Functional xf(x+f(y))=(y-x)f(f(x)) for all reals x,y

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cretanman

430 posts

cretanman

#1 h May 10, 2023, 3:50 PM • 4 Y

Y by liecheaper, Sedro, lian_the_noob12, ItsBesi

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

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Batapan

#2 h May 10, 2023, 4:02 PM • 6 Y

Y by Mai-san, DESTINY_55, dheerstar12, Elghassemmed, Ahsan_Khalid, NicoN9

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

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Assassino9931

1198 posts

Assassino9931

#3 h May 10, 2023, 4:06 PM • 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

1400 posts

grupyorum

#4 h May 10, 2023, 4:32 PM • 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

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VicKmath7

#5 h May 10, 2023, 6:37 PM

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Solution

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Mathlover_1

294 posts

Mathlover_1

#6 h May 10, 2023, 7:05 PM • 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

motannoir

#7 h May 10, 2023, 7:56 PM • 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 !

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hood09

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hood09

#8 h May 10, 2023, 9:01 PM

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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 $x$ .

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steppewolf

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steppewolf

#9 h May 11, 2023, 6:13 AM • 1 Y

Y by Tellocan

Authored by Nikola Velov

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Ubfo

15 posts

Ubfo

#10 h May 12, 2023, 10:33 AM

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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.

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F10tothepowerof34

195 posts

F10tothepowerof34

#11 h May 12, 2023, 10:22 PM

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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

Marinchoo

#12 h May 14, 2023, 8:20 AM

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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

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m.ljf2997

#13 h May 18, 2023, 7:44 PM

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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.

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ezpotd

1251 posts

ezpotd

#14 h May 19, 2023, 3:05 AM

Y by

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.

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Fibonacci_11235

43 posts

Fibonacci_11235

#15 h May 20, 2023, 3:51 PM

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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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KevinYang2.71

407 posts

KevinYang2.71

#46 h Jun 21, 2024, 11:49 PM

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We claim the only functions are $\boxed{f(x)\equiv 0}$ and $\boxed{f(x)\equiv c-x}$ for $c\in\mathbb{R}$ . It is easy to check that these work.

Let $P(x,y)$ denote the given assertion. $P(0,1)$ gives $f(f(0))=0$ . If $f(f(x))\equiv 0$ , $P(x-f(0),0)$ yields $f(x)=0$ if $x\neq f(0)$ . The case $x=f(0)$ was covered already. Hence $f(x)\equiv 0$ .

So assume there exists $k$ such that $f(f(k))\neq 0$ . Then $P(k,x)$ implies that $f$ is injective. Now $P(x,x)$ for $x\neq 0$ gives $f(x+f(x))=0=f(f(0))$ . By injectivity, we have $f(x)=f(0)-x$ . This also agrees with the $x=0$ case so we are done. $\square$

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g0USinsane777

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g0USinsane777

#47 h Jun 22, 2024, 7:53 PM

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Let $P(x,y)$ denote the assertion given in the question.

$P(x,x)$ gives us that $f(x+f(x)) = 0$ for all $x \ne 0$ $-------(*)$
$P(0,y)$ , where $y$ is positive, gives us $f(f(0))=0$ $-------(**)$

Now assume, $f(x_1) = f(x_2)$ for some $x_1,x_2$ , then for some $x \ne 0$ ,
$\Longrightarrow xf(x+f(x_1))=xf(x+f(x_2))$
$\Longrightarrow f(f(x))(x_1-x) = f(f(x))(x_2-x)$
which means that $f(f(x)) = 0 \ \ \forall x$ or $x_1 = x_2$

First lets consider the case when $f(f(x)) = 0$ for all $x$ , then from $P(x,y)$ , we have $xf(x+f(y)) = 0$
$\Longrightarrow x=0$ or $f(x+f(y)) = 0$ .
We can vary $x$ above to get all real values except $f(f(y))$ for a fixed $y$ and if $x=0$ , then we also get $f(f(y)) = 0$ by our assumption.
Hence, we get the solution $f(x) = 0 \ \ \forall x \in \mathbb{R}$

Considering the second case, $f$ becomes injective in this case.
From $(*)$ and $(**)$ , we get that $f(x) = c-x, c \in \mathbb{R}$ which satisfies our original equation.

Hence, the only solutions we get are $f(x)=0$ or $f(x) = c-x \ \ \forall x \in \mathbb{R}$ .

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A-_AR2_-A

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A-_AR2_-A

#49 h Jul 10, 2024, 5:40 PM

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My first comment

claim) $f$ is injective

proof) assume $a,b$ such that $a\neq b$ and $f(a)=f(b)$
$P(x,a)-P(x,b)$ will give that $a=b$ .

$P(x,x) \rightarrow xf(x+f(x))=0$ …. (1)
$P(0,x) \rightarrow xf(f(0)=0$ …. (2)

From (1,2) we get $xf(x+f(x))=xf(f(0) ) \rightarrow x+f(x)=f(0)$ $\rightarrow f(x)= c-x$

by plugging $P(x-f(x),x)$ in (1) we get $f(x)=0$

so we deduce that $f(x)=0$ and $f(x)=c-x$ are only solutions,

This post has been edited 7 times. Last edited by A-_AR2_-A, Jul 29, 2024, 2:37 PM

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SomeonesPenguin

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SomeonesPenguin

#50 h Aug 5, 2024, 3:22 PM • 1 Y

Y by zzSpartan

Cute FE.

$P(x,y):xf(x+f(y))=(y-x)f(f(x))$

Case 1. $f(f(x)) = 0\ \forall x\in \mathbb R$ . We have that $xf(x+f(y)) = 0$ . If $f$ is constant, then $f \equiv 0$ . If not, there exists $a\neq b \in \text{Im}f$ . Take $y$ such that $f(y) = a$ and vary $x$ such that $x+a$ covers $\mathbb R$ and for the case where $x=0$ just swap $a$ with $b$ and take $x=a-b \neq 0$ . Hence we get $f \equiv 0$ .

Case 2. there exists $a$ such that $f(f(a)) \neq 0$ . Suppose that there exists $y_{1}\neq y_{2}$ with $f(y_{1})=f(y_{2})$ . From $P(a,y_{1})$ and $P(a,y_{2})$ we get a contradiction. So $f$ is injective.

$P(0,1) \Rightarrow f(f(0)) = 0$ . And from $P(x,x)$ with $x\neq 0$ we get that $f(x+f(x)) = 0$ so from injectivity we have $x+f(x)=f(0)$ . Notice that $0$ also satisfies this relation so we get that $f(x)=c-x$ which does satisfy the FE.

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onyqz

195 posts

onyqz

#51 h Aug 6, 2024, 5:26 AM

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Let $P(x,y)$ denote the given assertion.
Now note that:
$P(x,x)$ gives $xf(x+f(x))=0$
$P(0,x)$ gives $f(f(0))=0$
Moreover, $P(f(0),x)$ and $P(0,x)$ give $f(0)f(f(0)+f(x))=(x-f(0))f(0)$ , therefore we check the two cases $f(0)=0$ and $f(0)\neq0$ .

Case 1: $f(0)=0$
$P(x,0)$ gives $f(x)=-f(f(x))$
$P(-f(x),x)$ and $P(x,0)$ give $0=(x+f(x))(f(-f(x))$
Case 1.1: $x+f(x)=0$ , hence $f(x)=-x, \forall x\in\mathbb{R}$
Case 1.2: $f(-f(x))=0$ . $P(x,0)$ and $P(x,-f(x))$ give $xf(x)=f(x)^2+xf(x)$ so $f(x)\equiv 0$ .

Case 2: $f(0)\neq 0$ and $f(f(0)+f(x))=x-f(0)$ (which in fact also shows surjectivity).
Let $f(0)=a$ , then from $f(f(0))=0$ we know $f(a)=0$ .

Claim: $f$ is injective.
Proof: Suppose there exist $b, c \in\mathbb{R}$ s.t. $f(b)=f(c)$ . Now look at $P(a,b): a(b-a)=af(a+f(b))=af(a+f(c))=a(c-a)$ , therefore $b=c$ and $f$ is injective.

Looking back at $P(x,x)$ and using injectivity finally shows $x+f(x)=a$ or $f(x)=a-x$ .
We conclude that $\boxed{f(x)=c-x}$ for some constant $c$ and $\boxed{f(x)\equiv 0}$ are the only solutions. $\square$

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Flint_Steel

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Flint_Steel

#52 h Aug 6, 2024, 7:38 AM

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Lunch break!
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Warideeb

59 posts

Warideeb

#53 h Aug 14, 2024, 2:50 PM

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Let $P(x,y)$ be the assertion of
$xf(x+f(y))=(y-x)f(f(x))$
$Our$ $claim$ $is$ $f \equiv 0$ or $f(x)=u-x$ for some constant $u$
Now $f \equiv 0$ is obvious
We assume $f \not\equiv 0$
Now $P(0,y)$ gives us
$y f(f(0))=0$ so $f(f(0))=0$ Let $f(0)$ $be$ $u$ then
$f(u)=0$ Now
$P(x,x)$ gives us
$xf(x+f(x))=0$ then $f(x+f(x))=0$
Now as $f \not\equiv 0$ there exists some $c \ne 0$ such $f(f(c)) \ne 0$ Now
$P(c,y)$ gives us
$cf(c+f(y))=(y-c)f(f(c))$ proving its bijectivity
Now then $f(x+f(x))=0=f(u)$
Then $x+f(x)=u$ then $f(x)=u-x$ for some constant $u$

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kinnikuma

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kinnikuma

#54 h Oct 3, 2024, 6:14 PM

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Let $f$ be a solution to this equation.

Letting $x = y$ gives $xf(x + f(x)) = 0$ .
Letting $x = 0$ and $y \neq 0$ gives $f(f(0)) = 0$ .
Let's then take $y$ such that $f(y) = 0$ : we get $xf(x) = (y-x)f(f(x))$ .

We are going to distinguish two cases :

- if there exists $x$ such that $f(f(x)) \neq 0$ , then $y = \frac{xf(x)}{f(f(x))} + x$ . So if there are two different $y$ such that $f(y) = 0$ then from this equation these two $y$ should be equal, absurd. Therefore there exists at most one $y$ such that $f(y) = 0$ : we have already established that $f(f(0)) = 0$ , so this $y$ must be $f(0)$ . For $x \neq 0$ , we have $f(x+f(x)) = 0$ from the $x=y$ equation. So $x+f(x) = f(0) \Longrightarrow f(x) = -x + f(0)$ . Combining with $f(0) = -0 + f(0)$ , we can affirm that $f$ is of the form $- x + k$ where $k$ is a constant.

- if not, this means that $f(f(x)) = 0$ for all $x$ . Therefore, $xf(x) = 0$ . Clearly for $x \neq 0$ , we necessarily have $f(x) = 0$ . By the way, with $y = 0$ , we have $xf(x+f(0)) = 0$ since $f(f(x)) = 0$ . Letting $x = -f(0)$ we have $-f(0)^2 = 0$ so $f(0) = 0$ . In summary, $f \equiv 0$ .

To conclude there are two potential solutions : $f(x) = 0$ and $f(x) = -x + k$ where $k$ is a constant. We easily check that there are indeed solutions.

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Acclab

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Acclab

#55 h Oct 7, 2024, 3:09 PM

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$P(x,x)$ and $P(0,x)$ gives that $x+f(x)$ is always annihilated by $f$ , then $P(x,x+f(x))$ yields that
$xf(x)=f(x)f(f(x))$ giving either $f(x)=0$ or $f(f(x))=x$ for each $x$ . If there is $x \neq 0$ such that the latter hold, $f$ is injective, giving $f(x) \equiv c-x$ , otherwise we have the constant $0$ function everywhere except $0$ and a quick check reveals at $0$ too, both of which works.

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mathematical717

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mathematical717

#56 h Oct 13, 2024, 9:33 AM

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Let the problem statement be denoted by $P(x,y)$ , we have that $P(x,x): xf(x+f(x))=0 \implies f(x+f(x))=0, \ \ \ \forall x \in \mathbb{R*}$ wherein $\mathbb{R*}$ is reals without $0$ . Now we have $P(-f(y),y): -f(y)f(0)=(y+f(y))f(f(-f(y)))$ where letting $y \ne 0$ we would have, $f(y)f(0)=0$ forcing either $f(0)=0$ otherwise $f(y) \equiv 0 \ \ \forall y \in \mathbb{R*}$ Now if we have $f \equiv 0$ that acts as a solution, assuming there are more we must have some $f(f(x)) \neq 0$ otherwise we have, $xf(x+f(y))=0 \ \forall (x,y) \in \mathbb{R} \implies xf(x+f(f(0)))=0 \implies xf(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R*}$ however for some $y \ne 0$ , we must have that $0=f(f(y))=f(0)$ which indicates identically $0$ a direct contradiction to our assumption. Now we have established our claim so let $f(y)=f(y'), \ y \ne y'$ then we compare $P(x,y)$ and $P(x,y')$ to obtain that $f$ is injective. However since $P(0,1): f(f(0))=0 \ \land \ f(x+f(x))=0 \ \ \forall x \in \mathbb{R*}$ so we must have $f(x)+x=f(0)$ for some constant $f(0)$ and thus $f(x)=f(0)-x \ \forall x \in \mathbb{R*}$ however we must also have that, $f(0)=f(0)-0$ and so we have $f \equiv f(0)-x \ \ \forall x$ .
Thus our solutions are $\boxed{f \equiv 0; f(x)=f(0)-x \ \ \forall x}$ which are trivial to check that they work.

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cosdealfa

27 posts

cosdealfa

#59 h Nov 11, 2024, 7:12 AM • 1 Y

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Denote by $P(x, y)$ the given assertion.
For an $y \neq 0$ $P(0, y): f(f(0)) = 0$
For an $x \neq 0$ $P(x, x): f(x + f(x))=0$

Case 1: $f$ is injective
$\Rightarrow x + f(x) = f(0)$ so $f(x) = f(0) - x, \forall x \in \mathbb{R}$

Case 2: $f$ is not injective
Then we can find $a, b \in \mathbb{R}$ such that $a \neq b$ and $f(a) = f(b)$

Then for any $x$ such that $x \neq a, x \neq b, x \neq 0$ we have:
$P(x, a) : xf(x + f(a)) = (a-x)f(f(x))$
$P(x, b): xf(x + f(b)) = (b-x)f(f(x))$

Assume LHS $\neq 0$ then RHS $\neq 0$ . It follows that $a-x = b-x$ which is false, by the way we picked $x$ .
So LHS $= 0$ therefore $f(x + f(a)) = 0$ . $\forall x$ such that $x \neq a, x \neq b, x \neq 0$ .
We are only left to check if $f(a + f(a)), f(b+ f(b)), f(f(a))$ are $0$ .
By $P(a,a)$ and $P(b,b)$ we get $af(a + f(a)) = bf(b + f(b))= 0$ If $a, b \neq 0$ we are done. So assume one of them is $0$ . If $a = 0$ :
$\Rightarrow f(b + f(0)) = 0$ and by the first observation $f(f(0)) =0$ . We also have $f(x + f(0)) = 0$ for all $x \neq 0, x\neq b$ . So basically, $f \equiv 0$ . $\blacksquare$

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iStud

257 posts

iStud

#60 h Dec 8, 2024, 1:08 PM

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nice problem for P1 level of BMO!

Clearly $\boxed{f(x)=0\quad\forall x\in\mathbb{R}}$ is a solution, so we'll now look for all nonzero solutions. Let $P(x,y)$ be the assertion of the functional equation given.

Take $x=y$ , so $xf(x+f(x))=0$ . For $x\ne 0$ , the equation yields $f(x+f(x))=0\dots(1)$ . Now taking $P(0,x)$ easily gives us that $xf(f(0))=0$ , which is $f(f(0))=0\dots(2)$ for $x\ne 0$ . Suppose there exists $a,b\in\mathbb{R}$ so that $f(a)=f(b)$ . Hence, by comparing $P(1,a)$ and $P(1,b)$ yields $a-1=b-1$ $\Longleftrightarrow$ $a=b$ $\Longleftrightarrow$ $f$ is injective. So by applying injectivity at $(1)$ and $(2)$ will implies $x+f(x)=f(0)$ $\Longleftrightarrow$ $f(x)=f(0)-x$ . Now by letting $f(0)=c$ for some $c\in\mathbb{R}$ , we easily get $\boxed{f(x)=c-x\quad\forall x\in\mathbb{R}}$ , which is indeed a solution. Therefore, we are done. $\blacksquare$

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Davud29_09

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Davud29_09

#61 h Jan 5, 2025, 11:10 AM

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P(1,0) implies that f(f(0))=0.if f(f(x)) isn't constant then f is injective .And P(x,x) implies f(x+f(x))=0=f(f(0)) and we get f(x)=c-x from injectivity.If f(f(x)) is constant f(f(x))=0 and y>>f(0) implies f(x)=0 Answers: f(x)=0 and f(x)=c-x.

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TestX01

329 posts

TestX01

#63 h Jan 14, 2025, 4:30 AM

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Suppose $f(f(x))=0$ holds for all $x$ . Then if $x\neq 0$ , we have $f(x+f(y))=0$ . Fix $y=y_0$ . Then $f(f(y_0))=0$ , and $f(f(y_0)+x)=0$ for $x\neq 0$ but $x+f(y_0)$ along with $f(y_0)$ produce entire set of reals. Hence $f$ is $0$ . We check easily that this works.

Now, subbing $x=0$ gives $0=yf(f(0))$ . Pick $y$ nonzero, then $f(f(0))=0$ . Hence, assume that $f(f(x_0))\neq 0$ for some $x_0\neq 0$ . Substituting that for $x$ , we have
$f(c+f(y))=c'(y-x_0)$ for constants $c',c,x_0$ all non-zero. $f$ is injective as there is an isolated $f(y)$ on the LHS, noting $c'\neq 0$ . $f$ is also surjective because shifting and scaling the reals still gives the reals on the RHS. Hence, there is an unique $k$ such $f(k)=0$ . Letting $x=y$ initially, we have $xf(x+f(x))=0$ , so when $x\neq 0$ , then $f(x+f(x))=0=f(k)$ , injectiveness yields $f(x)=k-x$ . Of course, when $x=0$ , then $f(0)=k=k-0$ too. One checks this works easily.

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awesomeming327.

1668 posts

awesomeming327.

#64 h Mar 19, 2025, 3:24 AM

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The answers are $f(x)=c-x$ and $f(x)=0$ . These clearly work, so we will now prove that they are the only functions. Let $P(x,y)$ denote the assertion.

Claim 1: $f(x+f(x))=0$ .

If $x\neq 0$ then use $P(x,x)$ to get $xf(x+f(x))=0$ . If $x=0$ then use $P(0,1)$ to get that $f(f(0))=0$ , which finishes the claim.

If $f(f(x))=0$ for all $x$ then $xf(x+f(y))=0$ for all $x$ . Then $P(x,f(x))$ gives $xf(x)=0$ for all $x$ . If $x\neq 0$ then $f(x)=0$ . If $x=0$ then take $P(-f(0),0)$ to get $-f(0)^2=0$ which implies $f(0)=0$ .

If $f(f(x))\neq 0$ for even a single value of $x$ then if $f(y_1)=f(y_2)$ , then comparing $P(x,y_1)$ and $P(x,y_2)$ immediately gives $y_1=y_2$ , so $f$ is injective. Then, by Claim $1$ , $x+f(x)$ is a constant. We are done.

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