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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
Yesterday at 11:16 PM
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.

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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
Yesterday at 11:16 PM
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
Austrian Regional MO 2025 P4
BR1F1SZ   2
N 13 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
Let $z$ be a positive integer that is not divisible by $8$. Furthermore, let $n \geqslant 2$ be a positive integer. Prove that none of the numbers of the form $z^n + z + 1$ is a square number.

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
13 minutes ago
Austrian Regional MO 2025 P3
BR1F1SZ   1
N 15 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
There are $6$ different bus lines in a city, each stopping at exactly $5$ stations and running in both directions. Nevertheless, for every two different stations there is always a bus line connecting these two stations. Determine the maximum number of stations in this city.

(Karl Czakler)
1 reply
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
15 minutes ago
Austrian Regional MO 2025 P2
BR1F1SZ   2
N 19 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
Let $\triangle{ABC}$ be an isosceles triangle with $AC = BC$ and circumcircle $\omega$. The line through $B$ perpendicular to $BC$ is denoted by $\ell$. Furthermore, let $M$ be any point on $\ell$. The circle $\gamma$ with center $M$ and radius $BM$ intersects $AB$ once more at point $P$ and the circumcircle $\omega$ once more at point $Q$. Prove that the points $P,Q$ and $C$ lie on a straight line.

(Karl Czakler)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
19 minutes ago
Austrian Regional MO 2025 P1
BR1F1SZ   2
N 20 minutes ago by NumberzAndStuff
Source: Austrian Regional MO
Let $n \geqslant 3$ be a positive integer. Furthermore, let $x_1, x_2,\ldots, x_n \in [0, 2]$ be real numbers subject to $x_1 + x_2 +\cdots + x_n = 5$. Prove the inequality$$x_1^2 + x_2^2 + \cdots + x_n^2 \leqslant 9.$$When does equality hold?

(Walther Janous)
2 replies
BR1F1SZ
Apr 18, 2025
NumberzAndStuff
20 minutes ago
positive integers forming a perfect square
cielblue   0
an hour ago
Find all positive integers $n$ such that $2^n-n^2+1$ is a perfect square.
0 replies
cielblue
an hour ago
0 replies
Function equation
LeDuonggg   6
N an hour ago by MathLuis
Find all functions $f: \mathbb{R^+} \rightarrow \mathbb{R^+}$ , such that for all $x,y>0$:
\[ f(x+f(y))=\dfrac{f(x)}{1+f(xy)}\]
6 replies
LeDuonggg
Yesterday at 2:59 PM
MathLuis
an hour ago
A nice and easy gem off of StackExchange
NamelyOrange   0
an hour ago
Source: https://math.stackexchange.com/questions/3818796/
Define $S$ as the set of all numbers of the form $2^i5^j$ for some nonnegative $i$ and $j$. Find (with proof) all pairs $(m,n)$ such that $m,n\in S$ and $m-n=1$.
0 replies
NamelyOrange
an hour ago
0 replies
at everystep a, b, c are replaced by a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)
NJAX   8
N 2 hours ago by Assassino9931
Source: 2nd Al-Khwarizmi International Junior Mathematical Olympiad 2024, Day2, Problem 8
Three positive integers are written on the board. In every minute, instead of the numbers $a, b, c$, Elbek writes $a+\gcd(b,c), b+\gcd(a,c), c+\gcd(a,b)$ . Prove that there will be two numbers on the board after some minutes, such that one is divisible by the other.
Note. $\gcd(x,y)$ - Greatest common divisor of numbers $x$ and $y$

Proposed by Sergey Berlov, Russia
8 replies
NJAX
May 31, 2024
Assassino9931
2 hours ago
Increments and Decrements in Square Grid
ike.chen   23
N 2 hours ago by Andyexists
Source: ISL 2022/C3
In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
[list]
[*] The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.
[*] The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.
[/list]
We say that a tree is majestic if its height is at least $10^6$. Determine the largest $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.
23 replies
+1 w
ike.chen
Jul 9, 2023
Andyexists
2 hours ago
4-var inequality
RainbowNeos   5
N 3 hours ago by RainbowNeos
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
5 replies
RainbowNeos
Yesterday at 9:31 AM
RainbowNeos
3 hours ago
Hard diophant equation
MuradSafarli   2
N 3 hours ago by MuradSafarli
Find all positive integers $x, y, z, t$ such that the equation

$$
2017^x + 6^y + 2^z = 2025^t
$$
is satisfied.
2 replies
MuradSafarli
3 hours ago
MuradSafarli
3 hours ago
An almost identity polynomial
nAalniaOMliO   6
N 3 hours ago by Primeniyazidayi
Source: Belarusian National Olympiad 2025
Let $n$ be a positive integer and $P(x)$ be a polynomial with integer coefficients such that $P(1)=1,P(2)=2,\ldots,P(n)=n$.
Prove that $P(0)$ is divisible by $2 \cdot 3 \cdot \ldots \cdot n$.
6 replies
nAalniaOMliO
Mar 28, 2025
Primeniyazidayi
3 hours ago
Euler's function
luutrongphuc   2
N 3 hours ago by KevinYang2.71
Find all real numbers \(\alpha\) such that for every positive real \(c\), there exists an integer \(n>1\) satisfying
\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]
2 replies
luutrongphuc
6 hours ago
KevinYang2.71
3 hours ago
Wot n' Minimization
y-is-the-best-_   25
N 4 hours ago by john0512
Source: IMO SL 2019 A3
Let $n \geqslant 3$ be a positive integer and let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to 2. Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of
\[
\left|1-\sum_{i \in X} a_{i}\right|
\]is minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ with sum equal to 2 such that
\[
\sum_{i \in X} b_{i}=1.
\]
25 replies
y-is-the-best-_
Sep 23, 2020
john0512
4 hours ago
Find all functions
WakeUp   21
N Apr 30, 2025 by CrazyInMath
Source: Baltic Way 2010
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
21 replies
WakeUp
Nov 19, 2010
CrazyInMath
Apr 30, 2025
Find all functions
G H J
G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2010
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WakeUp
1347 posts
#1 • 4 Y
Y by jhu08, Adventure10, Mango247, and 1 other user
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
This post has been edited 1 time. Last edited by WakeUp, Nov 19, 2010, 8:18 PM
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pco
23508 posts
#2 • 7 Y
Y by Abdollahpour, jhu08, Adventure10, Mango247, and 3 other users
WakeUp wrote:
Let $R$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$

$P(0,x)$ $\implies$ $f(0)(f(x)+x-2)$

If $f(0)\ne 0$, this implies $f(x)=2-x$ which indeed is a solution.

Let us from know consider that $f(0)=0$

$P(x,0)$ $\implies$ $f(x^2)=xf(x)$
Then : $P(x,y)$ $\implies$ $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
Same : $P(y,x)$ $\implies$ $yf(y)+f(xy)=f(x)f((y)+xf(y)+yf(x+y)$
Subtracting implies $(x-y)f(x+y)-f(x)-f(y))=0$

and so $f(x+y)=f(x)+f(y)$ $\forall x\ne y$

Plugging this in $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$, we get $f(xy)=f(x)f(y)+yf(x)+xf(y)$ $\forall x\ne y$

$\iff$ $f(xy)+xy=(f(x)+x)(f(y)+y)$

Let then $g(x)=f(x)+x$. We got :
$g(0)=0$
$g(x+y)=g(x)+g(y)$ $\forall x\ne y$
$g(xy)=g(x)g(y)$ $\forall x\ne y$

From the first, we get $g(-x)=-g(x)$ and so $g(2x+(-x))=g(2x)+g(-x)$ and so $g(2x)=2g(x)$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y$
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

So :
$g(x+y)=g(x)+g(y)$ $\forall x,y$
$g(xy)=g(x)g(y)$ $\forall x,y$
And so, very classical, $g(x)=x$ and $f(x)=0$

Hence the two solutions :
$f(x)=2-x$
$f(x)=0$
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WakeUp
1347 posts
#3 • 3 Y
Y by jhu08, Adventure10, Mango247
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution? Note also $f(x)=-x$ is a solution.
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Solving
185 posts
#4 • 3 Y
Y by jhu08, Adventure10, Mango247
SO im right?

x=y
2f(x^2)=f(x)^2+x(f(x)+f(2x))
2f(0)=f(0)^2
f(0)=0
or
f(0)=1/2
let x=x, y=0
f(x^2)+f(0)=f(x)f(0)+xf(x)
f(0)=1/2
f(x^2)+1/2=f(x)(1/2+x)
f(x)=ax+b
ax^2+b+1/2=(ax/2+ax^2+b/2+bx)
b/2=b+1/2
b=-1
a=2
f(x)=2x-1 is the solution
for
f(0)=0
f(x)=x
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pco
23508 posts
#5 • 3 Y
Y by jhu08, Adventure10, Mango247
WakeUp wrote:
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution?
Yes, :oops:, I wrote too quickly !
First we can see that $g(x)=0$ is a solution (and so $f(x)=-x$ is indeed !
If $g(x)$ is not the all zero function, let then $u$ such that $g(u)\ne 0$. If $u=1$, choose instead $u=-1$.
Then the second equation gives us $g(u)(g(1)-1)=0$ and so $g(1)=1$

Then $g(x(x+1))=g(x)g(x+1)$ (using second equation since $x\ne x+1$) $=g(x)(g(x)+g(1))=g(x)^2+g(x)$
But $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so, $g(x^2)=g(x)^2$ and so the second equation $g(xy)=g(x)g(y)$ is also true if $x=y$
...

WakeUp wrote:
Note also $f(x)=-x$ is a solution.
Yes, :oops: $g(x)=0$ is also a solution (I forgot it)

And so :
$f(x)=0$
$f(x)=-x$
$f(x)=2-x$
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borntobeweild
331 posts
#6 • 4 Y
Y by Jerry37284, jhu08, Adventure10, Mango247
This is just about as interesting as a FE can get while still dying to the standard strategies of plugging stuff in, taking cases, and testing. Nevertheless, it was a fun problem.

Reading this won't teach you anything except for what tricks you should have tried
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Onlygodcanjudgeme
18 posts
#7 • 3 Y
Y by jhu08, EHHSW, Adventure10
put $ (x,y) = (0,0) $ then we take that 1) $ f(0) =0 $ 2)$ f(0) =2 $
1) put $ (x,y) = (x,0) $ then we take that $ f(x^2) = x \cdot f(x) $ and this is odd function .
put $ (x,y) = (x,-x) $ then we take that f(-x) = x , f(x) =0
2)put $ (x,y) = (0,x) $ then we take that f(x) = 2-x
so answer is f(x)=0 , f(x) = -x ,f(x) = 2-x
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Jerry37284
46 posts
#8 • 3 Y
Y by jhu08, Adventure10, Mango247
@Onlygodcanjudgeme :I think you got "$\forall x$ : either $f(x)=0$, either $f(x)=-x$" and not "either $f(x)=0$ $\forall x$ , either $f(x)=-x$ $\forall x$ "
This post has been edited 1 time. Last edited by Jerry37284, Dec 10, 2018, 3:34 AM
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Keith50
464 posts
#9 • 3 Y
Y by jhu08, Mango247, Mango247
Let $P(x,y)$ be the given assertion, $P(0,0)\implies 2f(0)=f(0)^2\implies f(0)=0,2$.
If $f(0)=2$, then $P(0,x)\implies 4=2f(x)+2x\implies f(x)=2-x \ \ \forall x\in\mathbb{R}$.
So, if $f(0)=0,$ $P(x,0)\implies f(x^2)=xf(x) \ \ \ (1)$
$P(x,-x)\implies f(x^2)+f(-x^2)=f(x)f(-x)-xf(x) \ \ \ (2)$
Using $P(x,y)$ and $P(-x,-y)$, we can arrive at \[f(x)f(y)+yf(x)+xf(x+y)=f(-x)f(-y)-yf(-x)-xf(-x-y),\]letting $y=0$, we get $f(x)=-f(-x)$ where $x\ne0$, so $f$ is odd.
From $(2)$, using the fact that $f$ is odd, \[f(x)(f(x)+x)=0\implies f(x)=0,-x.\]Now, assume that there exists $a,b\in \mathbb{R}, a,b\ne 0$ such that $f(a)=0$ and $f(b)=-b$, using $(1)$, $f(a^2)=0, f(b^2)=-b^2$,
using $P(a,b)$ we get \[f(ab)=af(a+b).\]If $f(ab)=-ab$, \[-b=f(a+b).\]If $f(a+b)=0$, then $b=0$, a contradiction.
If $f(a+b)=-(a+b)$, then $a=0$, a contradiction. Thus, when $f(ab)=0$, $f(a+b)=0$ as $a=0$ is a contradiction.
Take $P(a,b)$ and $P(b,a)$, subtracting one from another gives \[f(a^2)-f(b^2)=bf(a)+af(a+b)-af(b)-bf(a+b)\]which simplifies to \[b^2=ab\implies b(a-b)=0\]and so $a=b$ since $b\ne 0$ but this means $f(b)=f(a)=0=-b$, a contradiction.
Hence, we have \[f(x)\equiv 0, 2-x, -x\]as solutions and plugging them into the equation, we see that they indeed satisfy. $\blacksquare$
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Kimchiks926
256 posts
#10 • 2 Y
Y by mkomisarova, jhu08
Solved together with @blastoor
Let $P(x,y)$ denote assertion of given functional equation.

Note that $P(0,0)$ gives us $2f(0) = f(0)^2$, which means that $f(0)=0$ or $f(0)=2$. If $f(0) =2$, then $P(0,x)$ gives us:
\begin{align*} 
 f(0) + f(0) = f(0)f(x) + xf(0) \\
4 = 2f(x) + 2x \\
f(x) = 2 -x 
\end{align*}It is easy to check that function $f(x) =2-x$ works.

From now we assume that $f(0) =0$. Note that $P(x,0)$ gives us:
$$ f(x^2) = xf(x) \qquad (1) $$In relation $(1)$ replacing $x$ by $-x$ yields:
$$ f(x^2) = xf(x) = -xf(-x) \implies -f(x) = f(-x) $$Also note that $P(x,-x)$ gives us:
\begin{align*}
f(x^2) + f(-x^2) = f(x)f(-x) -xf(x) \\
f(x^2) -f(x^2) = -f(x)^2 - xf(x) \\ 
f(x)^2 = -xf(x) 
\end{align*}We conclude that $f(x) = 0$ or $f(x) =-x$. Now we are left to escape pointwise trap. Assume that there exist nonzero real numbers $a, b$ such that $f(a) =-a$ and $f(b) = 0 $. Note that $f(b^2) = bf(b) = 0$ and that $P(b,a-b)$ gives us:
\begin{align*} 
f(b^2) +f(b(a-b))= f(b)f(a-b) + (a-b)f(b) + bf(a) \\ 
f(b(a-b)) = -ab 
\end{align*}If $f(b(a-b)) = b^2 -ab$, then $b^2 =0 $, which is contradiction since $b \ne 0$. On another hand id $f(b(a-b)=0=ab$, then one of the numbers $a,b$ is zero, which is again contradiction.

We conclude that $f(x)=0$, $f(x) = 2 -x$, $f(x) =-x$ are only solutions.
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508669
1040 posts
#11 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$. It is not hard to see they work. Now we show that these are the only such functions.

We see that $P(x, 0) \implies f(0)(f(x) + x - 2) = 0$, so if $f(0) \neq 0$, then $\boxed{f(x) = 2-x}$ which is indeed a solution.

Otherwise, let $f(0) = 0 \dots (1)$. Then by $P(x, 0)$, we yield that $f(x^2) = xf(x) = -xf(-x)$ (by replacing $x$ by $-x$) and so $f$ is odd function. Now $P(x, y) - P(y, x)$ along with $f(x^2) = xf(x)$ gives that $(x-y)(f(x+y)-f(x)-f(y)) = 0$ and so if $x \neq y$, definitely $f(x+y) = f(x) + f(y)$ and so $f$ is additive. Now, we re-arrange few terms in $P(x, y)$.

$P(x, y) \implies f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \implies xf(x) + yf(x) = f(x)f(y) + yf(x) + xf(x) + xf(y) \implies 0 = f(x)f(y) + xf(y) = f(y)(f(x) + x)$ which means that $f(x) = -x$ or $f(x) = 0$ for all reals $x$.

Let us say that $A = \{ x \lvert f(x) = -x, x \neq 0 \}$ and $A = \{ x \lvert f(x) = 0, x \neq 0 \}$. Let $a \in A, b \in B$. We see that $f(ab) = bf(a) + af(b)$, and so here in this case, $f(ab) = b \times -a = a \times 0 = 0$, so either of $a$ or $b$ is $0$, a contradiction to definition of elements belonging to sets $A$ and $B$. Hence, $\lvert A \rvert = 0$ or $\lvert B \rvert = 0$. We see that $f(0) = 0 = -0$. Therefore, we see that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all positive reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$
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jasperE3
11287 posts
#13 • 1 Y
Y by jhu08
Hint

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
$P(0,0)\Rightarrow 2f(0)=f(0)^2$
If $f(0)=2$ then:
$P(0,x)\Rightarrow\boxed{f(x)=2-x}$, which works.
Now assume $f(0)=0$.
$P(x,0)\Rightarrow f(x^2)=xf(x)$
$P(1,x)\Rightarrow f(x+1)=f(x)+f(1)-f(x)f(1)-xf(1)$
We use this recurrence to find $f(1)$.
$P(1,1)\Rightarrow f(2)=f(1)-f(1)^2$
$P(1,2)\Rightarrow f(3)=f(2)-f(1)-f(2)f(1)=f(1)^3-2f(1)^2$
$P(1,3)\Rightarrow f(4)=f(3)-2f(1)-f(3)f(1)=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$
But $f(4)=2f(2)=2f(1)-2f(1)^2$, so we find that $2f(1)-2f(1)^2=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$. Solving, we have $f(1)\in\{-1,0,2\}$.

$\textbf{Case 1: }f(1)=0$
$P(1,x)\Rightarrow f(x)=f(x+1)$
$P\left(x,\frac yx+1\right)-P\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)$ if $x\ne0$, but since it holds for $x=0$, $f$ is additive.
By USAMO 2002/4, since $f(x^2)=xf(x)$ and $f$ is additive, we must have $f(x)=xf(1)$, hence $\boxed{f(x)=0}$ which works.

$\textbf{Case 2: }f(1)=-1$
$P(1,x)\Rightarrow f(x+1)=2f(x)+x-1$
$P(x,1)\Rightarrow f(x^2)+f(x)=xf(x+1)\Rightarrow xf(x)+f(x)=2xf(x)+x^2-x\Rightarrow\boxed{f(x)=-x}$ since $f(1)=-1$, which works.

$\textbf{Case 3: }f(1)=2$
$P(1,x)\Rightarrow f(x+1)=-f(x)-2x+2$
$P(x,1)\Rightarrow(x-1)f(x)=-x^2+x\Rightarrow f(x)=\begin{cases}-x&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ which doesn't work.
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RopuToran
609 posts
#14 • 1 Y
Y by jhu08
Here is my way to solve the first case where $f(0)=0$ :D
$P(x,0)$ gives us $$f(x^2) = xf(x),  \forall x \quad (1)$$$P(y,x)$ gives us $$f(y^2) + f(xy)= f(x)f(y)+xf(y)+yf(x+y),  \forall x,y \quad (2)$$By pluging (1) into (2) then subtracting (2) and the original FE, we got $$(x-y) (f(x)+f(y)) = (x-y) f(x+y), \forall x,y$$, which implies $$ f(x) + f(y) = f(x+y), \forall x \neq y \quad (3)$$From $(1)$, we also have $f(x^2)= -x f(x)$ which leads to $$f(x)= f(-x), \forall x \quad (4)$$By $P(x,-x)$ and using $(4)$, we got $$f(x)^2 = -xf(x), \forall x (5)$$Using $(1)$ and $(3)$, from the origina FE, we have $$f(xy) = f(x)f(y)+xf(y) + yf(x), \forall x \neq y$$, which equivalent to $$ f(x)(f(y)+y) = f(xy) - xf(y), \forall x \neq y \quad (6)$$Case 1: There is a number $k \neq 0$ such that $f(k) = -k$. In $(6)$, let $y=k$, we have $f(kx)=-kx, \forall x \neq k$. Thus $f(x) = -x, \forall x \neq k^2$. In the other hand, $f(k^2) = kf(k) = -k^2$. So, $f(x)=-x, \forall x$.
Case 2: There is no number $k$ other than $0$ such that $f(k) = -k$, which means $f(x) \neq -x, \forall x \neq 0$. With $(5)$, we implies $f(x) = 0, \forall x \neq 0$. In the otherhand, $f(0)=0$, thus $f(x) = 0$.

P/S
This post has been edited 1 time. Last edited by RopuToran, Aug 8, 2021, 4:38 PM
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MathLuis
1522 posts
#15 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Case 1: $f$ is constant.
We will set $f(x)=c$ where $c$ is some real constant. Plugging this on the F.E. we get:
$$2c=c^2+cy+cx \implies c=0 \implies f(x)=0$$Cade 2: $f$ is non-constant.
Let $P(x,y)$ the assertion of the given F.E.
$P(0,0)$
$$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$$Case 2.1: $f(0)=2$
$P(0,x)$
$$4=2f(x)+2x \implies f(x)=2-x$$Case 2.2: $f(0)=0$
$P(x,0)$
$$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(x,-x)$ where $x$ is any non-cero real
$$f(x)^2+xf(x)=0 \implies f(x)=-x$$Since $f(0)=0$ we have that $f(x)=-x \; \forall x \in \mathbb R$
Thus the solutions are:

$\boxed{f(x)=0 \; \forall x \in \mathbb R}$

$\boxed{f(x)=2-x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

Thus we are done :blush:
This post has been edited 1 time. Last edited by MathLuis, Aug 8, 2021, 6:26 PM
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rama1728
800 posts
#16 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

A good problem for oddness of a function and how to tackle pointwise traps. Other steps are natural.

Solution
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JustKeepRunning
2958 posts
#17 • 1 Y
Y by jhu08
A nice exercise for pointwise trap!

The answers are $f\equiv 0, 2-x, -x$. These work.

Denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0,2$.

Case 1: $f(0)=2$.

$P(0,y)$ gives that $f(y)=2-y.$

Case 2: $f(0)=0$

$P(x,0)$ gives that $f(x^2)=xf(x),$ so we have that $f$ is odd. Then $P(x,-x)$ gives that $0=f(x)(f(x)+x),$ so $f(x)=0,-x$. To avoid pointwise trap, suppose that $f(x)=0$ and $f(y)=-y$ for some $x,y\neq 0$. Then from $P(x,y)$ and $P(y,x)$ and subtracting, we get that $xf(x)-yf(y)=yf(x)-xf(y)+(x-y)f(x+y)$. Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.
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RopuToran
609 posts
#18 • 1 Y
Y by jhu08
JustKeepRunning wrote:
Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

How did you simplify the equation into this?
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jasperE3
11287 posts
#19 • 1 Y
Y by jhu08
They used the properties $f(x)=0$ and $f(y)=-y$.
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ZETA_in_olympiad
2211 posts
#20
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Let $P(x,y)$ denote the assertion. Quickly $P(0,x)$ gives $f(x)\equiv 2-x$ or $f(0)=0.$ The former works, so we explore the latter.

$P(x,0)$ gives $f(x^2)=xf(x).$ And so comparing $P(x,y)$ with $P(y,x)$ shows that $f(x+y)=f(x)+f(y)$ for all $x\neq y.$ To conclude $P(x,-x)$ implies $f(x)\in \{0,-x\}$ but since additive $f\equiv 0$ or $f\equiv -x$ and both satisfy.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, Aug 1, 2022, 11:08 AM
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HamstPan38825
8857 posts
#22
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This dies to basically anything.

Setting $x=y=0$ yields $2f(0) = f(0)^2$. If $f(0) = 2$, setting $x=0$ yields $f(x) = 2-x$ immediately.

If $f(0) = 0$, setting $y=0$ yields $f(x^2) = xf(x)$, implying $f$ is odd. Setting $y=-x$ in the original, \[0=f\left(x^2\right) + f\left(-x^2\right) = f(x)f(-x)-xf(x) = -f(x)^2 - xf(x).\]So for each $x$, either $f(x) = 0$ or $f(x) = -x$. There are many ways to resolve the pointwise trap, but here is a really stupid way. By setting $y=x$ we get $f(x)^2 + xf(2x) = xf(x)$, i.e. $xf(2x) = 2xf(x)$ or $f(2x) = f(x)$. Furthermore, by swapping $x$ and $y$, we get \[(x-y)f(x) - xf(x+y)=(y-x)f(y) - yf(x+y)\]so $f(x+y) = f(x)+f(y)$ for all $x \neq y$. Combining this with the previous equation yields that $f$ is Cauchy and bounded below on $x \geq 0$, thus $f$ is linear. We can check that only $f \equiv 0$ works here.
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math-olympiad-clown
27 posts
#23
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case 1. f is a constant function :
c+c=c^2+cy+xc x=y=1 plug in we get c=0
so f(x)=0

case 2. f is not a constant function :
P(0,y) :2f(0)=f(0)f(y)+yf(0)

2-1. if f(0) is not 0 then y=1 plug in we get f(1)=1
P(1,y): 1+f(y)=f(y)+y+f(y+1) and we know that f(y)=2-y

2-2.f(0)=0 : P(0,0): f(x^2)=xf(x) this imply f is a odd function
P(x,-x) : 0=-(f(x)^2)-xf(x) we get f(x)=-x

so the answer is f(x)=0 or 2-x or -x
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CrazyInMath
457 posts
#24
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$f(x)=0$ works, now assume $f$ is not constant
$P(0,0)$ gives $f(0)=0, 2$

If $f(0)=2$, $P(0, x)$ gives $f(x)=2-x$ which works
If $f(0)=0$, $P(x, 0)$ gives $f(x^2)=xf(x)$ so $f$ is odd
then $P(x, -x)$ gives $-f(x)^2-xf(x)=0$ so $f(x)(f(x)-x)=0$, so $f(x)=-x, 0$.
If $f(a)=0$, $f(b)=-b$, by $P(a, b)$ and $P(b, a)$ we have contradiction.
So $f(x)=-x$ or $f(x)=0$ in this case.
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