour ago by jainam_luniya
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
5 replies
Sadigly
May 9, 2025
jainam_luniya
an hour ago
D'B, E'C and l are congruence.
cronus119   7
N 2 hours ago by Tkn
Source: 2022 Iran second round mathematical Olympiad P1
Let $E$ and $F$ on $AC$ and $AB$ respectively in $\triangle ABC$ such that $DE || BC$ then draw line $l$ through $A$ such that $l || BC$ let $D'$ and $E'$ reflection of $D$ and $E$ to $l$ respectively prove that $D'B, E'C$ and $l$ are congruence.
7 replies
cronus119
May 22, 2022
Tkn
2 hours ago
a set of $9$ distinct integers
N.T.TUAN   17
N 2 hours ago by hlminh
Source: APMO 2007
Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.
17 replies
N.T.TUAN
Mar 31, 2007
hlminh
2 hours ago
Asymmetric FE
sman96   13
N 2 hours ago by youochange
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
13 replies
sman96
Feb 8, 2025
youochange
2 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
Source: BMO 2023 Problem 1
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cretanman
430 posts
#1 • 5 Y
Y by liecheaper, Sedro, lian_the_noob12, ItsBesi, ehuseyinyigit
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
This post has been edited 4 times. Last edited by Amir Hossein, May 13, 2023, 1:00 AM
Reason: Fixed source
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Batapan
93 posts
#2 • 8 Y
Y by Mai-san, DESTINY_55, dheerstar12, Elghassemmed, Ahsan_Khalid, NicoN9, Rayanelba, ehuseyinyigit
Let $P(x,y)$ denote the given assertion.
$P(x,x) \implies f(x+f(x))=0$
$P(0,y) \implies f(f(0))=0$
$P(f(0),y) \implies f(0)f(f(0)+f(y))=(y-f(0))f(0)$

So $f(0)=0$ or $ f(f(0)+f(y))=(y-f(0))$

If $ f(f(0)+f(y))=(y-f(0))$, setting $y \rightarrow f(y)+y \implies f(y)=-y+f(0)$ and checking we see that $f(x)=-x+k$ truly satisfies the condition
Suppose now $f(0)=0$

$P(x,0) \implies f(f(x))=-f(x)$
So $P(x,y)$ becomes $xf(x+f(y))=(x-y)f(x)$
So $P(x,f(y)+y)$ gives us $f(x)=0$ or $f(x)=-x$

Finally $f(x)=0, f(x)=-x+k$ where $k$ is a constant
This post has been edited 3 times. Last edited by Batapan, May 10, 2023, 4:13 PM
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Assassino9931
1342 posts
#3 • 5 Y
Y by Mai-san, TheEmpress, KAME06, IchliebeMathematik, NicoN9
Suppose firstly that $f(f(x)) = 0$ for all $x$. Then $f(x+f(y)) = 0$ for all $x\neq 0$. If $f(0) = 0$, then $y=0$ gives $f(x) = 0$ for all $x$ and this is a solution; if $f(0) \neq 0$, then $x=-f(y))$ gives a contradiction.

So we may assume that $f(f(x)) \neq 0$ for some $x$. But then $f$ is injective, as if $f(y) = f(z)$, then $(y-x)f(f(x)) = (z-x)f(f(x))$ and so $y=z$.
Now $x=0$, $y=1$ gives $f(f(0)) = 0$ while $y=x \neq 0$ gives $f(x+f(x)) = 0$ and hence $f(x) = f(0) - x$ for $x\neq 0$ (but this also trivially holds for $x=0$). Conversely, direct substitution shows that $f(x) = k-x$ for any constant $k$ is a solution.
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grupyorum
1418 posts
#4 • 1 Y
Y by pro_dynamite
Let $P(x,y)$ denotes the given assertion. Then $P(x,x)$ yields $f(x+f(x))=0$ for all $x\ne 0$. Further, $P(0,x)$ gives with $x\ne 0$ that $f(f(0))=0$. Combining, $f(x+f(x))=0$ holds for all $x$. Now, let there is an $x_0$ such that $f(f(x_0))\ne 0$. If $f(y_1)=f(y_2)$ for some $y_1,y_2$, then considering $P(x_0,y_1)$ and $P(x_0,y_2)$ we obtain $y_1=y_2$: $f$ is injective. Using $f(x+f(x))=0$, we get $x+f(x)=c$ for some constant $c$, which yields $f(x)=-x+c$. Lastly, suppose $f(f(x))=0$ for all $x$, so that $xf(x+f(y)) = 0$ for all $x,y$. Taking $y=f(r)$, we get $xf(x) =0$ for all $x$, so $f(x)=0$ for $x\ne 0$. Lastly, taking $x\ne 0$ and using $f(f(x)) = 0$, we get $f(0)=0$, too. So $f\equiv 0$ identically.
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VicKmath7
1389 posts
#5
Y by
Solution
This post has been edited 2 times. Last edited by VicKmath7, May 11, 2023, 11:11 AM
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Mathlover_1
295 posts
#6 • 1 Y
Y by Math.1234
cretanman wrote:
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
North Macedonia
Let $P(x,y)$ denote the given assertion.
1)Let for all real $x$ $f(f(x))=0$ then from $P(x,y)$ we get $f(x+f(y))=0$ $x=a-f(y)$ $\rightarrow$ for all a $f(a)=0$ $f \equiv 0$
2)For same $x$ $f(f(x))$$\neq$$0$
Claim:Function is injective
Prove: Assume that for some $a$$\neq$$b$ $f(a)=f(b)$ then $( f(f(0))$$\neq$$0)$ from $P(x,a)$ and $P(x,b)$ we get $a=b$ contradiction
$P(y,0)$ $\rightarrow$ $f(f(0))=0$
$P(f(0)-f(x),x)$ $\rightarrow$ $(x+f(x)-f(0))×f(f(f(0)-f(x)))=0$ for $x$$\neq$$0$ $f(f(f(0)-f(x)))$$\neq$$0$ because from injectivety we get $f(0)=f(y)$ but it is impossible then for all $x$ $x+f(x)-f(0)=0$ $\rightarrow$ $f(x)=-x+c$ (there $c=f(0)$)
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motannoir
171 posts
#7 • 1 Y
Y by DanDumitrescu
My 100-th post !
The answers are $ f(x)=c-x$ for every x and $f(x)=0$ for every x which clearly work (c is an arbitrary constant)
Let's prove these are the only ones.
We have two cases
Case 1: $f(f(x))=0$ for every x.
Then our $P(x,y)$ becomes $f(x+f(y))=0 $ for every x not 0 and y real,but is easy to see it works also for x=0. But then let a be any real. Put $P(a-f(y),y)$ to get $f(a)=0$ so f is constant 0 function
Case 2: There exists a real such that $f(f(a))$ is not 0. If we put x=0 we get that $f(f(0))=0$ so a is not 0
Then put $P(a,y)$ to get that f is bijective
To finish put $P(x,x)$ to get that $f(x+f(x))=0$ for every x real( because for x=0 we proved already) and by injectivity we get f(x)=c-x and we are done !
This post has been edited 1 time. Last edited by motannoir, May 10, 2023, 7:58 PM
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hood09
117 posts
#8
Y by
Both $f(x) =-x+c$ and $f =0$ are solutions. Let's prove they are the only ones.
Let $f(0)=c$, by $P(0,y)$ we get $f(c)=0$ , and by $P(x,x)$ we get $f(x+f(x))=0$.
If there is a number $a \neq0$ such that $f(a)\neq0$ : applying $P(a,c)$ give us $f(f(a)) \neq 0$. then $P(a,y)$ give us that $f$ is injective so $x+f(x)=c$ thus $f(x)=-x+c.$
Else : for any $x\neq0$ we have $f(x)=0$, so if we suppose $c\neq0$ we can notice that $P(c,0)$ : $cf(2c)=-c^2$ then $c=0$ absurde, so $f(x) =0$ for any $x$.
This post has been edited 1 time. Last edited by hood09, May 10, 2023, 9:03 PM
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steppewolf
351 posts
#9 • 1 Y
Y by Tellocan
Authored by Nikola Velov
This post has been edited 1 time. Last edited by steppewolf, Feb 9, 2025, 10:40 PM
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Ubfo
16 posts
#10
Y by
Put $y=x\neq0$ gives $f(x+f(x))=0$. Let $f(c)=0$.
Case 1: $f(f(x))=0\forall x\in\mathbb{R}$. Then $xf(x+f(y))=0$ so $f(x)=0\forall x\neq -f(y)$. If f is nonconstant then $f(x)=0\forall x$, if constant then $f(x)=f(c)=0$
Case 2: $\exists x: f(f(x))\neq0$. $f(a)=f(b)$ by $P(x,a),P(x,b)$ imply $a=b$ so $f$ is injective. Assume $\exists t:f(t)\neq c-t$ then $$P(c-f(t),t)\Rightarrow(f(t)+t-c)f(f(c-f(t)))=0$$so there's at most one such $t=t_0$. $$P(\frac{t_0}2,t_0)\Rightarrow f(\frac{t_0}2+f(t_0))=f(f(\frac{t_0}{2}))\Rightarrow f(t_0)=c-t_0$$So either $f(x)=c-x$ or $f(x)=0$ for all $x\in \mathbb{R}$, both solutions fit.
This post has been edited 1 time. Last edited by Ubfo, May 12, 2023, 10:34 AM
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F10tothepowerof34
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#11 • 1 Y
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Let $P(x,y)=xf(x+f(y))=(y-x)f(f(x))$

$P(0,x)\Longrightarrow xf(f(0))=0 \Longleftrightarrow f(f(0))=0 \text{, for } x\neq0$ (1)
$P(x,x)\Longrightarrow xf(x+f(x))=0 \Longleftrightarrow f(x+f(x))=0 \text{, for } x\neq0$ (2)
Claim: The function is injective iff $f(f(x))\neq0$
Proof:
Let $f(a)=f(b)$
$P(x,a)$ yields: $xf(x+f(a))+xf(f(x))=af(f(x))$
$P(x,b)$ yields: $xf(x+f(b))+xf(f(x))=bf(f(x))$
Thus $af(f(x))=bf(f(x))$, which implies that the function is injective when $f(f(x))\neq0$ $\square$.

Case 1: $f(f(x))\neq0$
From (1) and (2) we have that: $f(x+f(x))=f(f(0))$, however since the function is injective when $f(f(x))\neq0$, thus $x+f(x)=f(0)\Longleftrightarrow f(x)=-x+f(0)\Longrightarrow f(x)=-x+c$ where $c$ is a constant.
when x is equal to 0

Case 2: $f(f(x))=0$
Since $f(f(x))=0$, the functional equation becomes $xf(x+f(y))=0$. Let $Q(x,y)=xf(x+f(y))$
$Q(x,f(x))$ yields: $xf(x)=0\Longleftrightarrow f(x)=0$ for $x\neq0$
$Q(-f(0),0)$ yields: $-f(0)^2=0\Longleftrightarrow f(0)=0$, thus $f(x)=f(0)=0$ which implies that $f(x)=0,    \forall x \in \mathbb{R}$

So to sum up, $\boxed{f(x)=-x+c} \text{ and } \boxed{f(x)=0},  \forall  x \in \mathbb{R}$
$\blacksquare$
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Marinchoo
407 posts
#12
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Case 1. $f$ isn't injective: $\exists a\neq b\in\mathbb{R}$ such that $f(a) = f(b)$. Comparing $(x,a)$ and $(x,b)$ yields that $f(f(x)) = 0$ $\forall x\in\mathbb{R}$. Plugging in $(x,f(y))$ now gives $xf(x) = 0$ for all $x$, hence $f(x) = 0$ $\forall x\neq 0$. However, $(x,y) = (-f(0),0)$ implies that $-f(0)^2 = 0$ so $f\equiv 0$, which is indeed a solution.

Case 2. $f$ is injective: Plugging in $(x,x)$ for $x\neq 0$ yields $f(x+f(x)) = 0$ and $(x,y) = (0,1)$ gives $f(f(0)) = 0$, so $f(x+f(x)) = 0$ for all $x$, hence $x+f(x) = y+f(y)$ $\forall x,y\in\mathbb{R}$ (due to the injectivity of $f$), whence $f(x) = c-x$ which works for all constants $c$.
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m.ljf2997
11 posts
#13
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First if F(a)=F(b) then a=b prove
P(x,a) and P(x,b) ==> a=b or F(F(x))=0
if F(F(x))=0 ==> xF(x+F(y))=0
P(x-F(y) , y)==> F(x)=0 ∀ x∈R
if a=b ==>
P(0,y) and P(x,x) ==> F(x+F(x))=F(F(0))=0 ==> x+F(x)=F(0)=K
F(x)=K-x
and it's easy to verify that F(x)=0 orF(x)=K-x ∀ K∈R is the Solution.
This post has been edited 2 times. Last edited by m.ljf2997, May 18, 2023, 8:01 PM
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ezpotd
1271 posts
#14
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why am i so scared of fes lmao...

Anyways, $P(0,0)$ gives $f(f(0)) = 0$, now if $f(f(x)) = 0$ for all $x$, we have $f(x) = 0$, otherwise we can take $P(x,a),P(x,b)$ with $f(f(x)) \neq 0, f(a) = f(b)$ to get $(a-x)f(f(x)) = (b-x)f(f(x))$, so $a=b$. Then $P(x,x)$ gives $f(x + f(x) ) = 0 = f(f(0))$, so $x + f(x) = f(0)$, so $f(x) = k - x$. All functions of this form work, so we are done.
This post has been edited 1 time. Last edited by ezpotd, May 19, 2023, 3:05 AM
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Fibonacci_11235
44 posts
#15
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I denote by $P(x, y)$ plugging some $x$ and $y$ into the given equation.
if $f(f(x)) = 0$ for all $x$ then:
$P(x, f(y)): xf(x) = 0 \implies f(x) = 0$ for all $x$.
now Suppose that there exists $a$ such that $f(f(a)) \neq 0$
$P(a, \frac{a+y}{f(f(a))}): af(...) = y$, in other words $f$ is surjective.
$P(0, y): yf(f(0)) = 0 \implies f(f(0))=0$
$P(f(0)-f(y), y): 0=(y+f(y)-f(0))f(f(f(0)-f(y)))$
Now suppose that $f(f(b)) = 0$ for some $b$.
$P(b, y): bf(b + f(y)) = 0$
Case 1: $b \neq 0$
$bf(b + f(y)) = 0 \implies f(x) = 0$ since $f$ is surjective.
Case 2: $b = 0$
$bf(b + f(y)) = 0 \implies 0=0$

Now let's focus on nonzero solutions, we know that for some $y$:
$y + f(y) - f(0) = 0 \Leftrightarrow f(y) = -y + f(0)$ or $f(0) - f(y) = 0 \Leftrightarrow f(y) = f(0)$

suppose that there exists some $c \neq 0$ for which $f(c) = f(0)$
$P(c, y): cf(c + f(y)) = 0 \implies f(x) = 0$ since c is not zero and $f$ is surjective.
But since we are looking for nonzero solutions, such c does not exist.
so $f(y) = -y + f(0)$ or $f(y) = 0$ for all $y$
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