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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Finding Max!
goldeneagle   4
N 28 minutes ago by aidan0626
Source: Iran 3rd round 2013 - Algebra Exam - Problem 2
Real numbers $a_1 , a_2 , \dots, a_n$ add up to zero. Find the maximum of $a_1 x_1 + a_2 x_2 + \dots + a_n x_n$ in term of $a_i$'s, when $x_i$'s vary in real numbers such that $(x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_{n-1} - x_n)^2 \leq 1$.
(15 points)
4 replies
goldeneagle
Sep 11, 2013
aidan0626
28 minutes ago
Inequalities
produit   0
29 minutes ago
Find the lowest value of C for which there exists such sequence
1 = x_0 ⩾ x_1 ⩾ x_2 ⩾ . . . ⩾ x_n ⩾ . . .
that for any positive integer n
x_{0}^2/x_{1}+x_{1}^{2}/x_{2}+ . . . +x_{n}^2/x_{n+1}< C.
0 replies
1 viewing
produit
29 minutes ago
0 replies
Easy Number Theory
math_comb01   38
N an hour ago by lakshya2009
Source: INMO 2024/3
Let $p$ be an odd prime and $a,b,c$ be integers so that the integers $$a^{2023}+b^{2023},\quad b^{2024}+c^{2024},\quad a^{2025}+c^{2025}$$are divisible by $p$.
Prove that $p$ divides each of $a,b,c$.
$\quad$
Proposed by Navilarekallu Tejaswi
38 replies
1 viewing
math_comb01
Jan 21, 2024
lakshya2009
an hour ago
number of positive divisors is equal to n/5
falantrng   4
N an hour ago by Adywastaken
Source: Azerbaijan NMO 2024. Senior P2
Let $d(n)$ denote the number of positive divisors of the natural number $n$. Find all the natural numbers $n$ such that $$d(n) = \frac{n}{5}$$.
4 replies
falantrng
Jul 8, 2024
Adywastaken
an hour ago
GHM and ABC are tangent to each other.
IndoMathXdZ   16
N 3 hours ago by Ilikeminecraft
Source: DeuX Mathematics Olympiad 2020 Shortlist G5 (Level II Problem 3)
Given a triangle $ ABC$ with circumcenter $O$ and orthocenter $H$. Line $OH$ meets $AB, AC$ at $E,F$ respectively.
Define $S$ as the circumcenter of $ AEF$. The circumcircle of $ AEF$ meets the circumcircle of $ABC$ again at $J$, $J \not= A$. Line $OH$ meets circumcircle of $JSO$ again at $D$, $D \not= O$ and circumcircle of $JSO$ meets circumcircle of $ABC$ again at $K$, $K \not= J$. Define $M$ as the intersection of $JK$ and $OH$ and $DK$ meets circumcircle of $ABC$ at points $K,G$.

Prove that circumcircle of $GHM$ and circumcircle of $ABC$ are tangent to each other.

Proposed by 郝敏言, China
16 replies
IndoMathXdZ
Jul 12, 2020
Ilikeminecraft
3 hours ago
junior perpenicularity, 2 circles related
parmenides51   3
N 4 hours ago by LeYohan
Source: Greece Junior Math Olympiad 2024 p2
Consider an acute triangle $ABC$ and it's circumcircle $\omega$. With center $A$, we construct a circle $\gamma$ that intersects arc $AB$ of circle $\omega$ , that doesn't contain $C$, at point $D$ and arc $AC$ , that doesn't contain $B$, at point $E$. Suppose that the intersection point $K$ of lines $BE$ and $CD$ lies on circle $\gamma$. Prove that line $AK$ is perpendicular on line $BC$.
3 replies
parmenides51
Mar 2, 2024
LeYohan
4 hours ago
A tangent problem
hn111009   0
4 hours ago
Source: Own
Let quadrilateral $ABCD$ with $P$ be the intersection of $AC$ and $BD.$ Let $\odot(APD)$ meet again $\odot(BPC)$ at $Q.$ Called $M$ be the midpoint of $BD.$ Assume that $\angle{DPQ}=\angle{CPM}.$ Prove that $AB$ is the tangent of $\odot(APD)$ and $BC$ is the tangent of $\odot(AQB).$
0 replies
hn111009
4 hours ago
0 replies
I accidentally drew a 200-gon and ran out of time
justin1228   24
N 4 hours ago by Ilikeminecraft
Source: USEMO P.5 2020
The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are colored red and blue in an alternating fashion.
Suppose the extensions of the red sides determine a regular $100$-gon, as do the extensions of the blue sides.

Prove that the $50$ diagonals $\overline{A_1A_{101}},\ \overline{A_3A_{103}},\ \dots,
\ \overline{A_{99}A_{199}}$ are concurrent.

Proposed by: Ankan Bhattacharya
24 replies
justin1228
Oct 25, 2020
Ilikeminecraft
4 hours ago
An I for an I
Eyed   67
N 5 hours ago by AR17296174
Source: 2020 ISL G8
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.
67 replies
Eyed
Jul 20, 2021
AR17296174
5 hours ago
IMO 2018 Problem 1
juckter   169
N Yesterday at 11:30 PM by Thelink_20
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
169 replies
juckter
Jul 9, 2018
Thelink_20
Yesterday at 11:30 PM
Lines concur on bisector of BAC
Invertibility   2
N Yesterday at 8:32 PM by NO_SQUARES
Source: Slovenia 2025 TST 3 P2
Let $\Omega$ be the circumcircle of a scalene triangle $ABC$. Let $\omega$ be a circle internally tangent to $\Omega$ in $A$. Tangents from $B$ touch $\omega$ in $P$ and $Q$, such that $P$ lies in the interior of $\triangle{}ABC$. Similarly, tangents from $C$ touch $\omega$ in $R$ and $S$, such that $R$ lies in the interior of $\triangle{}ABC$.

Prove that $PS$ and $QR$ concur on the bisector of $\angle{}BAC$.
2 replies
Invertibility
Yesterday at 8:19 PM
NO_SQUARES
Yesterday at 8:32 PM
angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N Yesterday at 7:22 PM by breloje17fr
Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
5 replies
parmenides51
Dec 11, 2018
breloje17fr
Yesterday at 7:22 PM
Problem 4 of Finals
GeorgeRP   2
N Yesterday at 7:00 PM by Assassino9931
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]
2 replies
GeorgeRP
Sep 10, 2024
Assassino9931
Yesterday at 7:00 PM
Interesting functional equation with geometry
User21837561   3
N Yesterday at 6:34 PM by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Yesterday at 8:14 AM
Double07
Yesterday at 6:34 PM
easy functional
B1t   13
N Apr 30, 2025 by AshAuktober
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
13 replies
B1t
Apr 26, 2025
AshAuktober
Apr 30, 2025
easy functional
G H J
G H BBookmark kLocked kLocked NReply
Source: Mongolian TST 2025 P1.
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B1t
24 posts
#1 • 1 Y
Y by farhad.fritl
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
This post has been edited 2 times. Last edited by B1t, Apr 26, 2025, 7:01 AM
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NicoN9
148 posts
#2
Y by
$f(x)f(x)$ means $f(x)^2$? or is it a typo?
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Ilikeminecraft
626 posts
#3
Y by
$x = y = z = 0$ implies $f(0) = 0$
$x = -y$ implies $f(z) = f(z) - xf(x) + f(x)^2,$ so $f\in\{0, x\}.$
assume $f(a ) = 0, f(b ) = b.$
take $y = 0,$ and we get $f(xf(x) + z) = f(z) + f(x)^2.$
if we take $x = b, y = a - b, z = a,$ we get $0 = ab.$ this implies $f \equiv 0, x.$
Z K Y
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B1t
24 posts
#5
Y by
NicoN9 wrote:
$f(x)f(x)$ means $f(x)^2$? or is it a typo?

I wrote it wrong. im sorry
Z K Y
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Ilikeminecraft
626 posts
#6
Y by
suppose $f$ is not identically $0$
suppose $x_0$ satisfies $f(x_0)\neq 0,$ and if we pick $x = x_0$ and range the values of $y,$ we obtain that $f$ is surjective. ignore the definition of $x_0$
take $y = 0$ and we get $f(xf(x) + z) = f(xf(x)) + f(z)$
take $x = - y$ yields $f(xf(x))=xf(x).$
thus, $f(f(xf(x)) + z) = f(z) + f(f(xf(x)))$ which implies that the function is cauchy
plugging back into original equation, one gets $f(xf(y)) = f(x)y$
pick $x = 1$ to get $f(f(y)) = y,$ which is well-known to imply that $|f(y)| = |y|.$

assume $A$ is the set of $x$ such that $f(x) = x$ and $B$ is the set of $x$ such that $f(x) = -x.$
if $x, y \in A,$ then $f(xy) = xy,$ so $xy \in A$
if $x\in A, y\in B,$ then $f(xy) = -xy,$ so $xy\in B$
if $x \in B, y \in A,$ then $f(xy) = -xy,$ so $xy \in B$
if $x \in B, y\in B,$ then $f(xy) = xy,$ so $xy\in A$

this implies $f$ is multiplicative
hence, since $f$ is both multiplicative and additive, $f$ is identity
This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 26, 2025, 3:22 PM
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Haris1
77 posts
#7
Y by
I wont write the solution , i will just write the steps.
$1.$ Prove that the function is additive
$2.$ Prove that its either constant or bijective
$3.$ Prove that its multiplicative and finish
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cazanova19921
552 posts
#8 • 1 Y
Y by farhad.fritl
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
P(x,y,z):\, f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
$P(0,0,0)$ $\implies$ $f(0)=0$.
$P(x,-x,0)$ $\implies$ $f(xf(x))=xf(x)$
So $P(x,y-x,z): \, f(xf(y)+z)=f(z)+yf(x)$ (new $P$)
- If $f(t)=0$ for some $t \neq 0$, then $P(x,t,0)$ $\implies$ $\boxed{f=0}$ which is a solution.
- Suppose $f(t)=0$ $\iff$ $t=0$.
$P(1,x,0)$ $\implies$ $f(f(x))=xf(1)$. hence $f$ is bijective, replace $x=1$ in this equation, we get $f(f(1))=f(1)$ so $f(1)=1$.
Therefore $f(f(x))=x$ for all $x$.
$P(x, f(y), 0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x, y$
$P(x, 1, y)$ $\implies$ $f(x+y)=f(x)+f(y)$ for all $x, y$
So $f$ is additive and multiplicative, hence $\boxed{f=\mathrm{Id}}$ which is also a valid solution.
This post has been edited 1 time. Last edited by cazanova19921, Apr 26, 2025, 1:55 PM
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MathLuis
1524 posts
#9
Y by
Denote $P(x,y,z)$ the assertion of the given F.E.
Notice from $P(0,y,z)$ we can inmediately get $f(0)=0$ and also $P(x,-x,z)$ gives that $f(xf(x))=xf(x)$ but also $P(x,y,0)$ now gives that $f(xf(x+y))=(x+y)f(x)$ so in fact shifting gives $Q(x,y)$ which is $f(xf(y))=yf(x)$ now notice if ther existed some $c \ne 0$ for which $f(c)=0$ then $Q(x,c)$ gives that $f(x)=0$ for all reals $x$ so its either that or $f$ is injective at zero, but now basically note that our F.E. may now by re-written as $R(x,y,z)$ to be $f(xf(y)+z)=f(z)+yf(x)$ for all reals $x,y,z$ but also take $y \ne 0$ and shift $x$ to get that $f$ is additive, but also from $Q(x, f(x))$ we get that $f(f(1)x^2)=f(x)^2$ and therefore shifting $x \to f(x)$ gives $f(f(1)f(x)^2)=f(1)^2 \cdot x^2$ so for example $f$ is surjective on all non-negative reals also from $Q(x,y)$ we have $f$ injective trivially when setting $x \ne 0$ and this take $f(d)=1$ and $Q(x,d)$ to get that $d=1$ and thus $f(1)=1$ so $Q(1,x)$ gives $f$ is an involution so by $Q(x,f(y))$ we get $f$ multiplicative so addivitive+multiplicative means $f$ is the identity function or constant (later case gives $f$ is zero everywhere so doesn't count), so $f(x)=0,x$ for all reals $x$ are the only two solutions that work thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 27, 2025, 3:32 PM
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jasperE3
11317 posts
#10
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B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]

Let $P(x,y)$ be the assertion $f(xf(x+y)+z)=f(z)+yf(x)+f(xf(x))$.
Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$.
$P(0,0,0)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,-j,0)\Rightarrow f(jf(j))=jf(j)$
$P(j,0,x)\Rightarrow f(x+jf(j))=f(x)+jf(j)$
$P(x,y+jf(j),z)\Rightarrow f(xf(x+y)+jf(j)x+z)=f(z)+yf(x)+f(xf(x))+jf(j)f(x)$
Comparing this last line with $P(x,y)$ we get:
$$f(xf(x+y)+jf(j)x+z)=f(xf(x+y)+z)+jf(j)f(x)$$and setting $z=-xf(x+y)$ and $x=1$ this is $jf(j)(f(1)-1)=0$, so $f(1)=1$.
$P(1,x-1,y)\Rightarrow f(f(x)+y)=x+f(y)$
Call this assertion $Q(x,y)$.
$Q(x,0)\Rightarrow f(f(x))=x$
$Q(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$
$P(x,0,y)\Rightarrow f(xf(x)+y)=f(xf(x))+f(y)$
$P(x,-x,0)\Rightarrow f(xf(x))=xf(x)$
Now $P(x,y)$ becomes:
\begin{align*}
xf(x)+f(z)+f(xf(y))&=f(xf(x))+f(z)+f(xf(y))\\
&=f(xf(x)+xf(y)+z)\\
&=f(xf(x+y)+z)\\
&=f(z)+yf(x)+f(xf(x))\\
&=f(z)+yf(x)+xf(x)
\end{align*}and so $f(xf(y))=yf(x)$. Taking $y\mapsto f(y)$ we have $f(xy)=f(x)f(y)$, well-known that the only additive and multiplicative functions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ which both work.
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GreekIdiot
221 posts
#11
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why is mongolian tst so easy?
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B1t
24 posts
#12
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GreekIdiot wrote:
why is mongolian tst so easy?

true
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MuradSafarli
109 posts
#13 • 2 Y
Y by B1t, Nuran2010
Interesting problem!

Let \( P(x, -x, y) \) denote the assertion of the functional equation:
\[
xf(x) = f(xf(x)) \tag{1}
\]
Now, consider the following:

- \( P(1, -1, x) \) gives:
\[
f(1) = f(f(1)).
\]- \( P(1, x-1, 0) \) gives:
\[
f(f(x)) = x \cdot f(1). \tag{2}
\]
Now, we consider two cases:

---

**Case 1:** \( f(1) = 0 \)

From equation (2), we have:
\[
f(f(x)) = 0 \quad \text{for all } x.
\]
Applying \( f \) to both sides of equation (1):
\[
xf(x) = f(xf(x)) = f(f(xf(x))) = 0,
\]thus implying:
\[
f(x) = 0 \quad \text{for all } x.
\]
---

**Case 2:** \( f(1) \neq 0 \)

From equation (2), we can conclude that \( f \) is bijective.

Suppose there exists some \( k \) such that \( f(k) = 1 \).
Applying \( P(k, -k, x) \) gives:
\[
k = 1,
\]thus \( k = 1 \).

Moreover, from (2), we obtain:
\[
f(f(x)) = x.
\]
Now, consider \( P(1, f(x) - 1, y) \). We get:
\[
f(x) + f(y) = f(x+y),
\]meaning \( f \) is **additive**.

Since \( f \) is additive and satisfies \( f(f(x)) = x \), we deduce that \( f(x) = cx \) for some constant \( c \). Substituting back into the original functional equation shows that \( c = 1 \), and thus:
\[
f(x) = x.
\]
---

**Final answer:**
1. \( f(x) = 0 \) for all real numbers \( x \), or
2. \( f(x) = x \) for all real numbers \( x \).
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complex2math
7 posts
#14
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Denote by $P(x, y, z)$ the assertion of the given functional equation. Then $P(0, 0, 0)$ and $P(x, -x, 0)$ gives the following

Claim 1. $f(0) = 0$ and $f(xf(x)) = xf(x)$.

Now we can rewrite the given functional equation as
\[
f(xf(x + y) + z) = f(z) + f(x)(x + y) \qquad (\heartsuit)
\]and let $P(x, y, z)$ re-denote the assertion of equation $(\heartsuit)$ and set $f(1) = a$.

Claim 2. If $a = 0$, then $f(x) \equiv 0$.

Proof. $P(x, 1 - x, z)$ implies $f(z) = f(ax + z) = f(z) + f(x)$, so $f(x) = 0$ for all $x \in \mathbb{R}$.

In what follows, we always assume $a \ne 0$.

Claim 3. $f(x)$ is bijective when $a \ne 0$.

Proof. From $P(1, y, 0)$ we get $f(f(1 + y)) = a(y + 1) \implies f(f(y)) = ay$. This is a bijection whenever $a \ne 0$.

Claim 4. $a = f(1) = 1$ and $f(x)$ is additive, i.e. $f(x + y) = f(x) + f(y)$ holds.

Proof. Since $xf(x) = f(xf(x))$, substituting $x = 1$ we obtain $f(1) = f(f(1)) \implies f(1) = 1$ as $f$ is injective. Then $P(x, 1 - x, z)$ gives $f(x + z) = f(z) + f(x)$.

Claim 5. $f(x)$ is multiplicative, i.e. $f(xy) = f(x)f(y)$ holds.

Proof. We have $f(f(y)) = ay = y$ so $P(x, y, 0)$ gives $f(xf(x + y)) = f(x)(x + y) = f(x)\cdot f(f(x + y))$. Then note that
\[
S_x := \{f(x + y): y \in \mathbb{R}\} = \mathbb{R}
\]for fixed $x \in \mathbb{R}$ since $f$ is surjective.

It's well-known that if $f(x)$ is both additive and multiplicative, then either $f(x) \equiv 0$ or $f(x) = x$.
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AshAuktober
1005 posts
#15
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The claims I made in order to solve:
1) $f(0) = 0$.
2) $f(xf(x)) = xf(x)$.
3) $f(xf(y)) = yf(x)$.
4) Either $f \equiv 0$ or $f(x) = 0 \implies x = 0$.
(Now onwards taking second case...)
5) $f$ is an involution
6) $f$ is additive
7) $f$ is multiplicative
And done!
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