International FE Olympiad P1

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Functional_equation

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Functional_equation

#1 h Feb 6, 2021, 5:43 AM • 4 Y

Y by Aritra12, whyareFEssofun, itslumi, bin_sherlo

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any reals $x,y$ ,
$f(x+f(y)+yf(x))=f(x+y)+xf(y)$
$\textit{Proposed by gghx}$

This post has been edited 2 times. Last edited by Functional_equation, Feb 6, 2021, 6:06 AM

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Pitagar

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Pitagar

#2 h Feb 6, 2021, 6:42 AM • 1 Y

Y by mijail

The solution that I submitted:

Problem 1-1: $f(x+f(y)+yf(x))=f(x+y)+xf(y)$
The only answers are $f(x)\equiv 0 , \forall x \in \mathbb{R}$ and $f(x)\equiv x, \forall x \in \mathbb{R}$ . Its easy to see that these functions satisfy the given equation. Now we'll prove that these are the only solutions.
Let $P(x,y)$ denote the given assertion $f(x+f(y)+yf(x))=f(x+y)+xf(y)$ .Now
$P(0,0)\Rightarrow f(f(0))=f(0); (1)$
$P(x,0)\Rightarrow f(x+f(0))=f(x)+xf(0); (2)$
$P(0,x)\Rightarrow f(f(y)+yf(0))=f(y); (3)$ . Now using $(1)$
$P(f(0),x)\Rightarrow f(f(0)+f(x)+xf(0))=f(f(0)+x)+f(0)f(x)=f(x)+xf(0)+f(0)f(x)$ from $(2)$ . But from $(2)$ applied for $f(x)+xf(0)$ instead of $x$ and then from $(3)$ we get consecutively
$f(f(0)+f(x)+xf(0))=f(f(x)+xf(0))+(f(x)+xf(0))f(0)=f(x)+f(0)f(x)+xf^2(0) \Rightarrow f(x)+xf(0)+f(0)f(x)=f(x)+f(0)f(x)+xf^2(0) \Rightarrow xf(0)=xf^2(0) \forall x \in \mathbb{R}$ , which is only possible when $f(0)=1$ or $f(0)=0$ . Now lets check the case $f(0)=1$
$P(0,y)\Rightarrow f(f(y)+y)=f(y)$ $(4)$
$P(x,0)\Rightarrow f(x+1)=f(x)+x$ $(5)$ .From (5) we get that $f(1)=f(0)+0=1$ From (5) again we get that $f(2)=f(1)+1=2$ , but from (4) we get that $f(2)=f(f(1)+1)=f(1)=1$ , which is a contradiction, so we get that $f(0)=0$
$P(0,k)\Rightarrow f(f(k))=f(k)$ $(*)$ .
Now lets assume that there exists a real number $a\neq 0$ , such that $f(a)=0$ . Now lets assume that there exists a real number $b$ , such that $f(b)\neq 0$ . Now using $(*)$ we get that
$P(a,f(b))\Rightarrow f(a+f(b))=f(a+f(b)) + af(b) \Leftrightarrow af(b)=0$ , which is impossible because of the assumption, thus when there exists such $a$ we get that $f\equiv 0$ , which is indeed a solution.
Now we are in the case where only for $k=0$ , $f(k)=0$ . Now again using $(*)$ we get
$P(1, -\frac{1}{f(1)} ) \Rightarrow f(f(-\frac{1}{f(1)}))= f(1-\frac{1}{f(1)} ) + f(-\frac{1}{f(1)}) \Leftrightarrow f(1-\frac{1}{f(1)}) = 0$ , so $1=\frac{1}{f(1)}$ , thus $f(1)=1$ . Now consecutively for every $x\in \mathbb{R}$ we get
$P(1,x)\Rightarrow f(1+f(x)+x)= f(1+x)+f(x)$ ;
$P(x,1)\Rightarrow f(x+1+f(x))=f(x+1)+x$ . $\Rightarrow f(x)=x$ and now we're done! $\blacksquare$

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gghx

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gghx

#3 h Feb 6, 2021, 6:54 AM • 3 Y

Y by mijail, Kyleray, Math_legendno12

Sketch: $f(0)=0, f$ injective at $0$ , $f(1)=1$ , then $P(x,1),P(1,x)$ .

Comments: Most found this hard for its position. Every solution submitted makes use of the fact that $f$ is injective at $0$ . Also, there are many different ways to find $f(0)$ , I think almost everyone has a different way.

Solution I submitted

This post has been edited 5 times. Last edited by gghx, Aug 10, 2021, 3:42 AM

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EpicNumberTheory

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EpicNumberTheory

#4 h Feb 6, 2021, 6:58 AM • 1 Y

Y by itslumi

My way of obtaining $f(0) = 0$ (Wasn't able to solve fully

)
$P(0,0) \Rightarrow f(f(0)) = f(0)$ $P(x,0) \Rightarrow f(x+f(0)) = f(x) + xf(0) \Rightarrow f(f(x+f(0))) = f(f(x) + xf(0))$ $P(0,x) \Rightarrow f(f(x) + xf(0)) = f(x)$ Combining $P(x,0)$ and $P(0,x)$ yeilds:
$\Rightarrow f(x) = f(f(x+f(0))$ Substituting $x = -f(0)$ yeilds:
$\Rightarrow f(-f(0)) = f(f(0)) = f(0)$ $P(-f(0),0) \Rightarrow f(0) = f(-f(0)) - f(0)^2 \Rightarrow -f(0)^2 = 0 \Rightarrow f(0) = 0$

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OlyMan

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OlyMan

#5 h Feb 6, 2021, 10:14 AM

Y by

@Pitagar I advance upto (3) but couldn't advance further! :oops:

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itslumi

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itslumi

#6 h Feb 6, 2021, 10:32 AM • 2 Y

Y by CANBANKAN, GorgonMathDota

I think it was too hard for P1 :wacko:

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EpicNumberTheory

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EpicNumberTheory

#7 h Feb 6, 2021, 10:36 AM

Y by

itslumi wrote:

I think it was too hard for P1 :wacko:

I don't think it's too hard. The hard part about it is to notice the "trick" to substitute in $P(1,-\frac{1}{f(1)})$ (which I failed to do). If you can find $f(0)$ in a sort of easy way, like I did for example, then its not too hard (Although it took me 2 days of thinking to get to $f(0) = 0$ :blush:

). Then $f(f(x)) = f(x)$ is trivial. Then its just about finding $f(1)$ and that "trick". Its a nice problem, perhaps a bit IMO style.

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itslumi

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itslumi

#8 h Feb 6, 2021, 11:55 AM • 4 Y

Y by CANBANKAN, Mango247, Mango247, Mango247

EpicNumberTheory wrote:

itslumi wrote:

I think it was too hard for P1 :wacko:

I don't think it's too hard. The hard part about it is to notice the "trick" to substitute in $P(1,-\frac{1}{f(1)})$ (which I failed to do). If you can find $f(0)$ in a sort of easy way, like I did for example, then its not too hard (Although it took me 2 days of thinking to get to $f(0) = 0$ :blush:

). Then $f(f(x)) = f(x)$ is trivial. Then its just about finding $f(1)$ and that "trick". Its a nice problem, perhaps a bit IMO style.

actually i got f(0)=0 very fast,but got stuck after that,i think the problem should be p2 at least

P.s(i didn't submit my Work for f(0)=0,i was frustrated)

This post has been edited 3 times. Last edited by itslumi, Feb 6, 2021, 11:58 AM

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Soundricio

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Soundricio

#9 h Feb 6, 2021, 4:29 PM

Y by

..............

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Reason: .

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CANBANKAN

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CANBANKAN

#10 h Feb 6, 2021, 10:32 PM • 1 Y

Y by MS_asdfgzxcvb

Took me 2.5h

The answer is $f(x)\equiv x, f(x)\equiv 0.$ They clearly work.

Let $P(x,y)$ denote the assertion $f(x+f(y)+yf(x))=f(x+y)+xf(y)$
Claim. $\forall n\in \mathbb{N}, f(-n)=nf(-1)$ . We induct on $n$ .

The base case, $n=1$ , is self-explanatory.

Suppose this is true for $n$ , then $P(-1,-n)$ gives $f(-1-f(-n)-nf(-1))=f(-1-n)-f(-n)$ .

Since $f(-n)=nf(-1), f(-1)=f(-n-1)-f(-n)$ , completing the induction.

Let $l=f(-1)$ Note $P(-2,-2)$ gives $f(-2l-2)=0$ .

Claim $f(0)=0$ .

Assume for contradiction that $f(0)\ne 0$ . Clearly, $-2l-2\ne 0$ .

Let $\alpha=f(0)$ .

$P(0,0)$ gives $f(\alpha)=\alpha$ .

$P(-2l-2,\alpha)$ gives $f(-2l-2+\alpha+0)=f(-2l-2+\alpha)+(-2l-2)f(\alpha)$ , which implies $(-2l-2)f(\alpha)=0$ , absurd.

Therefore, $f(0)=0$

Now, $P(0,x)$ gives $f(f(x))=f(x)$ for all $x.$

Let $S=\{ x|f(x)=0\}.$

Case 1: There exists a nonzero element of $S.$

Assume $f(x)=0, x\ne 0, f(y)=y$ . I claim $y=0.$ $P(x,y)$ yields $f(x+y)=f(x+y)+xf(y)$

Hence $f(y)=0, y=0$ .

If there exists $f(x)\ne 0,$ then $f(f(x))=f(x),$ contradiction. Therefore, $f(x)\equiv 0.$

Case 2: $f(x)=0 \leftrightarrow x=0$ .

Since $f(-2l-2)=0, 2l+2=0, l=-1$ . Therefore, $f(-x)=-x\forall x\in\mathbb{N}$

Now, $P(x,-1)$ gives $f(x-1-f(x))=f(x-1)-xf(-1)$ , and $P(-1,x)$ gives $f(x-1-f(x))=f(-1+x-xf(-1))=f(x-1)-f(x)$ . Therefore, $xf(-1)=-f(x),$ so $f(x)\equiv x,$ as desired.

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alexiaslexia

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alexiaslexia

#11 h Feb 9, 2021, 4:59 AM • 5 Y

Y by jasperE3, Functional_equation, whyareFEssofun, gghx, MS_asdfgzxcvb

Funniest, and hardest P1 I have done (during my 5-year olympiad career.)
For example, I consider this problem harder than this, as this took more than twice the time I needed for the USEMO problem.

Anyway, here's the Solution. Let $P(x,y)$ be the usual assertion.

$\color{green} \rule{25cm}{2pt}$
$\color{green} \textbf{Same Joke Twice: First Iteration.}$ We prove that $f(0)=0$ , and $f(f(x))=f(x)$ as in post $\#4$ .

$\color{green} \textbf{Proof 1.}$ Here $P(x,0)$ and $P(0,x)$ complements each other into
$f(f(x+f(0))) = f(x) \ - (\bigstar)$ Then, we apply what I'd call the method of $``\textit{\textbf{same joke twice}}"$ , that is, utilizing an equation two times or more, in an ingenious manner.

Firstly, $P(0,0)$ implies
$f(f(0))=f(0)$ $(\bigstar)$ with $x = -f(0)$ implies
$f(0) = f(f(0)) = f(-f(0))$ Then, a $\textit{same joke twice}$ with $P(x,0)$ implies
$f(x+f(0)) = f(x)+xf(0) \Rightarrow f(x) = f(x-f(0)) + (x-f(0))f(0)$ substituting $x = 0$ to this yields $f(0)^2 = 0$ , which yields the result $f(0) = 0$ . Landing a final blow, a $\textit{same joke twice}$ on $P(0,x)$ yields the second conclusion. $\blacksquare$

$\color{magenta} \rule{25cm}{3pt}$

In this Section, I will not use injectivity in 0 (explicitly --- unless you count the reasoning in $\color{magenta} \textbf{Part 2}$ to dismiss the trivial possibility $f(1) = 0$ as proving injectivity in $0$ ), but rely on the fact that $f(r \in \text{range}) = r$ .
$\color{magenta} \textbf{Same Joke Twice: Second Iteration.}$ We claim that if $f(1) = 0$ , then $f(x) = 0$ ; otherwise, $f(1) = 1$ .

$\color{magenta} \textbf{Proof 2.}$ $P\bigg(1,-\dfrac{1}{f(1)}\bigg)$ implies
$f\left(-\dfrac{1}{f(1)}\right) = f\left(-\dfrac{1}{f(1)}+1\right)+f\left(-\dfrac{1}{f(1)}\right)$ so we get that $f\left(-\dfrac{1}{f(1)}+1\right)=0$ .

Switching them a bit (surprisingly, this seems to be the key of the problem): $P\bigg(-\dfrac{1}{f(1)},1\bigg)$ implies
$f\left(-\dfrac{1}{f(1)}+f(1)+f\left(-\dfrac{1}{f(1)}\right)\right) = f\left(-\dfrac{1}{f(1)}+1\right)+ -\dfrac{1}{f(1)} \cdot f(1) = -1$ so, $f(-1) = -1$ , since $-1 \in \text{range}$ (this is the most surprising find, for me).

Then, from here, I will prove that $f(z \in \mathbb{Z}) = z$ . $P(x,-1)$ implies
$f(x-1-f(x)) = f(x-1)-x$ so, $f(x) = x$ will implies $f(x-1) = x-1$ . Chaining this down from $x = -1$ to $- \infty$ , this implies $f(-n) = -n$ . And then, applying $P(-n,-n)$ yields
$f(n^2-2n) = f(-2n) + (-n) \cdot f(-n) = n^2-2n$ which when chained down, yields $f(k) = k$ $\forall k \in \mathbb{N}$ .

Surely, we are almost done, right?

$\color{magenta} \textbf{Part 2.}$ However, we cannot substitute the value of $-\dfrac{1}{f(1)}$ whenver $f(1) = 0$ . In this case, another $\textit{same joke twice}$ on $P(1,y)$ yields
$f(f(y)+1) = f(y+1)+f(y)$ Putting $y = f(y')$ implies that $f(f(y')+1) = f(f(y')+1) + f(y')$ since $f(f(y')) = f(y')$ . As a result, $f(y') = 0$ .
So, the desired conclusion follows. $\blacksquare$ $\blacksquare$

$\color{blue} \rule{25cm}{2pt}$
$\color{blue} \textbf{Same Joke Twice: Third Iteration.}$ After $f(1) = 1$ , a third $\textit{same joke twice}$ on $P(x,1), P(1,y)$ yields
$f(x+1+f(x)) = f(x+1)+x; \quad f(1+f(y)+y) = f(y+1) + f(y)$ Regular comparison and change of variable yields $f(x+1)+x=f(x+1)+f(x)$ , finalizing the Solution. $\blacksquare$ $\blacksquare$ $\blacksquare$ .

Motivation: A very similar flavour to IMO 2020/2.

$\textbf{EDIT:}$
Reply to @2below

$\textbf{EDIT 2:}$
A new trick I learnt when posting this

This post has been edited 8 times. Last edited by alexiaslexia, Feb 12, 2021, 5:03 AM

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OlyMan

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OlyMan

#12 h Feb 9, 2021, 7:40 AM

Y by

EpicNumberTheory wrote:

itslumi wrote:

I think it was too hard for P1 :wacko:

I don't think it's too hard. The hard part about it is to notice the "trick" to substitute in $P(1,-\frac{1}{f(1)})$ (which I failed to do). If you can find $f(0)$ in a sort of easy way, like I did for example, then its not too hard (Although it took me 2 days of thinking to get to $f(0) = 0$ :blush:

). Then $f(f(x)) = f(x)$ is trivial. Then its just about finding $f(1)$ and that "trick". Its a nice problem, perhaps a bit IMO style.

Yes my thought process was a bit like your(not able to solve completely) but the guess $f(0)=0,$ I thought this very early but couldn't find a way to establish this! I also considered $P(1,x) \Rightarrow f(1+f(x)+f(1)x)=f(1+x)+f(x), \forall x\in\mathbb{R}.$ and $P(0,x)\Rightarrow f(f(x)+xf(0)=f(x), \forall x\in \mathbb{R}.$ So here I also made an attempt to show $f$ is additive (if $f(f(x))= f(x), \forall x\in\mathbb{R}$ ) but... couldn't find a way. I am quite unexperienced... I need a lot practice. Btw, did some one also got a solution by showing $f$ is addictive??

This post has been edited 1 time. Last edited by OlyMan, Feb 9, 2021, 7:42 AM
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gghx

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gghx

#13 h Feb 9, 2021, 9:44 AM • 2 Y

Y by whyareFEssofun, Mango247

alexiaslexia wrote:

Funniest, and hardest P1 I have done (during my 5-year olympiad career.)
For example, I consider this problem harder than this, as this took more than twice the time I needed for the USEMO problem.

Oh wow I expected this question to be hard, but not so hard!
When I created this problem, I was trying something where the equation is something in the form $f(stuff)=af(b)+f(c)$ where $a,b,c$ are functions in $x,y$ and then the substitution forcing $a=-1, b=c$ implies that $0$ is in the range of $f$ , and then slowly kill the question from there. But whatever I tried the problem always seemed to be to easy (like setting $x=c$ where $f(c)=0$ is an instant kill) or impossible to solve.

So then I was wondering what if we take $af(b)+f(c)$ such that it is impossible to force $a=-1,b=c$ ? The easiest way was the one in the problem, $xf(y)+f(x+y)$ , because taking $x=-1$ there is no solution for $y-1=y$ .
The inside of $f$ on the LHS was created by trying to prevent a situation where you can force the inside to be a certain value, like for example $0$ , but with not so many $f$ s as well else it was probably impossible to solve.

When trying this question I had just finished creating a few problems where the key is to bash a few values and then done. For example this, this, and this. So without any other ideas (since i created the problem to prevent them), I just started bashing. Took a while to get $f(0)$ but not that long either.

After that, I actually got $f(1-\frac{1}{f(1)})=0$ first before trying to prove injectivity at $0$
The subs $P(x,1),P(1,x)$ was really funny, because near that time i was trying to create an FE in the form $f(x^2+y^2x+y)=stuff$ , and then $P(x,1),P(1,x)$ yields the same left hand side. So when the same thing solves it i figured out i shouldn't post that FE, which may give unfair advantage to those who have seen that problem.

Took me about 45 minutes to solve, but i think the main point why this was such a hard q1 was because when i solved it I had the correct techniques at the back of my mind beforehand. (Im sorry to those who gave up on the entire competition after q1

)

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Functional_equation

530 posts

Functional_equation

#15 h Feb 13, 2021, 1:05 PM

Y by

This is IFEO SL A2

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hakN

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hakN

#16 h Apr 4, 2021, 2:03 PM • 2 Y

Y by Mango247, Mango247

Let $P(x,y)$ be the assertion.
$P(0,0) \implies f(0)=f(f(0))$ .
$P(x,0) \implies f(x+f(0))=f(x)+xf(0)$ . $(1)$
$P(x,f(0)) \implies f(x+f(0)+f(x)\cdot f(0)) = f(x)+2xf(0)$ . $(2)$
Multiplying $(1)$ by $2$ and subtracting from $(2)$ we get
$f(x+f(0)) - f(x+f(0)+f(x)\cdot f(0)) = f(x) - f(x+f(0))$ . $(3)$
Plugging $x=0$ in $(3)$ we get $f(f(0)^2 + f(0)) = f(0) = f(f(0)^2) + f(0)^3$ .
$P(0,x) \implies f(f(x) + xf(0)) = f(x)$ . $(4)$
Plugging $x=f(0)^2$ in $(4)$ we get $f(f(f(0)^2) + f(0)^3) = f(0) = f(f(0)^2) \implies f(0)^3 = 0 \implies f(0)=0$ .
So now we have $f(f(x))=f(x) \forall x\in \mathbb{R}$ .
$P(f(x),-1) \implies f(-1) = f(f(x)-1) + xf(-1)$ . Changing $x$ with $f(x)$ gives us $f(-1)(f(x)-x) = 0 \forall x\in \mathbb{R}$ .
Now if $f(-1)\neq 0$ we have $\boxed{f(x)=x} \forall x\in \mathbb{R}$ which is clearly a solution.
If $f(-1) = 0$ we have $f(f(x)-1) = 0$ .
Finally,
$P(-1,x) \implies 0=f(f(x)-1) = f(x-1)-f(x) \implies f(x)=f(x-1) \forall x\in \mathbb{R}$ . Again changing $x$ with $f(x)$ gives us that $\boxed{f(x) = 0} \forall x\in \mathbb{R}$ which is clearly a solution.

@below Yeah you are right

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gghx

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gghx

#17 h Jun 8, 2021, 4:18 AM

Y by

hakN wrote:

$P(f(x),-1) \implies f(-1) = f(f(x)-1) + xf(-1)$ .

Unfortunately this is wrong. The $x$ on the RHS should be $f(x)$ .

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megarnie

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megarnie

#18 h Nov 20, 2021, 8:28 PM

Y by

$P(0,0): f(f(0))=f(0)$ .

$P(x,0): f(x+f(0))=f(x)+xf(0)$ .

$P(0,x): f(f(x)+xf(0))=f(x)$ .

$P(f(0),x): f(f(0)+f(x)+xf(0))=f(x+f(0))+f(0)f(x)=f(x)+xf(0)+f(x)f(0)$ . But we also have $f(f(0)+f(x)+xf(0))=f(f(x)+xf(0))+f(x)f(0)+xf(0)^2=f(x)+f(x)f(0)+xf(0)^2$

So $f(x)+xf(0)+f(x)f(0)=f(x)+f(x)f(0)+xf(0)^2\implies xf(0)=xf(0)^2\implies f(0)=\{0,1\}$ .

AFTSOC $f(0)=1$ . So $f(1)=1$ .
$P(1,0): f(2)=2$ .
$P(0,1): f(2)=1$ , a contradiction.

So $f(0)=0$ .

Claim: If there exists a $k\ne0$ so that $f(k)=0$ , then we must have $\boxed{f\equiv0}$ , which is clearly a solution.
Suppose $f(a)\ne0$ .

$P(0,x): f(f(x))=f(x)$ .

$P(k,f(a)): f(k+f(a))=f(k+f(a))+kf(a)\implies kf(a)=0\implies f(a)=0$ , a contradiction.

Now for the case where if $x\ne0$ , then $f(x)\ne0$ .

Claim: It suffices to show $f(1)=1$ .
Proof:
$P(1,x): f(f(x)+x+1)=f(x+1)+f(x)$

$P(x,1): f(f(x)+x+1)=f(x+1)+x$ .

Subtracting the two equations gives $\boxed{f(x)=x\forall x\in\mathbb{R}}$ .

Claim: $f(1)=1$ , which finishes.

We will be using $f(f(x))=f(x)$ .

$P(1,-\frac{1}{f(1)}): f(-\frac{1}{f(1)})=f(-\frac{1}{f(1)}+1)+f(-\frac{1}{f(1)})\implies f(-\frac{1}{f(1)}+1)=0\implies -\frac{1}{f(1)}=-1\implies f(1)=1$

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ZETA_in_olympiad

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ZETA_in_olympiad

#19 h May 22, 2022, 2:00 PM • 1 Y

Y by Mango247

Nice problem. Going to rate in MOHS or PSC Scale after I solve it. BTW, is the IFEO online or something?

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ZETA_in_olympiad

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ZETA_in_olympiad

#20 h May 22, 2022, 3:02 PM

Y by

The problem is just substitution but lots of them, so a typical IMO P2 and ISL A4 IMHO. And for MOHS, its 20.

$P(0,0)\implies f(f(0))=f(0).$ Now compare $P(0,x)$ and $P(x,0).$ So $f(f(x+f(0)))=f(x).$ Now $x\mapsto -f(0)$ give $f(f(0))=f(-f(0)).$ It is not hard to get $f(0)=0$ from $P(-f(0),0).$ Also note that $f(f(x))=f(x).$

Now we do case-work on $f(x).$

Case 1. $f(x)\neq 0 ~\forall x\neq 0:$ Then $P(1,-f(1)^{-1})$ gives $f(1-f(1)^{-1})=0$ and thus $f(1)=1.$ Now we compare $P(1,x)$ and $P(x,1)$ and see that the identity satisfies.

Case 2. $f(x)=0$ for some $x\neq 0:$ Then let $m$ be such and let $n\neq 0$ and $f(n)\neq 0.$ Now by $P(m,f(n))$ we get $mf(n)=0,$ so a contradiction and the identically zero function satisfies.

And we're done exhausting the cases of $f(x).$

This post has been edited 2 times. Last edited by ZETA_in_olympiad, May 24, 2022, 1:42 PM

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CrazyInMath

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CrazyInMath

#21 h Aug 20, 2023, 2:11 PM

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$P(0,0)\Longrightarrow f(f(0))=f(0)$
$P(0,x)\Longrightarrow f(f(x)+xf(0))=f(x)$
$P(x,0)\Longrightarrow f(x+f(0))=f(x)+xf(0)$ so $f(x)=f(f(x+f(0))$
$x=-f(0)$ gives $f(-f(0))=f(f(0))=f(0)$ and so $P(-f(0),0)$ gives $f(0)=0$
Now we have $f(f(x))=f(x)$ , we want injectivity at $0$
Let $f(t)=0$ , $P(t,f(y))$ gives $f(t+f(y))=f(t+f(y))+tf(y)$ so $t=0$ unless $f(x)=0$
Now $P(1,\frac{-1}{f(1)})$ gives $f(1)=1$
$P(x,1)-P(1,x)$ yields $f(x)=x$ .

This post has been edited 1 time. Last edited by CrazyInMath, Aug 20, 2023, 2:12 PM

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flower417477

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flower417477

#22 h Oct 1, 2023, 5:47 AM

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storage

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bin_sherlo

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bin_sherlo

#23 h Dec 13, 2024, 8:11 AM

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$f(x+f(y)+yf(x))=f(x+y)+xf(y)$ The functions satisfying this equation are $f(x)=x$ and $f\equiv 0$ which clearly work. Let $P(x,y)$ be the assertion. If $f$ is constant, then $f\equiv 0$ . Suppose that it's nonconstant.

Claim: $f(0)=0$ .
Proof: $P(0,0)$ gives $f(f(0))=f(0)$ . Plug $y=0$ to get $f(x+f(0))=f(x)+xf(0)$ . Now, pick $x=0$ which yields
$f(f(y+f(0)))=f(f(y)+yf(0))=f(y)$ $y=-f(0)$ implies $f(0)=f(f(0))=f(-f(0))$ and $P(-f(0),0)$ gives $f(0)=f(-f(0))-f(0)^2$ hence $f(0)=f(-f(0))=f(0)^2-f(0)$ so $f(0)=0$ . $\square$ .

Claim: If $f$ is nonconstant, then $f(a)=0\iff a=0$ .
Proof: First, $P(0,y)$ implies $f(f(y))=f(y)$ . FTSOC assume $f(a)=0$ and $a\neq 0$ . $P(a,f(y))$ yields
$f(a+f(y))=f(a+f(f(y))+f(y)f(a))=f(a+f(y))+af(y)\implies af(y)=0$ Since $a\neq 0$ , we conclude that $f\equiv 0$ . $\square$

Claim: $f(1)=1$ .
Proof: $P(1,-\frac{1}{f(1)})$ implies
$f(-\frac{1}{f(1)})=f(f(-\frac{1}{f(1)}))=f(1+f(-\frac{1}{f(1)})-1)=f(1-\frac{1}{f(1)})+f(-\frac{1}{f(1)})\implies f(1-\frac{1}{f(1)})=0$ Thus, $f(1)=1$ . $\square$

By comparing $P(x,1)$ and $P(1,x)$ we observe that
$f(x+1)+x=f(x+1+f(x))=f(x+1)+f(x)$ So we have $f(x)=x$ as desired. $\blacksquare$

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jasperE3

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jasperE3

#24 h Apr 11, 2025, 6:32 AM

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Functional_equation wrote:

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for any reals $x,y$ ,
$f(x+f(y)+yf(x))=f(x+y)+xf(y)$
$\textit{Proposed by gghx}$

Note that $\boxed{f(x)=0}$ is a solution, else suppose there exists a $j$ with $f(j)\ne0$ .

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

$P(0,x)\Rightarrow f(f(x))=f(x)$

Suppose $f(k)=0$ for some $k$ .
$P(k,f(j))\Rightarrow k=0$

Now we have $f(1)\ne0$ , since $f(1)=0$ would lead to $1=0$ .
$P\left(1,-\frac1{f(1)}\right)\Rightarrow f\left(1-\frac1{f(1)}\right)=0\Rightarrow f(1)=1$
$P(x,1)\Rightarrow f(x+f(x)+1)=f(x+1)+x$
$P(1,x)\Rightarrow f(x+f(x)+1)=f(x+1)+f(x)$
Comparing, we get $\boxed{f(x)=x}$ which fits.

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