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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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0 replies
jlacosta
May 1, 2025
0 replies
Iran TST P6
luutrongphuc   4
N 43 minutes ago by AblonJ
Find all function $f: \mathbb{R^+} \to \mathbb{R^+}$ such that:
$$f \left( f(f(xy))+x^2\right)=f(y)f(x)-f(y)f(x+y)$$
4 replies
luutrongphuc
May 26, 2025
AblonJ
43 minutes ago
Product of opposite sides
nabodorbuco2   0
an hour ago
Source: Original
Let $ABCDEF$ a regular hexagon inscribed in a circle $\Omega$. Let $P_i$ be a point inside $\Omega$ and $P_e$ its polar reflection wrt $\Omega$. The rays $AP_i,BP_i,CP_i,DP_i,EP_i,FP_i$ meet $\Omega$ again at $A_i,B_i,C_i,D_i,E_i,F_i$. Call $Q_I$ the polygon formed by the vertices $A_i,B_i,C_i,D_i,E_i,F_i$. Similarly construct the polygon $Q_E$ using $P_e$ instead.

Show that $Q_I$ and $Q_E$ are congruent.
0 replies
nabodorbuco2
an hour ago
0 replies
Brilliant Problem
M11100111001Y1R   7
N an hour ago by flower417477
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
7 replies
M11100111001Y1R
May 27, 2025
flower417477
an hour ago
In Cyclic Quadrilateral ABCD, find AB^2+BC^2-CD^2-AD^2
Darealzolt   1
N an hour ago by Beelzebub
Source: KTOM April 2025 P8
Given Cyclic Quadrilateral \(ABCD\) with an area of \(2025\), with \(\angle ABC = 45^{\circ}\). If \( 2AC^2 = AB^2+BC^2+CD^2+DA^2\), Hence find the value of \(AB^2+BC^2-CD^2-DA^2\).
1 reply
Darealzolt
Today at 4:10 AM
Beelzebub
an hour ago
Ultra-hyper saddle with logarithmic weight
randomperson1021   0
Yesterday at 5:22 PM
Fix integers \(k\ge 3\) and \(1<r<k\), a parameter \(\lambda>0\), and a real log-exponent \(\beta\in\mathbb R\). For every real \(a\) define
$$
F_{a,\beta}^{(k,r)}(x)
  \;:=\;
  \sum_{n\ge 1}
       n^{\,a}\,(\log n)^{\beta}\,e^{\lambda n^{r}}\,x^{\,n^{k}},
  \qquad 0\le x<1.
$$
Put
$$
\Lambda_{k,r,\lambda}
   \;:=\;
   \lambda\!\left(1-\frac{r}{k}\right)
   \left(\frac{\lambda r}{k}\right)^{\!\frac{r}{\,k-r\,}},
   \qquad
   \gamma=\frac{r}{k-r}.
$$
(1) Show that there exists a real constant \(c=c(k,r)\) (independent of \(\lambda\) and of \(\beta\)) such that
$$
\lim_{x\to 1^{-}}
      F_{a,\beta}^{(k,r)}(x)\,
      e^{-\Lambda_{k,r,\lambda}\,(1-x)^{-\gamma}}
      \;=\;
      \begin{cases}
          0, & a<c,\\[6pt]
          \infty, & a>c.
      \end{cases}
$$
(2) Determine this critical value \(c\) explicitly and verify that it coincides with the classical case \(r=1\), namely \(c=-\tfrac12\).

(3) Evaluate the finite, non-zero limit that occurs at the borderline \(a=c\) (your answer may depend on \(k,r,\lambda\) but not on \(\beta\)).
0 replies
randomperson1021
Yesterday at 5:22 PM
0 replies
3rd AKhIMO for university students, P5
UzbekMathematician   1
N Yesterday at 3:53 PM by grupyorum
Source: AKhIMO 2025, P5
Show that for every positive integer $n$ there exist nonnegative integers $p, q$ and integers $a_1, a_2, ... , a_p, b_1, b_2, ... , b_q \ge 2$ such that $$ n=\frac{(a_1^3-1)(a_2^3-1)...(a_p^3-1)}{(b_1^3-1)(b_2^3-1)...(b_q^3-1)} $$
1 reply
UzbekMathematician
Wednesday at 2:10 PM
grupyorum
Yesterday at 3:53 PM
Sum of three squares
perfect_radio   9
N Yesterday at 1:36 PM by RobertRogo
Source: RMO 2004, Grade 12, Problem 4
Let $\mathcal K$ be a field of characteristic $p$, $p \equiv 1 \left( \bmod 4 \right)$.

(a) Prove that $-1$ is the square of an element from $\mathcal K.$

(b) Prove that any element $\neq 0$ from $\mathcal K$ can be written as the sum of three squares, each $\neq 0$, of elements from $\mathcal K$.

(c) Can $0$ be written in the same way?

Marian Andronache
9 replies
perfect_radio
Feb 26, 2006
RobertRogo
Yesterday at 1:36 PM
Prove the statement
Butterfly   12
N Yesterday at 10:55 AM by oty
Given an infinite sequence $\{x_n\} \subseteq  [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
12 replies
Butterfly
May 7, 2025
oty
Yesterday at 10:55 AM
Putnam 1981 A3
sqrtX   1
N Yesterday at 10:19 AM by Mathzeus1024
Source: Putnam 1981
Find
$$ \lim_{t\to \infty} e^{-t} \int_{0}^{t} \int_{0}^{t} \frac{e^x -e^y }{x-y} \,dx\,dy,$$or show that the limit does not exist.
1 reply
sqrtX
Mar 31, 2022
Mathzeus1024
Yesterday at 10:19 AM
Recurrence trouble
SomeonecoolLovesMaths   2
N Yesterday at 8:52 AM by SomeonecoolLovesMaths
Let $0 < x_0 < y_0$ be real numbers. Define $x_{n+1} = \frac{x_n + y_n}{2}$ and $y_{n+1} = \sqrt{x_{n+1}y_n}$.
Prove that $\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$ and hence find the limit.
2 replies
SomeonecoolLovesMaths
Wednesday at 11:27 AM
SomeonecoolLovesMaths
Yesterday at 8:52 AM
3rd AKhIMO for university students, P3
UzbekMathematician   1
N Yesterday at 3:07 AM by pineconee
Source: AKhIMO 2025, P3
Two points are chosen randomly - independently with uniform probability - from a semicircular arc with radius 1. A third point is chosen randomly - independently with uniform probability - from the diameter that connects the endpoints of the arc. What is expected value of the area of the triangle with the three chosen points as its vertices?
1 reply
UzbekMathematician
Wednesday at 1:57 PM
pineconee
Yesterday at 3:07 AM
D1038 : A generalization of Jensen
Dattier   4
N Wednesday at 11:21 PM by Dattier
Source: les dattes à Dattier
Let $f \in C^1([0,1]), g \in C^2(f([0;1]))$.

Is it true that

$$\min(|g''|)\times \min(|f'|^2) \leq 24 \times\left|\int_0^1g(f(x)) \text{d}x- g(\int_0^1 f(x) \text{d}x) \right| \leq \max(|g''|)\times \max(|f'|^2)$$?
4 replies
Dattier
Wednesday at 12:15 PM
Dattier
Wednesday at 11:21 PM
3rd AKhIMO for University Students, P2
UzbekMathematician   1
N Wednesday at 9:40 PM by grupyorum
Source: AKhIMO 2025, P2
Find all possible values of $gcd(a^{2m}+1, a^n+1)$, where $a, m, n$ are positive integers and $n$ is odd.
1 reply
UzbekMathematician
Wednesday at 1:48 PM
grupyorum
Wednesday at 9:40 PM
Problem 2, Grade 12th RMO Shortlist - Year 2002
sticknycu   5
N Wednesday at 4:07 PM by P_Fazioli
Let $A \in M_2(C), A \neq O_2, A \neq I_2, n \in \mathbb{N}^*$ and $S_n = \{ X \in M_2(C) | X^n = A \}$.
Show:
a) $S_n$ with multiplication of matrixes operation is making an isomorphic-group structure with $U_n$.
b) $A^2 = A$.

Marian Andronache
5 replies
sticknycu
Jan 3, 2020
P_Fazioli
Wednesday at 4:07 PM
Functional xf(x+f(y))=(y-x)f(f(x)) for all reals x,y
cretanman   59
N May 28, 2025 by math-olympiad-clown
Source: BMO 2023 Problem 1
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
Proposed by Nikola Velov, Macedonia
59 replies
cretanman
May 10, 2023
math-olympiad-clown
May 28, 2025
Functional xf(x+f(y))=(y-x)f(f(x)) for all reals x,y
G H J
G H BBookmark kLocked kLocked NReply
Source: BMO 2023 Problem 1
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cretanman
430 posts
#1 • 5 Y
Y by liecheaper, Sedro, lian_the_noob12, ItsBesi, ehuseyinyigit
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
Proposed by Nikola Velov, Macedonia
This post has been edited 4 times. Last edited by Amir Hossein, May 13, 2023, 1:00 AM
Reason: Fixed source
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Batapan
93 posts
#2 • 8 Y
Y by Mai-san, DESTINY_55, dheerstar12, Elghassemmed, Ahsan_Khalid, NicoN9, Rayanelba, ehuseyinyigit
Let $P(x,y)$ denote the given assertion.
$P(x,x) \implies f(x+f(x))=0$
$P(0,y) \implies f(f(0))=0$
$P(f(0),y) \implies f(0)f(f(0)+f(y))=(y-f(0))f(0)$

So $f(0)=0$ or $ f(f(0)+f(y))=(y-f(0))$

If $ f(f(0)+f(y))=(y-f(0))$, setting $y \rightarrow f(y)+y \implies f(y)=-y+f(0)$ and checking we see that $f(x)=-x+k$ truly satisfies the condition
Suppose now $f(0)=0$

$P(x,0) \implies f(f(x))=-f(x)$
So $P(x,y)$ becomes $xf(x+f(y))=(x-y)f(x)$
So $P(x,f(y)+y)$ gives us $f(x)=0$ or $f(x)=-x$

Finally $f(x)=0, f(x)=-x+k$ where $k$ is a constant
This post has been edited 3 times. Last edited by Batapan, May 10, 2023, 4:13 PM
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Assassino9931
1379 posts
#3 • 5 Y
Y by Mai-san, TheEmpress, KAME06, IchliebeMathematik, NicoN9
Suppose firstly that $f(f(x)) = 0$ for all $x$. Then $f(x+f(y)) = 0$ for all $x\neq 0$. If $f(0) = 0$, then $y=0$ gives $f(x) = 0$ for all $x$ and this is a solution; if $f(0) \neq 0$, then $x=-f(y))$ gives a contradiction.

So we may assume that $f(f(x)) \neq 0$ for some $x$. But then $f$ is injective, as if $f(y) = f(z)$, then $(y-x)f(f(x)) = (z-x)f(f(x))$ and so $y=z$.
Now $x=0$, $y=1$ gives $f(f(0)) = 0$ while $y=x \neq 0$ gives $f(x+f(x)) = 0$ and hence $f(x) = f(0) - x$ for $x\neq 0$ (but this also trivially holds for $x=0$). Conversely, direct substitution shows that $f(x) = k-x$ for any constant $k$ is a solution.
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grupyorum
1434 posts
#4 • 1 Y
Y by pro_dynamite
Let $P(x,y)$ denotes the given assertion. Then $P(x,x)$ yields $f(x+f(x))=0$ for all $x\ne 0$. Further, $P(0,x)$ gives with $x\ne 0$ that $f(f(0))=0$. Combining, $f(x+f(x))=0$ holds for all $x$. Now, let there is an $x_0$ such that $f(f(x_0))\ne 0$. If $f(y_1)=f(y_2)$ for some $y_1,y_2$, then considering $P(x_0,y_1)$ and $P(x_0,y_2)$ we obtain $y_1=y_2$: $f$ is injective. Using $f(x+f(x))=0$, we get $x+f(x)=c$ for some constant $c$, which yields $f(x)=-x+c$. Lastly, suppose $f(f(x))=0$ for all $x$, so that $xf(x+f(y)) = 0$ for all $x,y$. Taking $y=f(r)$, we get $xf(x) =0$ for all $x$, so $f(x)=0$ for $x\ne 0$. Lastly, taking $x\ne 0$ and using $f(f(x)) = 0$, we get $f(0)=0$, too. So $f\equiv 0$ identically.
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VicKmath7
1391 posts
#5
Y by
Solution
This post has been edited 2 times. Last edited by VicKmath7, May 11, 2023, 11:11 AM
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Mathlover_1
295 posts
#6 • 1 Y
Y by Math.1234
cretanman wrote:
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$,
\[xf(x+f(y))=(y-x)f(f(x)).\]
North Macedonia
Let $P(x,y)$ denote the given assertion.
1)Let for all real $x$ $f(f(x))=0$ then from $P(x,y)$ we get $f(x+f(y))=0$ $x=a-f(y)$ $\rightarrow$ for all a $f(a)=0$ $f \equiv 0$
2)For same $x$ $f(f(x))$$\neq$$0$
Claim:Function is injective
Prove: Assume that for some $a$$\neq$$b$ $f(a)=f(b)$ then $( f(f(0))$$\neq$$0)$ from $P(x,a)$ and $P(x,b)$ we get $a=b$ contradiction
$P(y,0)$ $\rightarrow$ $f(f(0))=0$
$P(f(0)-f(x),x)$ $\rightarrow$ $(x+f(x)-f(0))×f(f(f(0)-f(x)))=0$ for $x$$\neq$$0$ $f(f(f(0)-f(x)))$$\neq$$0$ because from injectivety we get $f(0)=f(y)$ but it is impossible then for all $x$ $x+f(x)-f(0)=0$ $\rightarrow$ $f(x)=-x+c$ (there $c=f(0)$)
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motannoir
171 posts
#7 • 1 Y
Y by DanDumitrescu
My 100-th post !
The answers are $ f(x)=c-x$ for every x and $f(x)=0$ for every x which clearly work (c is an arbitrary constant)
Let's prove these are the only ones.
We have two cases
Case 1: $f(f(x))=0$ for every x.
Then our $P(x,y)$ becomes $f(x+f(y))=0 $ for every x not 0 and y real,but is easy to see it works also for x=0. But then let a be any real. Put $P(a-f(y),y)$ to get $f(a)=0$ so f is constant 0 function
Case 2: There exists a real such that $f(f(a))$ is not 0. If we put x=0 we get that $f(f(0))=0$ so a is not 0
Then put $P(a,y)$ to get that f is bijective
To finish put $P(x,x)$ to get that $f(x+f(x))=0$ for every x real( because for x=0 we proved already) and by injectivity we get f(x)=c-x and we are done !
This post has been edited 1 time. Last edited by motannoir, May 10, 2023, 7:58 PM
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hood09
117 posts
#8
Y by
Both $f(x) =-x+c$ and $f =0$ are solutions. Let's prove they are the only ones.
Let $f(0)=c$, by $P(0,y)$ we get $f(c)=0$ , and by $P(x,x)$ we get $f(x+f(x))=0$.
If there is a number $a \neq0$ such that $f(a)\neq0$ : applying $P(a,c)$ give us $f(f(a)) \neq 0$. then $P(a,y)$ give us that $f$ is injective so $x+f(x)=c$ thus $f(x)=-x+c.$
Else : for any $x\neq0$ we have $f(x)=0$, so if we suppose $c\neq0$ we can notice that $P(c,0)$ : $cf(2c)=-c^2$ then $c=0$ absurde, so $f(x) =0$ for any $x$.
This post has been edited 1 time. Last edited by hood09, May 10, 2023, 9:03 PM
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steppewolf
351 posts
#9 • 1 Y
Y by Tellocan
Authored by Nikola Velov
This post has been edited 1 time. Last edited by steppewolf, Feb 9, 2025, 10:40 PM
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Ubfo
16 posts
#10
Y by
Put $y=x\neq0$ gives $f(x+f(x))=0$. Let $f(c)=0$.
Case 1: $f(f(x))=0\forall x\in\mathbb{R}$. Then $xf(x+f(y))=0$ so $f(x)=0\forall x\neq -f(y)$. If f is nonconstant then $f(x)=0\forall x$, if constant then $f(x)=f(c)=0$
Case 2: $\exists x: f(f(x))\neq0$. $f(a)=f(b)$ by $P(x,a),P(x,b)$ imply $a=b$ so $f$ is injective. Assume $\exists t:f(t)\neq c-t$ then $$P(c-f(t),t)\Rightarrow(f(t)+t-c)f(f(c-f(t)))=0$$so there's at most one such $t=t_0$. $$P(\frac{t_0}2,t_0)\Rightarrow f(\frac{t_0}2+f(t_0))=f(f(\frac{t_0}{2}))\Rightarrow f(t_0)=c-t_0$$So either $f(x)=c-x$ or $f(x)=0$ for all $x\in \mathbb{R}$, both solutions fit.
This post has been edited 1 time. Last edited by Ubfo, May 12, 2023, 10:34 AM
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F10tothepowerof34
195 posts
#11 • 1 Y
Y by Loro_
Let $P(x,y)=xf(x+f(y))=(y-x)f(f(x))$

$P(0,x)\Longrightarrow xf(f(0))=0 \Longleftrightarrow f(f(0))=0 \text{, for } x\neq0$ (1)
$P(x,x)\Longrightarrow xf(x+f(x))=0 \Longleftrightarrow f(x+f(x))=0 \text{, for } x\neq0$ (2)
Claim: The function is injective iff $f(f(x))\neq0$
Proof:
Let $f(a)=f(b)$
$P(x,a)$ yields: $xf(x+f(a))+xf(f(x))=af(f(x))$
$P(x,b)$ yields: $xf(x+f(b))+xf(f(x))=bf(f(x))$
Thus $af(f(x))=bf(f(x))$, which implies that the function is injective when $f(f(x))\neq0$ $\square$.

Case 1: $f(f(x))\neq0$
From (1) and (2) we have that: $f(x+f(x))=f(f(0))$, however since the function is injective when $f(f(x))\neq0$, thus $x+f(x)=f(0)\Longleftrightarrow f(x)=-x+f(0)\Longrightarrow f(x)=-x+c$ where $c$ is a constant.
when x is equal to 0

Case 2: $f(f(x))=0$
Since $f(f(x))=0$, the functional equation becomes $xf(x+f(y))=0$. Let $Q(x,y)=xf(x+f(y))$
$Q(x,f(x))$ yields: $xf(x)=0\Longleftrightarrow f(x)=0$ for $x\neq0$
$Q(-f(0),0)$ yields: $-f(0)^2=0\Longleftrightarrow f(0)=0$, thus $f(x)=f(0)=0$ which implies that $f(x)=0,    \forall x \in \mathbb{R}$

So to sum up, $\boxed{f(x)=-x+c} \text{ and } \boxed{f(x)=0},  \forall  x \in \mathbb{R}$
$\blacksquare$
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Marinchoo
407 posts
#12
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Case 1. $f$ isn't injective: $\exists a\neq b\in\mathbb{R}$ such that $f(a) = f(b)$. Comparing $(x,a)$ and $(x,b)$ yields that $f(f(x)) = 0$ $\forall x\in\mathbb{R}$. Plugging in $(x,f(y))$ now gives $xf(x) = 0$ for all $x$, hence $f(x) = 0$ $\forall x\neq 0$. However, $(x,y) = (-f(0),0)$ implies that $-f(0)^2 = 0$ so $f\equiv 0$, which is indeed a solution.

Case 2. $f$ is injective: Plugging in $(x,x)$ for $x\neq 0$ yields $f(x+f(x)) = 0$ and $(x,y) = (0,1)$ gives $f(f(0)) = 0$, so $f(x+f(x)) = 0$ for all $x$, hence $x+f(x) = y+f(y)$ $\forall x,y\in\mathbb{R}$ (due to the injectivity of $f$), whence $f(x) = c-x$ which works for all constants $c$.
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m.ljf2997
11 posts
#13
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First if F(a)=F(b) then a=b prove
P(x,a) and P(x,b) ==> a=b or F(F(x))=0
if F(F(x))=0 ==> xF(x+F(y))=0
P(x-F(y) , y)==> F(x)=0 ∀ x∈R
if a=b ==>
P(0,y) and P(x,x) ==> F(x+F(x))=F(F(0))=0 ==> x+F(x)=F(0)=K
F(x)=K-x
and it's easy to verify that F(x)=0 orF(x)=K-x ∀ K∈R is the Solution.
This post has been edited 2 times. Last edited by m.ljf2997, May 18, 2023, 8:01 PM
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ezpotd
1306 posts
#14
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why am i so scared of fes lmao...

Anyways, $P(0,0)$ gives $f(f(0)) = 0$, now if $f(f(x)) = 0$ for all $x$, we have $f(x) = 0$, otherwise we can take $P(x,a),P(x,b)$ with $f(f(x)) \neq 0, f(a) = f(b)$ to get $(a-x)f(f(x)) = (b-x)f(f(x))$, so $a=b$. Then $P(x,x)$ gives $f(x + f(x) ) = 0 = f(f(0))$, so $x + f(x) = f(0)$, so $f(x) = k - x$. All functions of this form work, so we are done.
This post has been edited 1 time. Last edited by ezpotd, May 19, 2023, 3:05 AM
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Fibonacci_11235
45 posts
#15
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I denote by $P(x, y)$ plugging some $x$ and $y$ into the given equation.
if $f(f(x)) = 0$ for all $x$ then:
$P(x, f(y)): xf(x) = 0 \implies f(x) = 0$ for all $x$.
now Suppose that there exists $a$ such that $f(f(a)) \neq 0$
$P(a, \frac{a+y}{f(f(a))}): af(...) = y$, in other words $f$ is surjective.
$P(0, y): yf(f(0)) = 0 \implies f(f(0))=0$
$P(f(0)-f(y), y): 0=(y+f(y)-f(0))f(f(f(0)-f(y)))$
Now suppose that $f(f(b)) = 0$ for some $b$.
$P(b, y): bf(b + f(y)) = 0$
Case 1: $b \neq 0$
$bf(b + f(y)) = 0 \implies f(x) = 0$ since $f$ is surjective.
Case 2: $b = 0$
$bf(b + f(y)) = 0 \implies 0=0$

Now let's focus on nonzero solutions, we know that for some $y$:
$y + f(y) - f(0) = 0 \Leftrightarrow f(y) = -y + f(0)$ or $f(0) - f(y) = 0 \Leftrightarrow f(y) = f(0)$

suppose that there exists some $c \neq 0$ for which $f(c) = f(0)$
$P(c, y): cf(c + f(y)) = 0 \implies f(x) = 0$ since c is not zero and $f$ is surjective.
But since we are looking for nonzero solutions, such c does not exist.
so $f(y) = -y + f(0)$ or $f(y) = 0$ for all $y$
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