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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.

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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
ISI UGB 2025 P4
SomeonecoolLovesMaths   5
N a few seconds ago by mqoi_KOLA
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
5 replies
1 viewing
SomeonecoolLovesMaths
Yesterday at 11:24 AM
mqoi_KOLA
a few seconds ago
hard inequality omg
tokitaohma   3
N 25 minutes ago by tokitaohma
1. Given $a, b, c > 0$ and $abc=1$
Prove that: $ \sqrt{a^2+1} + \sqrt{b^2+1} + \sqrt{c^2+1} \leq \sqrt{2}(a+b+c) $

2. Given $a, b, c > 0$ and $a+b+c=1 $
Prove that: $ \dfrac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}} + \dfrac{\sqrt{b^2+2bc}}{\sqrt{c^2+2a^2}} + \dfrac{\sqrt{c^2+2ca}}{\sqrt{a^2+2b^2}} \geq \dfrac{1}{a^2+b^2+c^2} $
3 replies
tokitaohma
Yesterday at 5:24 PM
tokitaohma
25 minutes ago
Divisibilty...
Sadigly   5
N 26 minutes ago by COCBSGGCTG3
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the sum of their squares divides the square of their product.
5 replies
Sadigly
Saturday at 9:07 PM
COCBSGGCTG3
26 minutes ago
Diophantine involving cube
Sadigly   1
N 40 minutes ago by mashumaro
Source: Azerbaijan Senior NMO 2020
$a;b;c;d\in\mathbb{Z^+}$. Solve the equation: $$2^{a!}+2^{b!}+2^{c!}=d^3$$
1 reply
Sadigly
5 hours ago
mashumaro
40 minutes ago
Old hard problem
ItzsleepyXD   2
N an hour ago by ItzsleepyXD
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
2 replies
ItzsleepyXD
Apr 25, 2025
ItzsleepyXD
an hour ago
The Return of Triangle Geometry
peace09   9
N 2 hours ago by mathfun07
Source: 2023 ISL A7
Let $N$ be a positive integer. Prove that there exist three permutations $a_1,\dots,a_N$, $b_1,\dots,b_N$, and $c_1,\dots,c_N$ of $1,\dots,N$ such that \[\left|\sqrt{a_k}+\sqrt{b_k}+\sqrt{c_k}-2\sqrt{N}\right|<2023\]for every $k=1,2,\dots,N$.
9 replies
peace09
Jul 17, 2024
mathfun07
2 hours ago
Set Partition
Butterfly   0
2 hours ago
For the set of positive integers $\{1,2,…,n\}(n\ge 3)$, no matter how its elements are partitioned into two subsets, at least one of the subsets must contain three numbers $a,b,c$ ($a=b$ is allowed) such that $ab=c$. Find the minimal $n$.
0 replies
Butterfly
2 hours ago
0 replies
Points Lying on its Cevian Triangle's Thomson Cubic
Feuerbach-Gergonne   1
N 2 hours ago by golue3120
Source: Own
Given $\triangle ABC$ and a point $P$, let $\triangle DEF$ be the cevian triangle of $P$ with respect to $\triangle ABC$. Let $H$ be the orthocenter of $\triangle ABC$, and denote the isotomic conjugate of $H, P$ with respect to $\triangle ABC$ by $X, Q$, respectively. Let the centroid of $\triangle DEF$ be $M$, and denote the isogonal conjugate of $P$ with respect to $\triangle DEF$ by $R$. Prove that
$$
P, Q, X \text{ are collinear} \iff P, R, M \text{ are collinear}. 
$$or in brief
$$
P \in \text{ K007 of } \triangle ABC \iff P \in \text{ K002 of } \triangle DEF. 
$$
1 reply
Feuerbach-Gergonne
Jul 19, 2024
golue3120
2 hours ago
Areas of triangles AOH, BOH, COH
Arne   71
N 3 hours ago by EpicBird08
Source: APMO 2004, Problem 2
Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Prove that the area of one of the triangles $AOH$, $BOH$ and $COH$ is equal to the sum of the areas of the other two.
71 replies
Arne
Mar 23, 2004
EpicBird08
3 hours ago
Problem 6
termas   68
N 3 hours ago by HamstPan38825
Source: IMO 2016
There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.

(a) Prove that Geoff can always fulfill his wish if $n$ is odd.

(b) Prove that Geoff can never fulfill his wish if $n$ is even.
68 replies
termas
Jul 12, 2016
HamstPan38825
3 hours ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   2
N 3 hours ago by Assassino9931
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
2 replies
bin_sherlo
Yesterday at 7:13 PM
Assassino9931
3 hours ago
combi/nt
blug   2
N 3 hours ago by aaravdodhia
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
2 replies
blug
May 9, 2025
aaravdodhia
3 hours ago
System of equations in juniors' exam
AlperenINAN   2
N 3 hours ago by Assassino9931
Source: Turkey JBMO TST 2025 P3
Find all positive real solutions $(a, b, c)$ to the following system:
$$
\begin{aligned}
a^2 + \frac{b}{a} &= 8, \\
ab + c^2 &= 18, \\
3a + b + c &= 9\sqrt{3}.
\end{aligned}
$$
2 replies
AlperenINAN
Yesterday at 7:41 PM
Assassino9931
3 hours ago
Trigo relation in a right angled. ISIBS2011P10
Sayan   10
N 3 hours ago by mqoi_KOLA
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
10 replies
+1 w
Sayan
Mar 31, 2013
mqoi_KOLA
3 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
23511 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
Z K Y
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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
23511 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 ...
Z K Y
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NathalieShwarz
51 posts
#9 • 1 Y
Y by Adventure10
But I think that we need to prove the injectivity ?
Z K Y
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pco
23511 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
1525 posts
#15
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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
11321 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
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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
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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
11321 posts
#19
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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
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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
11321 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
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strange solution
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Fibonacci_11235
44 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
1271 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
1007 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
146 posts
#30
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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
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#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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