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Source: IFEO Day 1 P3

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Functional_equation

530 posts

Functional_equation

#1 h Feb 6, 2021, 5:57 AM • 4 Y

Y by Aritra12, Mathematicsislovely, ywq233, megarnie

Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that $f(f(x)f(f(x))+y)=xf(x)+f(y)$ for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$

This post has been edited 2 times. Last edited by Functional_equation, Feb 6, 2021, 6:15 AM

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Functional_equation

530 posts

Functional_equation

#2 h Feb 6, 2021, 9:43 AM • 1 Y

Y by megarnie

Note: We will not share the official solutions yet. We will share when the Solution Bookled is ready.

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23512 posts

pco

#3 h Feb 7, 2021, 9:44 AM • 10 Y

Y by Pitagar, Functional_equation, Mathematicsislovely, EmilXM, Atpar, ywq233, Supercali, OlympusHero, megarnie, terg

Functional_equation wrote:

Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that $f(f(x)f(f(x))+y)=xf(x)+f(y)$ for all $x,y\in \mathbb R^+$

Let $P(x,y)$ be the assertion $f(f(x)f(f(x))+y)=xf(x)+f(y)$

Simple induction implies
New assertion $Q(x,y,n)$ : $f(y+nf(x)f(f(x)))=f(y)+nxf(x)$ $\forall x,y>0$ , $\forall n\in\mathbb Z_{\ge 0}$

1) New assertion $R(x,y)$ : $f(y)\ge \frac x{f(f(x))}y-xf(x)$ $\forall x,y>0$
Proof

2) $f(f(x))=x\quad\forall x>0$ (and so $f(x)$ is bijective)
Proof

3) $\boxed{f(x)=x\quad\forall x>0}$
Proof

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Functional_equation

530 posts

Functional_equation

#4 h Feb 7, 2021, 12:01 PM

Y by

@PCO congratulations! :first:

:first:

Note: None of the participants in the contest was able to solve this problem.

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Functional_equation

530 posts

Functional_equation

#5 h Feb 13, 2021, 1:09 PM • 3 Y

Y by ywq233, Aritra12, megarnie

This is IFEO SL A7

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Functional_equation

530 posts

Functional_equation

#6 h Jun 20, 2021, 11:57 AM • 1 Y

Y by MeowX2

Official Solution

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114 posts

#7 h Jun 20, 2021, 12:20 PM • 1 Y

Y by Mango247

I think that there is a simple method
Juste set x=0 and since f(0) exist we can also set y=-f(0)f(f(0))
Now we get f(0)=-f(0)f(f(0)) and then we have f(0)=0 or f(f(0))=-1
And the remainder follow from both cases
And we find that the only solution is f(x)=x

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114 posts

#8 h Jun 20, 2021, 12:23 PM • 1 Y

Y by Mango247

Functional_equation wrote:

@PCO congratulations! :first:

:first:

Note: None of the participants in the contest was able to solve this problem.

Are u sure?
Cause it seems like obvious

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DottedCaculator

7354 posts

DottedCaculator

#9 h Jun 20, 2021, 12:24 PM

Y by

Hello, $f(0)$ does not exist, as the domain of $f$ is all positive real numbers.

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114 posts

#10 h Jun 20, 2021, 12:59 PM • 1 Y

Y by Mango247

DottedCaculator wrote:

Hello, $f(0)$ does not exist, as the domain of $f$ is all positive real numbers.

Why do you except 0 to positive real number?

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DottedCaculator

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DottedCaculator

#11 h Jun 20, 2021, 1:01 PM • 1 Y

Y by Mango247

0 is not positive.

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114 posts

#12 h Jun 20, 2021, 1:05 PM • 3 Y

Y by Mango247, Mango247, Mango247

DottedCaculator wrote:

0 is not positive.

Is it negative?

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372 posts

#13 h Jun 20, 2021, 1:08 PM • 2 Y

Y by The_Musilm, megarnie

$\mathbb R^+$ does not contain a zero. Unless you define it differently.

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114 posts

#14 h Jun 20, 2021, 1:09 PM

Y by

JuniorPerelman wrote:

DottedCaculator wrote:

0 is not positive.

Is it negative?

I see know
Excuse me french maths and English maths are different

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800 posts

#15 h Jul 16, 2021, 10:46 AM

Y by

JuniorPerelman wrote:

JuniorPerelman wrote:

DottedCaculator wrote:

0 is not positive.

Is it negative?

I see know
Excuse me french maths and English maths are different

Wdym french maths and english maths?

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2507 posts

#16 h Dec 25, 2021, 5:30 PM • 1 Y

Y by rama1728

rama1728 wrote:

Wdym french maths and english maths?

I think JuniorPerelman means French math terms are differently defined than English math terms. Though the overall essence is the same.

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#17 h May 27, 2022, 5:27 AM • 2 Y

Y by rama1728, Mango247

JuniorPerelman wrote:

DottedCaculator wrote:

0 is not positive.

Is it negative?

It's non-negative and non-positive.

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1526 posts

#18 h Jun 4, 2022, 11:12 PM

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Let $P(x,y)$ the assertion of the given F.E.
Claim 1: $f$ is an involution.
Proof: By this lemma:

Lemma wrote:

If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
$f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0$ then $\frac{g}{h}$ is a constant function.

We have that $f(f(x))=cx$ so re-write the F.E. as
$f(cxf(x)+y)=xf(x)+f(y)$ Now take $f$ in both sides of $P(f(x),y)$ and then use the result of $P(x,f(y))$ to get that
$c^3xf(x)+cy=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+cy \implies c=1$ Hence $f(f(x))=x$ so $f$ is involutive.
Claim 2: $f(y)+xf(x) \ge y$ for all $x,y$
Proof: Assume that $y>f(y)+xf(x)$ for some $x,y$ then we have $\frac{y}{xf(x)}>\frac{f(y)}{xf(x)}+1$ now since for any positive real $r$ there exists a positive integer $n$ such that $r+1>n \ge r$ we have that there exists $n$ positive integer such that $\frac{y}{xf(x)}>\frac{f(y)}{xf(x)}+1>n \ge \frac{f(y)}{xf(x)}$ which means that $y>nxf(x) \ge f(y)$ . Now by easy induction we get that
$f(nxf(x)+y)=nxf(x)+f(y)$ And on this equation do $y=y-nxf(x)$ to get that
$f(y-nxf(x))=f(y)-nxf(x) \le 0 \; \text{contradiction!!}$ Hence our claim is true.
Finishing: Call the assertion of Claim 2 $Q(x,y)$ so now by $Q(x,f(y))$ we get that $y+xf(x) \ge f(y)$ and multpliying by $y$ in both sides $y^2+yxf(x) \ge yf(y)$ and by letting $y \to 0$ we have that $yf(y)$ is as smaller as we want so on $Q(x,y)$ set $xf(x)$ as smaller as we want so we get $f(y) \ge y$ but by setting $y=f(y)$ we get $y \ge f(y)$ hence $f(y)=y$ .
Hence $\boxed{f(x)=x \; \forall x \in \mathbb R^+}$ is a solution thus we are done

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ZETA_in_olympiad

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ZETA_in_olympiad

#19 h Jun 5, 2022, 9:06 AM

Y by

MathLuis wrote:

Lemma wrote:

If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
$f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0$ then $\frac{g}{h}$ is a constant function.

You have to prove the lemma (as it's not obvious nor trivial) if you can't then you should show where it's proved i.e \ cite where it can be found.

This post has been edited 2 times. Last edited by ZETA_in_olympiad, Jun 5, 2022, 9:32 AM
Reason: i.e

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5608 posts

#20 h Jun 5, 2022, 12:05 PM • 1 Y

Y by ZETA_in_olympiad

ZETA_in_olympiad wrote:

MathLuis wrote:

Lemma wrote:

If $f,g,h:\mathbb{R}^+ \to\mathbb{R}^+$ be functions such that
$f(g(x)+y)=h(x)+f(y) \quad \forall x,y>0$ then $\frac{g}{h}$ is a constant function.

You have to prove the lemma (as it's not obvious nor trivial) if you can't then you should show where it's proved i.e \ cite where it can be found.

The lemma can be found https://artofproblemsolving.com/community/c6h2807267p24753821

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480 posts

#21 h Dec 2, 2023, 4:27 PM

Y by

Solution with the Malaysian team:

The main idea (almost ubiquitous with R+ FEs) is to estimate $f(small)$ as negligible, and this is often done by induction on $\mathbb{Z}$ , or by introducing a new variable - similar to IMOSL 2007 A4.

$\textbf{Part 1.}$ Observe that we have for all $m, n\in \mathbb{Z}$ , $f(mf(x)f(f(x))+nf(y)f(f(y))+z)=mxf(x)+nyf(y)+f(z)$ Let $t=mf(x)f(f(x))+nf(y)f(f(y))+z$ and vary $z>0$ (intuitively, this $z$ is small), this implies that $t>mf(x)f(f(x))+nf(y)f(f(y)) \Rightarrow f(t)>mxf(x)+nyf(y)$ For simplicity let $a=f(x)f(f(x)), b=f(y)f(f(y)), c=xf(x), d=yf(y)$ , so that we get $t>ma+nb \Rightarrow f(t)>mc+nd$ If $\frac{a}{b}<\frac{c}{d}$ then let $n=-\left\lfloor\frac{ma}{b}\right\rfloor$ we get $b+1>ma-\left\lfloor\frac{ma}{b}\right\rfloor b \Rightarrow f(b+1)>mc-\left\lfloor\frac{ma}{b}\right\rfloor d>m\left(c-\frac{a}{b} d\right) - d$ which is a contradiction by taking $m$ arbitrarily huge.

So $\frac{a}{b}\ge\frac{c}{d}$ . By swapping $x$ and $y$ , we get that $\frac{a}{b}=\frac{c}{d}$ must hold, that is $\frac{f(f(x))}{x}=\frac{f(f(y))}{y}$ for all $x$ and $y$ . Hence $f(f(x))=cx$ for some $c>0$ .

$\textbf{Part 2.}$ The original equation becomes $f(cxf(x)+y)=xf(x)+f(y)$ Consider $P(x,f(y))$ and $P(f(x),y)$ , then $f(cxf(x)+f(y))=xf(x)+cy$ $f(c^2xf(x)+y)=cxf(x)+f(y)$ So this gives $c^3xf(x)+cy=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+cy$ which gives $c=1$ . So $f(f(x))=x$ , and we get $f(xf(x)+y)=xf(x)+f(y)$ .

$\textbf{Part 3.}$ With the same idea as above (to invoke inequalities), we see that $f(nxf(x)+y)=nxf(x)+f(y)$ for all $n\in \mathbb{Z}$ , so $t>nxf(x) \Rightarrow f(t)>nxf(x)$ Replace $t$ by $f(t)$ , then $f(t)>nxf(x) \Rightarrow t>nxf(x)$ This immediately implies $nxf(x)<t\le (n+1)f(x) \iff nxf(x)<f(t)\le (n+1)xf(x)$ $\Rightarrow |f(t)-t|\le xf(x)$ for all $t$ and $x$ . It suffices to prove that $xf(x)$ is arbitrarily small. Suppose for some $C>0$ , $xf(x)\ge C$ for all $x$ , then $f(x)\ge\frac{C}{x}$ . Then take $t=\frac{C}{n}, x=1$ for large integers $n>C$ , so that $f(t)\ge n>1>t$ , so $n-\frac{C}{n}=\frac{C}{t}-t\le f(t)-t\le f(1)$ but take $n\rightarrow +\infty$ gives a contradiction.

So $xf(x)$ is arbitrarily small, implying $f(t)=t$ for all $t>0$ .

QED

This post has been edited 8 times. Last edited by navi_09220114, Sep 9, 2024, 5:14 PM

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#22 h Apr 21, 2025, 10:27 AM

Y by

By Lemma :
$\frac{f(f(x))}{x}$ is constant. So for some $c \in \mathbb{R^+} , f(f(x))=cx$ $\forall x \in \mathbb{R^+}$
$P(f(x),y): f(c^2xf(x)+y) = cxf(x)+f(y).$
so $c^3xf(x)+y=f(f(c^2xf(x)+y))=f(cxf(x)+f(y))=xf(x)+xy$ .
implies that $c^3=1 \rightarrow c=1 \rightarrow f(f(x))=x$
$P(x,y) : f(xf(x)+y)=xf(x)+f(y) , P(x,f(y)) : f(xf(x)+f(y)) = xf(x)+y$

Claim $f(y)+xf(x) \geq y$
Proof : if $y>f(y)+xf(x)$ ,by $f(f(y))=y \rightarrow f(y) > y+xf(x)$
so $y-xf(x) > f(y) > y+xf(x)$ contradiction.

so $f(y)+xf(x) \geq y , y+xf(x) \geq f(y)$
implies that $y+xf(x) \geq f(y) \geq y-xf(x)$

Claim there is no $C \in \mathbb{R^+}$ such that $C \leq xf(x)$ $\forall x \in \mathbb{R^+}$
Proof : if there exist $C \leq xf(x)$ $\forall x \in \mathbb{R^+}$
from $y+xf(x) \geq f(y)$ implies that $y^2+yxf(x) \geq yf(y) \geq C$ but $y \rightarrow 0$ lead to a contradiction.
So $xf(x)$ is arbitrary small , thus $f(y) = y$ $\forall y \in \mathbb{R^+}$ . done $\square$

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11334 posts

#23 h Apr 28, 2025, 4:35 PM

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Functional_equation wrote:

Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that $f(f(x)f(f(x))+y)=xf(x)+f(y)$ for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$

Claim 1: $f(f(x))=cx$ for some constant $c\in\mathbb R^+$
True by the $fgh$ lemma (link).

Claim 2: $c=1$
Let $P(x,y)$ be the assertion $f(cxf(x)+y)=xf(x)+f(y)$ . Note that $f(f(x))=cx$ implies:
$f(cx)=f(f(f(x)))=cf(x)$ for all $x\in\mathbb R^+$ . Then, suppose $c\ne1$ , we have:
$P(x,cyf(y))\implies f(cxf(x)+cyf(y))=xf(x)+f(cyf(y))=xf(x)+cf(yf(y))$ .
Swapping $x,y$ and comparing, we get that:
$cf(xf(x))=xf(x)+d$ for some constant $d\in\mathbb R$ . Then:
$c^2d=c^3f(xf(x))-c^2xf(x)=cf(cxf(cx))-cxf(cx)=d,$ and so $d=0$ (as $c\ne1$ ).
The equation above turns into $cf(xf(x))=xf(x)$ now, plugging in $x=c$ we have $cf(cf(c))=cf(c)$ , and by cancellation and injectivity $f(c)=1$ .
Now spamming $f(f(x))=cx$ will give us our result: we have (applying $f$ to both sides) $f(1)=f(f(c))=c^2$ , so $f\left(c^2\right)=f(f(1))=c$ , so $f(c)=f\left(f\left(c^2\right)\right)=c^3$ , so $c^3=1$ , contradiction. Now that this claim has been proven, Claim 1 simplifies to $f(f(x))=x$ .

Let $S=\{a\in\mathbb R^+\mid f(x+a)=f(x)+a\forall x\in\mathbb R^+\}$ .
Claim 3: $\mathbb Q^+\subseteq S$
$P(1,1)\Rightarrow f(2f(1))=2f(1)$
From Claim 2 we have $xf(x)\in S$ for all $x\in\mathbb R^+$ . In particular, $\frac1nf\left(\frac1n\right)\in S$ for all $n\in\mathbb N$ and $f(1)\in S$ . Because of how $S$ is defined, if $a\in S$ then any natural multiple of $a$ must also be in $S$ (it's closed under addition), so $f\left(\frac1n\right)\in S$ and $2f(1)\in S$ . Then, we have:
$2f(1)+\frac1n=2f(1)+f\left(f\left(\frac1n\right)\right)=f\left(2f(1)+f\left(\frac1n\right)\right)=f(2f(1))+f\left(\frac1n\right)=2f(1)+f\left(\frac1n\right),$ so $f\left(\frac1n\right)=\frac1n$ . Then $\frac1nf\left(\frac1n\right)\in S$ becomes $\frac1{n^2}\in S$ , but remember all integral multiples of this figure are also in $S$ , so $\frac{mn}{n^2}=\frac mn\in S$ for any $m,n\in\mathbb N$ , hence proven.

Finish: $f(x)\ge x$
Suppose there is some $u\in\mathbb R^+$ with $f(u)<u$ . Since $\mathbb Q^+$ is dense in $\mathbb R^+$ , we can choose a rational $q$ such that $f(u)<q<u$ . Recall that $q\in S$ , so:
$f(u)=f(u-q)+q>q,$ a contradiction.
Therefore $f(x)\ge x$ for all $x\in\mathbb R^+$ . From $f(f(x))=x$ we obtain:
$x=f(f(x))\ge f(x)\ge x$ with equality everywhere, hence $\boxed{f(x)=x}$ is the only solution (we can easily check that it fits).

This post has been edited 1 time. Last edited by jasperE3, Apr 28, 2025, 4:38 PM

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