Romania TST 2021 Day 1 P4

Art of Problem Solving

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oVlad

#1 h May 15, 2021, 4:53 PM • 5 Y

Y by centslordm, tiendung2006, itslumi, Mango247, ItsBesi

Determine all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following relationship for all real numbers $x$ and $y$ $f(xf(y)-f(x))=2f(x)+xy.$

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pco

#2 h May 15, 2021, 5:13 PM • 5 Y

Y by centslordm, Mango247, Mango247, ATM_, AlexCenteno2007

oVlad wrote:

Determine all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the relationship $f(xf(y)-f(x))=2f(x)+xy,\text{ for any }x,y\in\mathbb{R}.$

Let $P(x,y)$ be the assertion $f(xf(y)-f(x))=2f(x)+xy$
Let $a=f(0)$

$P(1,x)$ $\implies$ $f(f(x)-f(1))=2f(1)+x$ and $f(x)$ is bijective
Let $u=f^{-1}(0)$

$P(u,0)$ $\implies$ $f(ua)=0=f(u)$ and so since injective, $ua=u$ And so $u=0$ or $a=1$

If $u=0$ (and so $a=0$ ) : $P(x,0)$ $\implies$ $f(-f(x))=2f(x)$ and so, since surjective, $f(x)=-2x\quad\forall x$
Which unfortunately is not a solution.

If $a=1$ : $P(u,u)$ $\implies$ $u^2=1$ and so $u=-1$ (since $u\ne a$ )
$P(x,-1)$ $\implies$ $f(-f(x))=2f(x)-x$
$P(-1,x)$ $\implies$ $f(-f(x))=-x$
And so $2f(x)-x=-x$ and $f\equiv 0$ , which is not a solution.

Hence $\boxed{\text{No such function}}$

This post has been edited 1 time. Last edited by pco, May 15, 2021, 5:13 PM
Reason: Typo

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9029 posts

#3 h May 15, 2021, 5:27 PM • 3 Y

Y by centslordm, Mango247, alexanderhamilton124

Solution

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9029 posts

#4 h May 15, 2021, 5:28 PM • 2 Y

Y by pco, centslordm

pco wrote:

...and so $u=-1$ (since $u\ne a$ )

Why $u \ne a$ ? In fact, I think that $u=a=1$ leads to the solution $f(x)=1-x$ .

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pco

#5 h May 15, 2021, 5:49 PM • 1 Y

Y by centslordm

You're right!
I decided $u\ne a$ wrongly using (badly) injectivity

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#6 h May 16, 2021, 7:40 AM • 1 Y

Y by centslordm

I guess this is the outcome when I try and do FE's past midnight, here is an unnecessarily lengthy solution which is motivated by $P(x,2)$ after some "initial work".
Solution

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#9 h May 16, 2021, 10:15 AM • 1 Y

Y by centslordm

Cute

Let $P(x,y)$ be the assertion.
Firstly, we do the following observations:

$P\left(x,-\frac{2f(x)}{x}\right)\implies f(xf\left(-\frac{2f(x)}{x}\right)-f(x))=0$ .
$f$ is injective from $P(1,a_1)$ and $P(1,a_2)$ if $f(a_1)=f(a_2)$ .

Hence, $\exists a$ so that $f(a)=0$ .
$P\left(x,-\frac{f(x)}{x}\right)\implies xf\left(-\frac{f(x)}{x}\right)-f(x)=x$ .
If $a\neq 0$ , then we get plugging $x=a$ into the equation above, we obtain $f(0)=1$ . Here from $P(1,1)$ we obtain $1=2f(1)+1\implies f(1)=0$ (and thus $a$ is actually $1$ from injectivity). Now we break our proof into two cases:
Case 1. $f(1)=0$ and $f(0)=1$ .
$P(0,y)\implies f(-1)=2$ .
$P(1,y)\implies f(f(y))=y$ and therefore $f(2)=f(f(-1))=-1$ .
$P(x,2)\implies f(-x-f(x))=2(x+f(x))$ .
On the other hand,
$P(f(x+f(x)),1)\implies f(-x-f(x))=2(x+f(x))+f(x+f(x))$ , therefore $f(x+f(x))=0=f(1)$ and from the injectivity, we get solution $f(x)=1-x\forall x\in\mathbb R$ .

Case 2. $f(0)=0$ . Here we bring out a weird contradiction argument to nuke this case.
$P(x,0)\implies f(-f(x))=2f(x)$ . Taking $x\rightarrow -f(x)$ , we have $f(-2f(x))=4f(x)$ .
From $P\left(x,-\frac{2f(x)}{x}\right)$ by injectivity, $f\left(-\frac{2f(x)}{x}\right)=\frac{f(x)}{x}$ .
Taking here $x\rightarrow -f(x)$ , we obtain $f(4)=-2$ .
Notice that from $f(-2f(x))=4f(x)$ , taking $x=4$ , we have $-2=f(4)=-8$ , contradiction, get nuked.

We conclude that our only solution is $\boxed{f(x)=1-x\forall x\in\mathbb R}$ . This obviously satisfies the given.

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#10 h May 16, 2021, 12:30 PM • 1 Y

Y by centslordm

$P(1,x)\Rightarrow f(f(x)-f(1))=2f(1)+x\Rightarrow f$ is bijective. Let $k=f^{-1}(0)$ .
$P(k,x)\Rightarrow f(kf(x))=kx$
$P(k,kf(x))\Rightarrow f(k^2x)=k^2f(x)$
Now we can say that $0\in\{f(0),f(-1),f(1)\}$ . There are many ways to prove this ( $x=1/k,0,k$ ), but we will take $x=k$ , so $f(k^3)=k^2f(k)=0=f(k)$ , hence $k\in\{0,-1,1\}$ .

$\textbf{Case 1: }f(0)=0$
$P(x,0)\Rightarrow f(-f(x))=2f(x)$ , by surjectivity now we have $f(x)=-2x$ , which doesn't work.

$\textbf{Case 2: }f(1)=0$
$P(1,1)\Rightarrow f(0)=1$
$P(0,20)\Rightarrow f(-1)=2\Rightarrow f(2)=f(f(-1))=-1$
$P(1,x)\Rightarrow f(f(x))=x$
$P(f(x),1)\Rightarrow f(-x)=2x+f(x)$
$P(x,2)\Rightarrow f(-x-f(x))=2f(x)+2x f(x+f(x))=0\Rightarrow\boxed{f(x)=1-x}$ , which works (using $P(f(x),1)$ ).

$\textbf{Case 3: }f(-1)=0$
$P(-1,x)-P(x,-1)\Rightarrow 2f(x)-x=-x\Rightarrow f(x)=0$ , which doesn't work.

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oVlad

#11 h May 16, 2021, 3:06 PM • 3 Y

Y by centslordm, Mango247, Mango247

My sol: (a bit over–complicated lol)

Observe that if $f(a)=f(b)$ then $P(1,a)$ and $P(1,b)$ directly give us $a=b$ , so $f$ is injective. Moreover, $P(1,c-2f(1))$ gives us $f(f(c-2f(1))-f(1))=c$ so $f$ is surjective too. Therefore, $f$ is bijective.

Now, let $c\in\mathbb{R}$ such that $f(c)=0$ . $P(c,c)$ gives us $f(0)=c^2$ and, using this, $P(c,0)$ gives us $f(c^3)=0$ . Because $f$ is bijective, we can conclude that $c^3=c$ so $c\in\{0,1,-1\}$ .

Assume $c=0$ . Using the fact that $f$ is bijective, $P(f^{-1}(-x),0)$ gives us $f(x)=-2x$ . Therefore, for any $x$ we have $f(x)=-2x$ , but a simple check will reveal the fact that this does not work.

Assume $c=-1$ . Then $P(-1,x)$ gives us $f(-f(x))=-x$ and $P(x,-1)$ gives us $f(-f(x))=2f(x)-x$ and combining these two equations, $f(x)=0$ for any $x$ . However, this is also a contradiction.

Therefore, $c=1$ . So $f(1)=0$ and using the fact that $f(0)=c^2$ we get $f(0)=1$ . Moreover, $P(0,0)$ gives us $f(-1)=2$ and $P(1,x)$ results in $f(f(x))=x$ . Therefore, $-1=f(f(-1))=f(2)$ .

Therefore, $P(x,2)$ gives us $f(-(f(x)+x))=2(f(x)+x)$ . Let $f(x)+x=k$ , so for this $k, \ f(-k)=2k$ .

Now, observe that $P(x,0)$ gives us $f(x-f(x))=2f(x)$ . Combining $f(x-f(x))=2f(x)$ with the fact that $f$ is bijective, we get that $g:\mathbb{R}\to\mathbb{R}, \ g(x)=x-f(x)$ is surjective. Now, let $x=\alpha$ such that $g(\alpha)=-k$ . Therefore, $f(\alpha-f(\alpha))=2f(\alpha)\iff f(-k)=2f(\alpha)\iff 2k=2f(\alpha)$ . So $f(\alpha)=k$ but we chose $\alpha$ such that $g(\alpha)=-k\iff \alpha-f(\alpha)=-k\iff\alpha-k=-k$ so $\alpha=0$ . Therefore, $0-f(0)=-k$ so $k=1$ !

But remember that we chose $x+f(x)$ to be $k$ . Therefore, for any $x$ we have $x+f(x)=1$ which gives us the desired function, $f(x)=1-x$ .

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#12 h May 16, 2021, 5:27 PM • 1 Y

Y by centslordm

For instance $f(x)=x$ is bijective but $x-f(x)$ is not.

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oVlad

#13 h May 16, 2021, 6:04 PM • 1 Y

Y by centslordm

I know. Let me prove the assertion for you. Assume $g(x)$ is not surjective. Therefore, there exists some $k$ such that $g(x)\neq k$ for any $x$ . Therefore, because of the fact that $f$ is bijective, $f(x-f(x))\neq f(k)$ for any $x$ . However, just take $x=f^{-1}\bigg(\frac{f(k)}{2}\bigg)$ and observe that plugging this in $f(x-f(x))=2f(x)$ gives us the fact that, indeed, for some $x$ we have $f(x-f(x))=f(k)$ , contradiction. Therefore, $g(x)$ is surjective.

Indeed, if $f(x)$ is bijective it does not solely imply $g(x)$ is surjective, but we also have the condition that $f(x-f(x))=2f(x)\iff f(g(x))=2f(x)$ .

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#14 h May 17, 2021, 8:36 AM • 1 Y

Y by centslordm

$f$ is bijective. Let $f(a)=0$ and $f(b)=1$ . $P(a,b)$ gives $ab=0$ .

1) $a=0$ , $P(x,0) \implies f(x)= -2x$ , not a solution.

2) $b=0$ , $P(a,a) \implies a=1$ or $a=-1$ .

$i)$ $a=-1$ , $P(-1,x)$ and $P(x,-1) \implies f(x)=0$ , not a solution.

$ii)$ $a=1$ , $P(1,x) \implies f(f(x))=x$ ,

$+P(x,1) \implies f(-f(x))=2f(x)+x$ ,
$+P(f(x),1) \implies *f(-x)=f(x)+2x \implies f(-1)=2$ ,
$+P(-1,x) \implies f(-f(x)-2)=4-x$ .

$+$ Using * with the last equation gives $f(f(x)+2)=-x-2f(x)=-f(-f(x))$ . From surjectivity we have $f(-x)+f(x+2) \implies f(x)+f(2-x)=0$ .

$+P(x,2-y)$ and using * gives $f(xf(y)+f(x))+2xf(y)+xy=2x$ . Plugging $y=0$ in the last equation we have $f(x+f(x))=0 \implies f(x)=1-x$ which is a solution.

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#15 h May 19, 2021, 4:29 PM • 1 Y

Y by centslordm

Problem Statement
Answer
Let the assertion to the given Functional Equation be $P(x,y)$ .
Claim 1: $f$ is surjective.
$\text{Proof:}$ $P(1,y)$ gives $f(f(y)-f(1))=2f(1)+y$ and as $y$ is surjective in $\mathbb{R} \to \mathbb{R}$ , so $f$ must be surjective. $\square$

Now, as $f$ is surjective, so $\exists \,\, \alpha : f(\alpha)=0$

Claim 2: $f(0)=\alpha^2$ .
$\text{Proof:}$ $P(\alpha , \alpha)$ gives $f(0)=\alpha ^2$ This proves our claim. $\square$

Claim 3: $f$ is injective.
$\text{Proof:}$ $P(\alpha,y)$ gives $f(\alpha f(y))=\alpha y$ This is immediately shows injectivity. $\square$
Remark

Claim 4: Only possible values of $\alpha$ are $1,0,-1$ .
$\text{Proof:}$ $P(\alpha,0)$ gives $f(\alpha f(0))=0$ By Claim 2 and Claim 3, we have: $f(\alpha^3)=f(\alpha)=0$ By bijectivity, we get $\alpha^3=\alpha \implies \alpha \in \{1,-1,0\}$ . $\square$

Claim 5: $\alpha$ can neither be $-1$ nor $0$ .
$\text{Proof:}$ We will split it into two cases:

Case 1: $f(-1)=0$ is impossible.
$\text{Proof:}$ Trying $P(0,y)$ gives $f(-f(0))=2f(0)$ but by Claim 2 putting $f(0)=\alpha^2=1$ we get:
$0=f(-1)=2 \times 1=2$ which is absurd. $\square$

Case 2: $f(0)=0$ is impossible.
$\text{Proof:}$ Trying $P(x,0)$ we get:
$f(-f(x))=2f(x)$ Using surjectivity, set $f(x)=z$ for all real $z$ . We get $f(z)=-2z$ which does satisfy the original equation. $\square$

$\textcolor{red}{\textbf{\textsf{Claim 6:}}}$ If $f(1)=0$ then, $f(x)=1-x$ is the only solution.
Credits We will firstly show some lemmas:

Lemma 1: $f(f(x))=x$ if $f(1)=0$
$\text{Proof:}$ $P(x,1)$ gives it immediately with $f(1)=0$ or put $\alpha=1$ in Claim 3. $\square$

Lemma 2: $f(-x-f(x))=2(x+f(x))$ if $f(1)=0$ .
$\text{Proof:}$ $P(x,2)$ gives the desired as $f(2)=-1$ . $f(2)=-1$ can be obtained by $P(0,y)$ and by Lemma 1.

Lemma 3: (Killer) $f(-x-f(x))=2(f(x)+x)+f(f(x)+x)$ if $f(1)=0$ .
$\text{Proof:}$ $P(f(f(x)+x),1)$ along with Lemma 1 gives the desired. $\square$ Spider Sense

$\text{Proof of\,\,}\textcolor{red}{\textbf{\textsf{Claim 6}}}\text{:}$ By combining Lemma 2 and Lemma 3 we get that $f(f(x)+x)=0=f(1)$ and by bijectivity of $f$ , we finally conclude that $\boxed{f(x)=1-x}$ is the only solution to the Functional Equation. $\blacksquare$

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DakuMangalSingh

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DakuMangalSingh

#16 h Jul 4, 2021, 6:04 PM • 4 Y

Y by itslumi, Mango247, Mango247, Mango247

Answer

Solution

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#18 h Apr 1, 2022, 1:58 AM

Y by

Neat F.E.
Let $P(x,y)$ the assertion of the given F.E.. We claim that $f(x)=1-x$ is the only solution to this F.E.
Claim 1: $f$ is bijective
Proof: Set $x \ne 0$ and then by $P \left(x,\frac{y-2f(x)}{x} \right)$
$f \left(xf \left(\frac{y-2f(x)}{x} \right)-f(x) \right)=y \implies f \; \text{surjective}$ Now assume that $f(a)=f(b)$ then by comparing $P(x,a)$ and $P(x,b)$ for $x \ne 0$
$ax=bx \implies a=b \implies f \; \text{injective}$ Hence $f$ is bijective aa desired.
Claim 2: $f(1)=0$ and $f(0)=1$
Proof: Using Claim 1 we can let $c,d$ such that $f(c)=0$ and $f(d)=1$ , then by $P(c,d)$
$cd=0 \implies c \; \text{or} \; d =0$ Case 1: $c=0$
Then by $P(x,0)$ and surjectivity
$f(-f(x))=2f(x) \implies f(t)=-2t \; \forall t \in \mathbb R$ Which doesnt work hence we got a contradiction.
Case 2: $d=0$
Then by $P \left(c,\frac{1}{c} \right)$ and Claim 1
$cf \left(\frac{1}{c} \right)=0 \implies c^2=1 \implies c= \pm 1$ Case 2.1: $c=-1$
Now by $P(0,y)$
$0=f(-1)=2 \; \text{contradiction!!}$ Case 2.2: $c=1$
So we got $f(1)=0$ and $f(0)=1$ so our claim is complete.
Finishing: First we will get $f$ involutive by $P(1,x)$
$f(f(x))=x \implies f \; \text{involution}$ Now the brute force part begins, as at this moment i have no idea what to do i will spam until i get something that looks promicing, which is afterall probably the main idea lol.
$P(x,0)$
$f(x-f(x))=2f(x)$ $P(x,2)$ (becuase $f(2)=-1$ as $f$ is involutive)
$f(-x-f(x))=2x+2f(x)$ $P(f(x),0)$
$f(f(x)-x)=2x$ $P(f(x),1)$
$f(-x)=2x+f(x)$ Now this along with a previous equation and the injectivity yeld
$f(x+f(x))=-2x-2f(x)+2x+2f(x)=0 \implies x+f(x)=1 \implies f(x)=1-x$ Hence $\boxed{f(x)=1-x \; \forall x \in \mathbb R}$ is the only function that works, so we are done

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#19 h Jun 5, 2022, 6:29 PM

Y by

Clearly $f$ is bijective. Let $f(n)=0$ then $P(n,0)$ and $P(n,n)$ gives $f(-1)=0$ or $f(0)=0$ or $f(1)=0.$

A. $f(-1)=0:$ Then compare $P(x,-1)$ with $P(-1,x)$ to get $f(x)=0,$ which fails.
B. $f(0)=0:$ Then $P(x,0)$ gives $f(x)=-2x,$ also fails.
C. $f(1)=0:$ Then $f(0)=1.$ And $P(1,x)$ gives $f$ is an involution. Moreover $P(0,2)$ provides $f(2)=-1.$ Now injectivity and $P(x,2)$ compared with $P(f(x),1)$ forces $f(x)=1-x,$ which infact works.

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Iora

#20 h Nov 8, 2022, 4:43 PM

Y by

Let's hope I didn't make a silly mistake again

Let $P(x,y)$ be the asssertion of the given f.e. Note that by $P(1,y)$ we have $f(f(y)-f(1))=2f(1)+y$ , since RHS varies in real numbers, $f$ is surjective, and since if $\exists a \neq b: f(a)=f(b)$ , using the latter equation we have $2f(1)+a=2f(1)+b$ , contradiction, hence $f$ is bijective.

Since function is surjective, $\exists c: f(c)=0$ . Then
$P(c,0):f(cf(0))=0$ But since $f$ is injective, we must have $cf(0)=c$ , so either $f(0)=0$ or $f(0)=1$ . If $f(0)=1$ , we have:

$P(1,1): f(0)=2f(1)+1 \Rightarrow f(1)=0 \Leftrightarrow c=1$ Then going back to $P(1,y)$ , we have $f(f(y))=y$ , which means $f(x)=f^{-1}(x)$ , which means $f$ is involution function.
Note that $P(0,y)$ gives $f(-1)=2$ , and $f(2)=-1$ by involution. Now we spam $f(x)$ as variable and hope we get something useful.
$\begin{align*} &P(f(x),1): f(-x) = 2x+ f(x) \\ &P(f(x),2):f(-f(x)-x)=2x+2f(x) \end{align*}$ Let $x+f(x)=y$ . Notice from $P(f(x),2)$ that $f(-y)=2y$
But from $P(f(x),1)$ we know that $f(-y)=2y+f(y)$ , hence $2y+f(y)=2y \leftrightarrow f(y)=0$ . Hence $y=1$ , which gives us $\boxed{f(x)=1-x}$ which is satisfying our problem.

For the case $f(0)=0$ , we have:
$\begin{align*} &P(x,0):f(-f(x))=2f(x) \Rightarrow f(y)=-2y \end{align*}$ But it does not satisfy the equation, hence contradiction.

Therefore $\boxed{f(x)=1-x}$ is the only solution.

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#21 h Feb 8, 2023, 10:39 AM

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This lemma also kills the problem
https://artofproblemsolving.com/community/c3037128h2864292_cde_lemma

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SomeonesPenguin

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SomeonesPenguin

#22 h Sep 22, 2024, 5:14 PM

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.

Let $P(x,y)$ denote the given assertion. And let $f(1)=c$ . One can also quickly note that $f$ is injective.

$\bullet$ $P(1,y)\Rightarrow f(f(y)-c)=2c+y$ $(1)$

$\bullet$ $P(x,f(y)-c)\Rightarrow f(xy+2cx-f(x))=2f(x)+xf(y)-cx$ $(2)$

Putting $x=1$ in this equation yields $f(y+c)=f(y)+c$ . Suppose that $c\neq 0$ , plugging $y\mapsto y+c$ in $(2)$ gives $f(xy+3cx-f(x))=2f(x)+xf(y)=f(xf(f(y))-f(x))$
Since $f$ is injective, this gives $y+3c=f(f(y))$ , in particular $f$ is surjective. Plugging $y\mapsto f(y)$ in $(2)$ gives $f(xf(y)+2cx-f(x))=2f(x)+xy+2cx=f(xf(y)-f(x))+2cx$
Since $f$ is surjective, this gives $f(xy+2cx-f(x))=f(xy-f(x))+2cx$ . Plugging $x=1$ into the last equation gives $f(y+c)=f(y-c)+2c=y+4c$
Hence $f$ is linear. This gives the solution $f(x)=1-x$ .

Otherwise, if $c=0$ , $(2)$ becomes $f(xy-f(x))=2f(x)+xf(y)$ so plugging $y=-1$ in here gives $f(-x-f(x))=2f(x)+xf(-1)$ . Now $x\mapsto f(x)$ yields $2f(x)+xf(-1)=2x+f(x)f(-1)$ so $f(-1)=2$ or $f$ is linear. Since we have dealt with the latter, suppose that $f(-1)=2$ . Also $P(1,y)$ gives $f(f(y))=y$ .

$\bullet$ $P(f(x),f(1))\Rightarrow f(f(x)-x)=2x=f(f(2x))$ so $f(2x)=f(x)-x$

$\bullet$ $P(2,f(x/2))\Rightarrow f(x+1)=2f(x/2)-2=2f(x)+x-2$

Now let $g:\mathbb R\to \mathbb R$ , $g(x)=f(x)+x$ . From the above, we get $g(2x)=g(x)$ and $g(x+1)+1=2g(x)\iff g(x+1)-1=2(g(x)-1)$ so inducting gives $g(x+n)=2^ng(x)-2^n+1$ . Notice that $(2)$ turns into $g(xy-g(x)+x)=g(x)+xg(y)-x$ so plugging $y\mapsto 2y$ gives $f(2xy-g(x)+x)=g(xy-g(x)+x)$ . Now letting $x=2$ and $y\mapsto \frac{y}{2}$ gives $g(2y+3)=g(y+3)=g(2y+6)=8g(2y+3)-7$ so $g(x)=1$ for all $x$ . Therefore, $f(x)=x-1$ $\forall x\in\mathbb R$ which does satisfy the given FE.

This post has been edited 2 times. Last edited by SomeonesPenguin, Sep 22, 2024, 7:00 PM

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#23 h Sep 22, 2024, 8:23 PM

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Denote with $P(x,y)$ the assertation into $f(xf(y)-f(x))=2f(x)+xy$
$P(1,y)\Rightarrow$ $f$ is surjective
If $f(a)=f(b)$ , then comparing $P(1,a), P(1,b)$ gives us $a=b \Rightarrow$ $f$ is bijective.
Let $t,r\in\mathbb{R}$ be s.t. $f(t)=0,f(r)=1$ .
$P(t,r)\Rightarrow 0=rt$
If $f(0)=0$ , then $P(x,0)$ gives us $f(-f(x))=2f(x)$ .
$P(-f(x),3)\Rightarrow f(-f(x)f(3)-2f(x))=f(x)\Rightarrow (-f(3)-2)f(x)=x \Rightarrow f(x)=cx$
Checking the last we see that it's impossible, hence $f(0)=1$ .
$P(0,0)\Rightarrow f(-1)=2$
$P(t,t)\Rightarrow 1=f(0)=t^2 \Rightarrow f(1)=0$
$P(1,-1)\Rightarrow f(2)=-1$
$P(1,y)\Rightarrow f(f(y))=y$
Let $A=\{x \mid f(x)=-2x\}$ . Now let $a\in A$ :
$P(f(a),1)\Rightarrow f(-a)=2a+f(a)=0 \Rightarrow a=-1$
$P(x,2)\Rightarrow f(-x-f(x))=2f(x)+2x \Rightarrow -x-f(x)\in A \Rightarrow -x-f(x)=-1 \Rightarrow \boxed{f(x)=1-x}$

This post has been edited 2 times. Last edited by GeorgeRP, Sep 22, 2024, 8:27 PM

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#24 h Nov 8, 2024, 7:44 AM

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$f(xf(y)-f(x))=2f(x)+xy$ Only function is $f(x)=1-x$ . Let $P(x,y)$ be the assertion. By changing $y$ , we see that $f$ is surjective. Also $f(a)=f(b)$ would imply $ax=bx$ or $a=b$ hence $f$ is also injective.
Claim: $f(1)=0$ .
Proof: Let $f(1)=c$ and $f(a)=0$ . $P(a,y)$ yields $f(af(y))=ay$ . Thus, $f(0)=f(af(a))=a^2$ and $f(a^3)=f(af(0))=0=f(a)$ hence $a\in \{-1,0,1\}$ .
$P(1,-c): \ \ f(f(-c)-c)=2c-c=c=f(1)\implies f(-c)=c+1$ $P(x,\frac{y-2f(x)}{x}): \ \ f(xf(\frac{y-2f(x)}{x})-f(x))=y$ So left hand side does not depend on $x$ . Hence
$f(xf(\frac{y-2f(x)}{x})-f(x))=f(f(y-2c)-c)\implies xf(\frac{y-2f(x)}{x})-f(x)=f(y-2c)-c$ Subsituting $x,y+2c$ yields $xf(\frac{y+2c-2f(x)}{x})=f(x)+f(y)-c$ . If we choose $f(y)=c-f(x)$ , then $y+2c-2f(x)=ax$ thus, $f(ax+2f(x)-2c)+f(x)=c$ for $x\neq 0$ . If $a=0$ , then $f(2f(x)-2c)+f(x)=c$ . Since $f$ is surjective, we can replace $x$ with $f(x)$ which implies $f(2x-2c)+x=c$ or $f(x)=c-\frac{x+c}{2}=\frac{c-x}{2}$ which is not a solution. Now suppose that $f(-1)=0$ . Choose $x=a$ to conclude that $f(a^2-2c)=c$ hence $a^2=2c+1$ . But $a=-1$ yields $c=0$ which contradicts with the injectivity. Thus, $f(1)=0$ . $\square$
Plugging $x=1$ gives $f(f(y))=y$ .
$f(xy-f(x))=2f(x)+xf(y)=f(f(2f(x)+xf(y)))\implies f(2f(x)+xf(y))=xy-f(x)$ $f(2f(x)+xy)=xf(y)-f(x)=f(xy-f(x))-3f(x)$ Replace $y$ with $yx$ to see that $f(2f(x)+y)=f(y-f(x))-3f(x)$ . Choosing $f(x),y$ yields
$f(2x+y)=f(y-x)-3x\iff f(y+3x)=f(y)-3x\iff f(x+y)+x=f(y)$ Picking $y=0$ implies $f(x)+x=1\iff f(x)=1-x$ as desired. $\blacksquare$

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#25 h Mar 31, 2025, 3:35 PM • 1 Y

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Denote by $P(x,y)$ the given assertion.
$P(1,x): f(f(x)-f(1))=2f(1)+x$
Easily, the function $f$ is surjective(in the above relation, take $x\rightarrow x-2f(1)$ ). Moreover, if there exist two reals $a<b$ for which $f(a)=f(b)$ then $P(1,a)$ and $P(1,b)$ imply $a=b$ , a contradiction, so the function $f$ is injective too.
Since the function $f$ is surjective there exist $k, l \in \mathbb{R}$ such that $f(k)=1$ and $f(l)=0$ .
$P(l,k): f(lf(k)-f(l))=2f(l)+kl \Rightarrow kl=f(lf(k))=f(l)=0$ .

Suppose we had: $l=0 \iff f(0)=0$ . Then:
$P(1,1): f(f(1)-f(1))=f(0)=0=2f(1)+1 \Rightarrow f(1)=-\frac{1}{2}$ .
$P(1,0):f(-f(1))=2f(1) \Rightarrow f(\frac{1}{2})=-1$ .
$P(\frac{1}{2}, \frac{1}{2}): f(\frac{f(\frac{1}{2})}{2}-f(\frac{1}{2}))=2f(\frac{1}{2})+\frac{1}{4}\Rightarrow 1=\frac{1}{4}$ , contradiction.
Therefore $l\neq 0$ so $k=0 \iff f(0)=1$ .
$P(1,1):f(0)=1=2f(1)+1 \Rightarrow f(1)=0$
$P(1,x): f(f(x))=x$
$P(0,0):f(-f(0))=2f(0)\Rightarrow f(-1)=2 \Rightarrow f(2)=-1$

$P(f(x),y): f(f(x)f(y)-x)=2x+f(x)y$
Plugging $y\rightarrow 1$ in the above, we get $f(-x)=2x+f(x), \forall x \in \mathbb{R}$ . Plugging in $x\rightarrow f(x)$ yields: $f(-f(x))=2f(x)+x$ .
Now, $P(2, x): f(2f(x)+1)=-2+2x=2(x-1)$ .
However $P(f(x), 0): f(f(x)-x)=2x, \forall x\in \mathbb{R}$ so by setting $x \rightarrow x-1$ we get $f(2f(x)+1)=f(f(x-1)-x+1)$ so by injectivity: $2f(x)=f(x-1)-x \Rightarrow 2f(x)+x=f(x-1)$
But $2f(x)+x=f(-f(x))$ therefore $f(x-1)=f(-f(x))$ and by injectivity, $f(x)=1-x, \forall x\in \mathbb{R}$ , which trivially works, so we are done $\blacksquare$

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