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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
Central sequences
EeEeRUT   13
N 3 minutes ago by v_Enhance
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
13 replies
EeEeRUT
Apr 16, 2025
v_Enhance
3 minutes ago
Interesting inequality
sqing   0
10 minutes ago
Source: Own
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+  c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
0 replies
sqing
10 minutes ago
0 replies
IMO Shortlist 2014 C7
hajimbrak   19
N 22 minutes ago by quantam13
Let $M$ be a set of $n \ge 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $AB$ and $CD$ sharing a common interior point and replace them by the segments $AC$ and $BD$ if none of them is present at this moment. Prove that it is impossible to perform $n^3 /4$ or more such moves.

Proposed by Vladislav Volkov, Russia
19 replies
hajimbrak
Jul 11, 2015
quantam13
22 minutes ago
<BAC = 2 <ABC wanted, AC + AI = BC given , incenter I
parmenides51   3
N an hour ago by LeYohan
Source: 2020 Dutch IMO TST 1.1
In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.
3 replies
parmenides51
Nov 21, 2020
LeYohan
an hour ago
China South East Mathematical Olympiad 2014 Q3B
sqing   5
N an hour ago by MathLuis
Source: China Zhejiang Fuyang , 27 Jul 2014
Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$
5 replies
sqing
Aug 17, 2014
MathLuis
an hour ago
Parallelograms and concyclicity
Lukaluce   32
N an hour ago by v_Enhance
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
32 replies
Lukaluce
Apr 14, 2025
v_Enhance
an hour ago
Gcd of N and its coprime pair sum
EeEeRUT   18
N an hour ago by lksb
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
18 replies
EeEeRUT
Apr 16, 2025
lksb
an hour ago
Easy right-angled triangle problem
gghx   7
N 2 hours ago by LeYohan
Source: SMO open 2024 Q1
In triangle $ABC$, $\angle B=90^\circ$, $AB>BC$, and $P$ is the point such that $BP=BC$ and $\angle APB=90^\circ$, where $P$ and $C$ lie on the same side of $AB$. Let $Q$ be the point on $AB$ such that $AP=AQ$, and let $M$ be the midpoint of $QC$. Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$.
7 replies
gghx
Aug 3, 2024
LeYohan
2 hours ago
ai+aj is the multiple of n
Jackson0423   2
N 2 hours ago by Jackson0423
Consider an strictly increasing sequence of integers \( a_n \).
For every positive integer \( n \), there exist indices \( 1 \leq i < j \leq n \) such that \( a_i + a_j \) is divisible by \( n \).
Given that \( a_1 \geq 1 \), find the minimum possible value of \( a_{100} \).
2 replies
Jackson0423
Yesterday at 12:41 AM
Jackson0423
2 hours ago
Triple mixtilinear then sum of segments
Noob_at_math_69_level   5
N 3 hours ago by awesomeming327.
Source: DGO 2023 Team P4
Let $\triangle{ABC}$ be an acute triangle with the $A,B,C-$mixtilinear incircles are $\Omega_A,\Omega_B,\Omega_C$ respectively. $\Omega_A$ is tangent to the circumcircle of $\triangle{ABC}$ at $X$. $O_2,O_3$ are the centers of circles $\Omega_B,\Omega_C$ respectively. Suppose the reflection of line $BX$ over $BO_3$ intersects the reflection of line $CX$ over $CO_2$ at point $S.$ Prove that: $BS+BX=CS+CX.$

Proposed by many authors
5 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
3 hours ago
Inspired by a cool result
DoThinh2001   0
3 hours ago
Source: Old?
Let three real numbers $a,b,c\geq 0$, no two of which are $0$. Prove that:
$$\sqrt{\frac{a^2+bc}{b^2+c^2}}+\sqrt{\frac{b^2+ca}{c^2+a^2}}+\sqrt{\frac{c^2+ab}{a^2+b^2}}\geq 2+\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}.$$
Inspiration
0 replies
DoThinh2001
3 hours ago
0 replies
Basic ideas in junior diophantine equations
Maths_VC   4
N 3 hours ago by TopGbulliedU
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
4 replies
Maths_VC
May 27, 2025
TopGbulliedU
3 hours ago
Iran TST Starter
M11100111001Y1R   8
N 3 hours ago by flower417477
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
8 replies
M11100111001Y1R
May 27, 2025
flower417477
3 hours ago
diophantine with factorials and exponents
skellyrah   2
N 5 hours ago by maromex
find all positive integers $a,b,c$ such that $$ a! + 5^b = c^3 $$
2 replies
skellyrah
Yesterday at 7:56 PM
maromex
5 hours ago
Functional equation on the set of reals
abeker   26
N Apr 29, 2025 by Bardia7003
Source: MEMO 2017 I1
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$.
26 replies
abeker
Aug 25, 2017
Bardia7003
Apr 29, 2025
Functional equation on the set of reals
G H J
G H BBookmark kLocked kLocked NReply
Source: MEMO 2017 I1
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abeker
35 posts
#1 • 3 Y
Y by Amir Hossein, Adventure10, ItsBesi
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$.
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pco
23515 posts
#3 • 5 Y
Y by FC_YangGuifei, vsathiam, Amir Hossein, Adventure10, Mango247
abeker wrote:
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$.
The only constant solution is $\boxed{\text{S1 : }f(x)=0\quad\forall x}$.
So let us from now look only for nonconstant solutions.

Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$

If $\exists u$ such that $f(u)=0$
$P(u,x-u)$ $\implies$ $f(u^2)=uf(x)$
And so $u=0$, else $f(x)=\frac{f(u^2)}u$ is constant.

And since $P(0,0)$ implies $f(f(0)^2)=0$, we get that $f(x)=0$ $\iff$ $x=0$

Comparing $P(x,0)$ with $P(-x,0)$, we get $f(-x)=-f(x)$ $\forall x$
$P(x,-x)$ $\implies$ $f(x^2-f(x)^2)=0$ and so $f(x)=\pm x$ $\forall x$

Suppose now that $\exists x,y$ such that $f(x)=x$ and $f(y)=-y$
$P(x,y)$ $\implies$ $x^2-xy=\pm(x^2+xy)$ which implies $xy=0$ and so

Either $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution

Either $\boxed{\text{S3 : }f(x)=-x\quad\forall x}$ which indeed is a solution
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DerJan
407 posts
#4 • 2 Y
Y by Adventure10, Mango247
Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$.
$P(0,0) \Rightarrow f(f(0)^2)=0$
$P(0,f(0)^2) \Rightarrow f(0)=0$
Now, let $u,v$ satisfy $f(u)=f(v)$ and $v\neq 0$. We have
$$-vf(-v+u)=f(v^2+f(-v)f(u))=f(v^2+f(-v)f(v))=-vf(-v+v) \Rightarrow f(u-v)=f(0)=0$$Note that $f(x)=0\quad\forall x$ is a solution. Assume there exists an $a$, such that $f(a) \neq 0$. We get
$P(u-v,0) \Rightarrow f((u-v)^2)=(u-v)f(u-v)=0$
$P(u-v,a-(u-v)) \Rightarrow 0=(u-v)f(a) \Rightarrow u=v$
Thus, $f$ is injective.
$P(1,y) \Rightarrow f(1+f(1)f(y))=f(1+y) \Rightarrow f(y)=\frac{y}{f(1)}$
From here we easily get the other two solutions.
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MilosMilicev
241 posts
#5 • 1 Y
Y by Adventure10
For $x=0$, we get that $f$ has a zero point. Let $f(c)=0$. Plugging in $x=0, y=c$, we get $f(0)=0$.
Now for $y=0$, for all $x$ we have $f(x^2)=xf(x)$, and by changing $x$ by $-x$ we get that $f$ is odd.
1) There exists t different from $0$ s. t. $f(t)=0$.
For $x=t$, for all $y$, we have $f(t^2)=t*f(y+t)=t*f(t)=0$. So for all $y, f(y+t)=0$, but $y+t$ can get every real value while $y$ varying, so $f$ is a zero function.
2) $f(x)=0 => x=0$:
Take $x,-x$:
$f(x^2+f(x)f(-x))=0$, so $x^2=-f(x)*f(-x)=f(x)^2$, so for all $x, f(x)$ is either $x$ or $-x$.
We can easily check that $f(x)=x, f(y)= -y$ implies $xy=0$.
So $f$ is zero, fixed or $f(x)=-x$ for all $x$. These three functions satisfy the condition.
This post has been edited 3 times. Last edited by MilosMilicev, May 15, 2018, 1:12 PM
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Sylvestra
38 posts
#6 • 2 Y
Y by Adventure10, Mango247
How can I enter submitter information on the problems?
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NathalieShwarz
51 posts
#7 • 1 Y
Y by Adventure10
pco wrote:
abeker wrote:
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$.
The only constant solution is $\boxed{\text{S1 : }f(x)=0\quad\forall x}$.
So let us from now look only for nonconstant solutions.

Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$

If $\exists u$ such that $f(u)=0$
$P(u,x-u)$ $\implies$ $f(u^2)=uf(x)$
And so $u=0$, else $f(x)=\frac{f(u^2)}u$ is constant.

And since $P(0,0)$ implies $f(f(0)^2)=0$, we get that $f(x)=0$ $\iff$ $x=0$

Comparing $P(x,0)$ with $P(-x,0)$, we get $f(-x)=-f(x)$ $\forall x$
$P(x,-x)$ $\implies$ $f(x^2-f(x)^2)=0$ and so $f(x)=\pm x$ $\forall x$

Suppose now that $\exists x,y$ such that $f(x)=x$ and $f(y)=-y$
$P(x,y)$ $\implies$ $x^2-xy=\pm(x^2+xy)$ which implies $xy=0$ and so

Either $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution

Either $\boxed{\text{S3 : }f(x)=-x\quad\forall x}$ which indeed is a solution

I didn't understand the passage from $f(x^2-f(x)^2)=0$ to $f(x)=\pm x$
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pco
23515 posts
#8 • 2 Y
Y by Adventure10, Mango247
NathalieShwarz wrote:
I didn't understand the passage from $f(x^2-f(x)^2)=0$ to $f(x)=\pm x$
I first proved that $f(u)=0$ implies $u=0$
I then proved that $f(x^2-f(x)^2)=0$

So ...
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NathalieShwarz
51 posts
#9 • 1 Y
Y by Adventure10
But I think that we need to prove the injectivity ?
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pco
23515 posts
#10 • 2 Y
Y by Adventure10, Mango247
NathalieShwarz wrote:
But I think that we need to prove the injectivity ?
Have you just read me ? :

(1) I proved first that $f(u)=0$ implies $u=0$
Do you understand this phrase ?
Do you agree the proof ?

If so :
(2) I proved then that $f(x^2-f(x)^2)=0$ $\forall x$
Do you understand this phrase ?
Do you agree the proof ?

If so:
Let $x\in\mathbb R$
(3) Let $u=x^2-f(x)^2$
Then, using (2) above : $f(u)=0$
Then, using (1) above : $u=0$
Then, using (3) above : $f(x)=\pm x$

Is it OK now ?
And no need for injectivity.
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NathalieShwarz
51 posts
#11 • 3 Y
Y by mistakesinsolutions, Adventure10, Mango247
pco wrote:
NathalieShwarz wrote:
But I think that we need to prove the injectivity ?
Have you just read me ? :

(1) I proved first that $f(u)=0$ implies $u=0$
Do you understand this phrase ?
Do you agree the proof ?

If so :
(2) I proved then that $f(x^2-f(x)^2)=0$ $\forall x$
Do you understand this phrase ?
Do you agree the proof ?

If so:
Let $x\in\mathbb R$
(3) Let $u=x^2-f(x)^2$
Then, using (2) above : $f(u)=0$
Then, using (1) above : $u=0$
Then, using (3) above : $f(x)=\pm x$

Is it OK now ?
And no need for injectivity.

hhhhh thank you, I'm kind of stupid
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rkm0959
1721 posts
#13 • 2 Y
Y by Adventure10, Mango247
Denote $P(x,y)$ as the assertion $f(x^2+f(x)f(y)) = xf(x+y)$ and assume that such function $f$ exists.
$f \equiv 0$ is a solution, and it is the only constant solution. Let's look at nonconstant solutions now.

Since $f$ is nonconstant, we may take $\alpha$ such that $f(\alpha)$ is nonzero.
We will use this to show that $f$ is surjective. For any $T \in \mathbb{R}$, take $P \left( \frac{T}{f(\alpha)} , \alpha - \frac{T}{f(\alpha)}. \right)$.

Next, we will show that $f$ is injective. First, we will show that if $f(u)=0$ if and only if $u=0$.
First, take $P(0,y)$ to have $f(f(0)f(y))=0$ for all $y$.
If $f(0) \neq 0$, by the fact that $f$ is surjective we get $f(x) \equiv 0$ for all $x$.
This is an obvious contradiction, so $f(0)=0$. Now if $f(u)=0$ and $u \neq 0$, by $P(u,y)$ we have $f(u^2) = uf(u+y)$, so again, contradiction to surjective condition. We now have $f(u)=0 \iff u=0$.

To improve this to injectivity, assume $f(u)=f(v)$ and $u \neq v$. Then by $P(x,u)$ and $P(x,v)$, it is easy to note that $f$ is a periodic function with period $|u-v|$. Since $f(|u-v|)$ is nonzero, this cannot hold. Therefore, $f$ is injective.

Now $P(x,-x)$ gives $f(x^2+f(x)f(-x))=0$ and $P(x,0)$ gives $f(x^2)=xf(x)$.
So by injectivity we have $x^2+f(x)f(-x)=0$, and $-f(-x)=f(x)$.
This gives us that $f(x)$ is equal to either $x$ or $-x$. Now we take care of the "point-wise trap".

Assume nonzero reals $u, v$ exist such that $f(u)=u$ and $f(v)=-v$. $P(u,v)$ gives the desired contradiction.

So in the end we have three solutions, $f \equiv 0$, $f \equiv x$, and $f \equiv -x$. These clearly work, done.
This post has been edited 3 times. Last edited by rkm0959, Feb 19, 2018, 1:15 PM
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Mathematicsislovely
245 posts
#14 • 1 Y
Y by ismayilzadei1387
Denote $P(x,y)$ the assertion of the question.Observe that $f\equiv 0$ is the only constant solution of $f$.Now assume that $f$ is non-constant.

$P(0,0)\implies f(f(0)^2)=0$.So assume for some $t_0, f(t_0)=0$.

$P(t_0,x-t_0)\implies f(t_0^2)=t_0f(x)$.If $t_0\ne 0$ then $f(x)=\frac{f(t_0^2)}{t_0}=\text {constant}$.Which is a contradiction.
Hence we get $f(x)=0\iff x=0$.

Now we claim that $f$ is injective function.Suppose for some $a,b\in \mathbb R,f(a)=f(b)$.
Then $P(x,a)$ and $P(x,b)$ implies,
$f(x^2+f(x)f(a))=f(x^2+f(x)f(b))\\
\iff xf(x+a)=xf(x+b)\\
\iff f(x+a)=f(x+b)$.

Putting, $x=-a$ in the last equation we get $f(b-a)=0\iff b=a$.

Now, $P(1,y)$ implies $f(1+f(1)f(y))=f(1+y)\iff f(y)=cy$ where $c=\frac{1}{f(1)}$.
Putting it in the main equation we get $c=1$ or $c=-1$.

Hence (1)$f\equiv 0$, (2)$f(x)=x\forall x\in \mathbb R$ and (3)$f(x)=-x\forall x\in \mathbb R$ are all the solutions.$\blacksquare$
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MathLuis
1557 posts
#15
Y by
abeker wrote:
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
$$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$.

Assume that $f(x)=c$ for $c$ any constant.
$$c=cx \; \forall x \in \mathbb R \implies c=0 \implies f(x) \equiv 0 \; \forall x \in \mathbb R$$Now take that $f$ is not constant.
Let $P(x,y)$ the assertion of the given FE.
$P(0,x)$
$$f(f(x)f(0))=0  \; \forall x \in \mathbb R \implies f(0)=0$$$P(x,0)$
$$f(x^2)=xf(x) \; \forall x \in \mathbb R$$$P(x,-x)$
$$f(x^2+f(x)f(-x))=0 \; \forall x \in \mathbb R$$Now assume that exists more than 1 cero no the function.
Let $\mathbb C$ the set of the ceros of $f$ and take $c \in \mathbb C$ then $f(c)=0$.
$P(x,c)$
$$f(x^2)=xf(x+c) \; \forall x \in \mathbb R$$Then $f$ is periodic at $c$.
Note that $c^2$ is also a cero by $f(x^2)=xf(x)$.
Then:
$$cf(2c)=0 \implies c=0 \; \text{or} \; f(2c)=0$$Now take $f(2c)=0$ then $f(4c^2)=0$
In fact if we skip the case when $c=0$ then $f(n \cdot c)=0 \; \forall n \in \mathbb Z$
$P(-x,0)$
$$f(x^2)=xf(x)=-xf(-x) \implies f(-x)=-f(x)$$Assume that exists $a,b \ne n \cdot c$ such that $f(a)=f(b)$ then i will prove that $a=b$.
$P(a,-b)+P(-b,a)$
$$f(a^2-f(a)^2)+f(b^2-f(b)^2)=(a-b)f(a-b) \implies a=b$$Then $f$ is injective.
That means exists an only $0$ and hence $c=0$ (contradiction!!) (We assumed that exists more than 1 cero, thats why contradiction)
Hence $f(0)=0$ has an unique $0$
The same proof for get $f$ injective.
Then:
$$x^2-f(x)^2=0 \implies f(x)= \pm x$$Then the solutions are:
$\boxed{f(x)=x \; \forall x \in \mathbb R}$

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

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

Thus we are done :blush:
This post has been edited 1 time. Last edited by MathLuis, May 4, 2021, 1:01 PM
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jasperE3
11395 posts
#16 • 1 Y
Y by Mango247
$\boxed{f(x)=0}$ works, assume now that $\exists j:f(j)\ne0$. Let $P(x,y)$ denote the given assertion.

$P(0,x)\Rightarrow f(f(0)f(x))=0$
$P(0,f(0)f(x))\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(x^2)=xf(x)\Rightarrow f$ is odd

If $\exists k:f(k)=0$:
$P(k,j-k)\Rightarrow kf(j)=f(k^2)=kf(k)=0\Rightarrow k=0$

$P(x,-x)\Rightarrow f(x^2-f(x)^2)=0\Rightarrow f(x)^2=x^2$

The assertion becomes $f(x^2+f(x)f(y))^2=x^2f(x+y)^2$, or $2x^2f(x)f(y)=2x^3y$. Then $f(x)f(y)=xy\forall x\ne0$, but since it also holds for $x=0$, we set $y=1$ to get that $f(x)=cx$ for some constant $c$. Testing, we have that either $\boxed{f(x)=x}$ or $\boxed{f(x)=-x}$.
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hakN
429 posts
#17
Y by
Note that $\boxed{f(x) = 0 \ \forall x \in \mathbb{R}}$ is the only constant solution. Now we look for non-constant solutions.
Let $P(x,y)$ be the assertion.
If there exist a $u\neq 0$ such that $f(u) = 0$, then
$P(u,y-u) \implies \frac{f(u^2)}{u} = f(y)$ and so $f$ is constant, contradiction.
$P(0,x) \implies f(f(0)f(x)) = 0$ so $f(0)f(x) = 0 \implies f(0) = 0$.
$P(x,0) \implies f(x^2) = xf(x) \implies f$ is odd.
$P(x,-x) \implies f(x^2 - f(x)^2) = 0 \implies f(x)^2 = x^2$ so $f(x) \in \{x,-x\}$ for all $x\in \mathbb{R}$.
So we have $\boxed{f(x) = x \ \forall x\in \mathbb{R}}$ and $\boxed{f(x) = -x \ \forall x\in \mathbb{R}}$ as solutions.
Let for some $a , b \neq 0$ we have $f(a) = a$ and $f(b) = -b$.
$P(a,b) \implies f(a^2 - ab) = af(a+b) \in \{a^2 + ab , -a^2 - ab\} \implies ab = 0$, contradiction.
So we have our three solutions.
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logrange
120 posts
#18
Y by
I have a much simpler solution but I don't know that it is correct or not, kindly check.
Substituting $x=0$ in the original equation gives $f(f(0)f(y))=0$
$f(0)f(y)$ can be any real number (except if either $f(0)=0$ or $f(y)=0$) and it is not possible that $f(R)=0$ (R represents every real number/$f(0)f(y)$) and so the above equation is true if and only if either $f(0)=0$ or $f(y)=0$.
Case 1 - $f(x)=0$, it is indeed a solution.
Case 2 - $f(0)=0$
$P(x,-x) -     x^2+f(x)f(-x)=0$
$f(x^2)=xf(x)=-xf(-x)$ which gives $f(-x)=-f(x)$
Combining, we get $f(x)=x$ and $f(x)=-x$
This post has been edited 1 time. Last edited by logrange, May 9, 2021, 7:03 PM
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jasperE3
11395 posts
#19
Y by
Quote:
$f(0)f(y)$ can be any real number
What if $f(x)=x^2$ or $f(x)=\operatorname{sgn}(x)$?
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logrange
120 posts
#20 • 1 Y
Y by Mango247
jasperE3 wrote:
Quote:
$f(0)f(y)$ can be any real number
What if $f(x)=x^2$ or $f(x)=\operatorname{sgn}(x)$?

Nice! Thank you for checking.
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logrange
120 posts
#21
Y by
What if I write finitely many, will my solution become wrong?
This post has been edited 1 time. Last edited by logrange, May 9, 2021, 7:09 PM
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jasperE3
11395 posts
#22 • 1 Y
Y by Mango247
For $f(x)=x^2+1$, there are infinitely many possible values for $f(0)f(y)$ and in the case of $f(x)=1+(-1)^{|x|+2}$, there are finitely many (both cases have $f(0)\ne0$ and $f\not\equiv0$).
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llplp
191 posts
#23
Y by
strange solution
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Fibonacci_11235
45 posts
#24
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I denote by $P(x, y)$ plugging in some $x$ and $y$ into the given equation.
Firstly, suppose that there exists a real number $a$ such that $f(a) \ne 0$, otherwise $f(x) = 0$ for all $x \in \mathbb{R}$
$P(\frac{x}{f(a)}, a-\frac{x}{f(a)}):f(...) = x$ and thus f is bijective.
$P(1, y): f(1 + f(1)f(y)) = f(1+y)$
since f is bijective and therefore injective, we know:
$1+f(1)f(y) = 1+y$
Case 1: $f(1) = 0$
then we have $1 + 0 = 1 + y \implies y = 0$ but that does not hold true for all $y \in \mathbb{R}$ so $f(1) \ne 0$
$1 + f(1)f(y) = 1+y \implies f(y) = \frac{y}{f(1)} = cy$ for some $c \ne 0$
Plugging back in the original equation, we get $c=1$ or $c=-1$ So...
$f(x) = 0$, $f(x) = x$, $f(x)=-x$ are all solutions.
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ezpotd
1306 posts
#25
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Consider $P(x,-x)$ and $P(-x,x)$, this gives $f(x^2 + f(x)f(-x)) = xf(0)= -xf(0)$, thus $f(0) = 0$. Then $P(x,0)$ gives $f(x^2) = xf(x)$, and $f(x) = -f(-x)$. Now consider $a$ with $f(a) = 0$. $P(a,y)$ gives $f(a^2 + f(a)f(y)) = af(a + y)$. The left hand side evaluates to $f(a^2 + f(a)f(y)) = f(a^2) = af(a)  =0$. Thus we are either forced $a = 0$ or $f$ is all zero. Assume the former. Now we show $f$ injective, consider $a,b$ with $a < b$, $f(a) = f(b)$, then $P(x,a), P(x,b)$ forces $xf(x  + a) = xf(x+b)$, so either $f(x + a) = f(x + b)$ or $x = 0$(this second case doesn't matter because $f(a)  = f(b)$ by definition). Thus we can conclude $f(x + a) = f(x + b)$ for all $x$. Then consider $x = -a$, this gives $f(0)=  f(b)$, which is a contradiction. Thus $f$ is injective. Then take $P(1,a)$ to get $f(1)f(a) = a$, so $f$ is linear going thru the origin, testing functions we get $f(x) = x,-x$, along with our original solution of $f = 0$
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rafigamath
57 posts
#26
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First assume that there exists $u\ne 0$ such that $f(u)=0$.$P(u,0)$ implies $f(u^2)=0$.So either 1.$f(x)=0$ for all $x$ or 2.$u=0$.So we proved that $f(0)=0$ and $f$ is injective at $0$(since $f(u^2)=uf(u+y)$ and $f(u^2)=0$ at $y=0$ we can say this).$P(x,0)$ implies $f(x^2)=xf(x)$ so $f$ is odd,and $P(x,-x)$ gives $f(x^2+f(x)f(-x))=0$ $\implies$ $f(x)=x$ or $f(x)=-x$.Plug in $f(x)=x,f(y)=-y$ and this will cause a contradiction so $f(x)=0$;or $f(x)=x$; or $f(x)=-x$ for all $x$.
This post has been edited 2 times. Last edited by rafigamath, Feb 26, 2024, 1:00 PM
Reason: typo
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AshAuktober
1013 posts
#27
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We claim the only solutions are $f(x) \equiv 0  \forall  x$, $f(x) = x  \forall  x$ and $f(x) = -x  \forall  x$. Clearly all of these functions work. Now we prove these are the only such functions.
Let the above assertion be represented by $P(x, y)$.
Clearly the only constant solution is $f(x) \equiv 0$. Now assume $f$ to be nonconstant.
Claim 1: $\boxed{f(0) = 0}$
Proof: Note that $P(0,0)$ gives us $f(f(0)^2) = 0$.
Now $P(0, f(0)^2)$ gives us $f(0) = 0 \square$.
Claim 2: $\boxed{f \text{ is odd}}$
Proof: Observe that $P(x, 0)$ gives us $f(x^2) = xf(x) = -xf(-x) \implies f(-x) = -f(x) \square$
Claim 3: $\boxed{\text{ There exists no nonzero } a \text{ such that } f(a) = 0}$
Proof: Assume FtSoC that such $a$ exists. Then $P(a, y-a)$ gives us $f(a^2) = af(y) \implies f(y) \equiv \frac{f(a^2)}{a}$. However we have assumed $f$ to be nonconstant. Contradiction. $\square$
Claim 4: $\boxed{f(x) = x \text{ or } -x \forall x}$
Proof: $P(x, -x)$ yields $f(x^2 + f(x)f(-x)) = 0 \implies x^2 + f(x)f(-x) = x^2 - f(x)^2 = 0 \implies f(x) = x \text{ or } -x \forall x \square$
Claim 5: $\boxed{ f(x) = x \forall x \text{ or } f(x) = -x \forall x}$
Proof: Assume FtSoC that for some nonzero $m$, $n$ we have $f(m) = m, f(n) = -n$.
Then $f(m^2 - mn) = mf(m+n)$. If $f(m^2 - mn) = m^2 - mn, f(m+n) = m-n, \text{else} f(m+n) = n-m$, both of which lead to a contradiction. $\square$
Therefore we are done.
Q. E. D.
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ItsBesi
147 posts
#30
Y by
Nice one :)

Answer: $f \equiv 0 , f(x)=\pm x \forall x \in \mathbb{R}$

It's easy to see that these work hence we move on.

Solution: Let $P(x,y)$-denote the given assertion.

Note that the only constant solution is $f \equiv 0$ hence assume $f$-is not constant

Claim: $f(0)=0$

Proof:

$P(0,0) \implies f(f(0)^2)=0 ...(1)$

$P(0,f(0)^2) \implies f(0)=0$ $\square$

Claim: $f(x^2)=xf(x)$

Proof:

$P(x,0) \implies f(x^2)=xf(x) \forall x \in \mathbb{R} ...(*)$ $\square$

Claim: $f-\text{odd}$

Proof:

From $(*) \implies f(x^2)=xf(x) , x \rightarrow -x \implies xf(x) \stackrel{(*)}{=} f(x)^2=-xf(x) \implies f(x)=-f(-x) \forall x \in \mathbb{R} / \{0\}.$
Combining with $f(0)=0 \implies f(x)=-f(x) \forall x \in \mathbb{R} \square$.

Claim: $\exists \alpha \in \mathbb{R} : f(\alpha)=0 \implies \alpha=0$ or $f-\text{injective on} 0$

Proof: FTSOC assume $\alpha \neq 0$

From $(*) \implies f(x^2)=xf(x) , x \rightarrow \alpha \implies f(\alpha^2)=\alpha \cdot f(\alpha)=0 \implies f(\alpha^2)=0$

$P(\alpha,1) \implies \alpha (\alpha+1)=0 \implies \alpha=0 \vee f(\alpha+1)=0$. Since we assumed $\alpha \neq 0 \implies f(\alpha+1)=0$

$P(1,\alpha) \implies f(1)=0$

$P(1,x) \implies f(f(x)f(1)+1)=f(x+1) \implies f(1)=f(x+1) \implies f(x+1)=0 , x \rightarrow x-1 \implies f(x)=0 \iff f(x) \equiv 0$.
which is a contradition since we assumed $f$-is not constant.

Hence our assumption is wrong so $\alpha=0$ $\square$

Claim: $f(x)=\pm x$

Proof: $P(x,-x) \implies f(x^2+f(x)f(-x))=0=f(0) \implies f(x^2+f(x)f(-x)=f(0)$
combining with our previous claim we have:

$x^2+f(x)f(-x)=0 \implies f(x)f(-x)=-x^2 \stackrel{f-\text{odd}}{\implies} -f(x)f(x)=-x^2 \implies f(x)^2=x^2 \implies f(x)=\pm x$

DON'T FORGET POINTWISE TRAP
This post has been edited 1 time. Last edited by ItsBesi, Jan 24, 2025, 11:44 AM
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Bardia7003
22 posts
#31
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Let $P(x,y)$ denote the given assertion.
$f \equiv 0$ is a solution, we want to find the other solutions, so we can assume there exists $k$ such that $f(k) \neq 0$
$P(x, k - x): f(x^2 + f(x)f(k-x)) = xf(k)$ so $f$ is surjective.
$(0, y): f(f(0)f(y)) = 0$. If $f(0) \neq 0$, then by surjectivity there exists $y$ such that $f(y) = \frac{k}{f(0)}$ so $f(k) = 0$, contradiction. Therefore, $f(0) = 0$.
$P(x, 0): f(x^2) = xf(x) \quad x:= -x \to f(x^2) = xf(x) = -xf(-x) \to f(-x) = -f(x)$.
$P(x, -x): f(x^2 + f(x)f(-x)) = 0 \to f(x^2 - f(x^2)) = 0$.
Suppose $f(t) = 0$. Then $P(x, k): f(x^2) = xf(x+t) = xf(x) \to \forall x \neq 0: f(x) = f(x+t), f(0) = f(t) \to \forall x: f(x) = f(x+a)$
$P(t, y): f(t^2) = tf(t+y) = tf(y)$. Now if $t \neq 0$ then $f$ is constant, but $f$ is surjective, contradiction. As a result, $t = 0$.
Now, $f(x^2 - f(x^2)) = 0 \to f(x^2) = x^2 \to f(x) = \pm x$
Now, as we don't fall into the pointwise trap :), we assume there exists $a, b \neq 0$ such that $f(a) = a, f(b) = -b$.
$P(a, b) \pm(a^2 - ab) = a. \pm(a+b)$.
We write all the four possibilities:
- $a^2 - ab = a^2 + ab \to 2ab = 0$, contradition.
- $a^2 - ab = -a^2 - ab \to 2a^2 = 0$, contradition.
- $-a^2 + ab = a^2 + ab \to 2a^2 = 0$, contradition.
- $-a^2 + ab = -a^2 - ab \to 2ab = 0$, contradition.
So, no such $a, b$ exists. This means $f(x) = -x \forall x$ or $f(x) = x \quad  \forall x$, which we can check and see that both are indeed solutions.
To sum up, we proved the only solutions are $\boxed{f(x) = 0 \quad \forall x \in \mathbb{R}}$, $\boxed{f(x) = x \quad \forall x \in \mathbb{R}}$, and $\boxed{f(x) = -x \quad \forall x \in \mathbb{R}}$. $\blacksquare$
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