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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
< KCE = < LCP , 4 circles related, hard version
parmenides51   4
N 12 minutes ago by Sivege
Source: 2019 RMM Shortlist G4, version 2 , generalized
Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. A point $D$ is chosen on the internal bisector of $\angle ACB$ so that the points $D$ and $C$ are separated by $AB$. A circle $\omega$ centered at $D$ is tangent to the segment $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $\Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$.

Poland
4 replies
parmenides51
Jun 18, 2020
Sivege
12 minutes ago
Hard inequality
ys33   1
N 12 minutes ago by sqing
Let $a, b, c, d>0$. Prove that
$\sqrt[3]{ab}+ \sqrt[3]{cd} < \sqrt[3]{(a+b+c)(b+c+d)}$.
1 reply
ys33
an hour ago
sqing
12 minutes ago
Find (a,n)
shobber   71
N 14 minutes ago by MATHS_ENTUSIAST
Source: China TST 2006 (1)
Find all positive integer pairs $(a,n)$ such that $\frac{(a+1)^n-a^n}{n}$ is an integer.
71 replies
shobber
Mar 24, 2006
MATHS_ENTUSIAST
14 minutes ago
too many equality cases
Scilyse   18
N 17 minutes ago by mathfun07
Source: 2023 ISL C6
Let $N$ be a positive integer, and consider an $N \times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.

Prove that the cells of the $N \times N$ grid cannot be partitioned into less than $N$ right-down or right-up paths. For example, the following partition of the $5 \times 5$ grid uses $5$ paths.
IMAGE
Proposed by Zixiang Zhou, Canada
18 replies
Scilyse
Jul 17, 2024
mathfun07
17 minutes ago
Surjective number theoretic functional equation
snap7822   2
N 25 minutes ago by shanelin-sigma
Source: 2025 Taiwan TST Round 3 Independent Study 2-N
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
[list=i]
[*] For all $m, n \in \mathbb{N}$, if $m > n$ and $f(m) > f(n)$, then $f(m-n) = f(n)$;
[*] $f$ is surjective.
[/list]
Find the maximum possible value of $f(2025)$.

Proposed by snap7822
2 replies
snap7822
Yesterday at 12:18 PM
shanelin-sigma
25 minutes ago
Bigger Cyclic Sets Exist?
FireBreathers   0
26 minutes ago
Define the set of numbers $a_1, . . . , a_m$ is $bigger$ than the set of numbers $b_1, . . . , b_n$ if among all inequalities of the form $a_i > b_j$ the number of true inequalities is at least $2$ times greater than the number of false ones. Prove that there do not exist three sets $X, Y, Z$ such that $X$ is $bigger$ than $Y$, $Y$ is $bigger$ than $Z$, $Z$ is $bigger$ than $X$.
0 replies
FireBreathers
26 minutes ago
0 replies
Inequality with 3 variables and a special condition
Nuran2010   8
N 39 minutes ago by sqing
Source: Azerbaijan Al-Khwarizmi IJMO TST 2024
For positive real numbers $a,b,c$ we have $3abc \geq ab+bc+ca$.
Prove that:

$\frac{1}{a^3+b^3+c}+\frac{1}{b^3+c^3+a}+\frac{1}{c^3+a^3+b} \leq \frac{3}{a+b+c}$.

Determine the equality case.
8 replies
Nuran2010
Apr 29, 2025
sqing
39 minutes ago
D1024 : Can you do that?
Dattier   2
N an hour ago by sansgankrsngupta
Source: les dattes à Dattier
Let $x_{n+1}=x_n^2+1$ and $x_0=1$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
2 replies
Dattier
Apr 29, 2025
sansgankrsngupta
an hour ago
4-var inequality
RainbowNeos   3
N an hour ago by RainbowNeos
Given $a,b,c,d>0$, show that
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]
3 replies
RainbowNeos
Yesterday at 9:31 AM
RainbowNeos
an hour ago
4 lines concurrent
Zavyk09   7
N an hour ago by bin_sherlo
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
7 replies
Zavyk09
Apr 9, 2025
bin_sherlo
an hour ago
Generalized mirror problem
Taha1381   8
N an hour ago by Lemmas
Source: Iranian second round/day1/problem1
We have a rectangle with it sides being a mirror.A light Ray enters from one of the corners of the rectangle and after being reflected several times enters to the opposite corner it started.Prove that at some time the light Ray passed the center of rectangle(Intersection of diagonals.)
8 replies
Taha1381
May 2, 2019
Lemmas
an hour ago
4 variables with quadrilateral sides 2
mihaig   5
N an hour ago by mihaig
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}=2.$$Prove
$$\left(a+b+c+d-2\right)^2+8\geq3\left(abc+abd+acd+bcd\right).$$
5 replies
mihaig
Apr 29, 2025
mihaig
an hour ago
Consecutive sum of integers sum up to 2020
NicoN9   1
N an hour ago by Mathzeus1024
Source: Japan Junior MO Preliminary 2020 P2
Let $a$ and $b$ be positive integers. Suppose that the sum of integers between $a$ and $b$, including $a$ and $b$, are equal to $2020$.
All among those pairs $(a, b)$, find the pair such that $a$ achieves the minimum.
1 reply
NicoN9
5 hours ago
Mathzeus1024
an hour ago
IMO 2023 P2
799786   91
N 2 hours ago by ND_
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
91 replies
799786
Jul 8, 2023
ND_
2 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
11282 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
1521 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
This post has been edited 3 times. Last edited by rama1728, Aug 8, 2021, 7:14 PM
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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
11282 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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