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Source: Iran 2nd round 2022 P2

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Pmshw

#1 h May 8, 2022, 3:57 PM • 6 Y

Y by megahertz13, tiendung2006, itslumi, maolus, ItsBesi, lksb

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for any real value of $x,y$ we have:
$f(xf(y)+f(x)+y)=xy+f(x)+f(y)$

This post has been edited 1 time. Last edited by Pmshw, May 8, 2022, 3:58 PM

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gghx

#2 h May 8, 2022, 4:17 PM • 4 Y

Y by megahertz13, itslumi, timgu, Infinityfun

IFEO P1 vibes?

$P(x,y): f(xf(y)+f(x)+y)=xy+f(x)+f(y)$
$P(0,-f(0)): f(-f(0))=0$

If $f(c)=0$ , then $P(c,c): c^2=0$ .
Hence, $f(0)=0$ and $f$ is injective at $0$ .
$P(x,0): f(f(x))=f(x)$

For any $y\ne 0$ , $P(-\frac{y}{f(y)},y): f(f(-\frac{y}{f(y)}))=-\frac{y^2}{f(y)}+f(-\frac{y}{f(y)})+f(y)$ .
Hence, $f(y)=\frac{y^2}{f(y)}$ or $f(y)^2=y^2$ , obviously still true when $y=0$ .

$P(-1,-1): f(-1)=-1$ .
$P(-1,y): f(y-f(y)-1)=f(y)-y-1$ .
Thus, if $f(a)=-a$ , then $f(2a-1)=-2a-1$ , so either $2a-1=2a+1$ (rej) or $2a-1=-2a-1\implies a=0$ .

Thus, $f(x)=x\forall x \in \mathbb{R}$ , which obviously works.

This post has been edited 1 time. Last edited by gghx, May 8, 2022, 4:17 PM

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Pmshw

#3 h May 8, 2022, 4:20 PM • 1 Y

Y by megahertz13

Solved by eser43
Define $P(x,y)$ as the equation below: $f(xf(y)+f(x)+y)=xy+f(x)+f(y)$
$P(0,-f(0))$ : $f(0)=f(0)+f(-f(0))$
so we have: $f(-f(0))=0$
$P(-f(0),-f(0))$ : $f(0)^2=0$
so $f(0)=0$
$P(x,0)$ : $f(x)=f(f(x)) (\star)$
Assume that there is a real $t$ such that $f(t) \neq t$

We want to prove that $f(x)-x$ is surjective.

$P(\frac{x}{t-f(t)}+1,t)$ : $f(Z)-Z=x$
when $Z$ is whats inside the $f(xf(y)+f(x)+y)$ after setting $P(\frac{x}{t-f(t)}+1,t)$ .

so $f(x)-x$ is indeed surjective.

now we have there exists $r_x$ for all $x$ such that $f(r_x)-r_x=x-f(-1)$ .
$P(r_x,-1)$ : $f(r_xf(-1)+f(r_x)-1)=f(r_x)-r_x + f(-1)=x$
this proves that $f(x)$ is surjective .
So there exists a $t'$ such that $f(t')=t$ .
In $(\star)$ we put $x\rightarrow t'$ : $f(t)=t$ Contraction.
so the assumption about the existence of $t$ is wrong.
so we have for all real $x$ : $f(x)=x$

This post has been edited 2 times. Last edited by Pmshw, May 8, 2022, 4:24 PM

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VZH

#4 h May 9, 2022, 4:00 PM • 1 Y

Y by megahertz13

My continuation from $f(0)=0:$
$P(-1,y+f(x)+xf(y))\implies f(-xy-f(y)+y+xf(y)+f(-1))=x(y-f(y))-y+f(y)+f(-1)$
Easily, $f(x)\equiv x$ is a solution.
Otherwise, there exists $y_0$ such that $f(y_0)\ne y_0$ . Then, taking $y=y_0$ and varying $x$ in the above equation, we have $f$ surjective.
$P(x,0)\implies f(f(x))=f(x)$ .
$P(f(x),f(y))\implies f(f(x)f(y)+f(x)+f(y))=f(x)f(y)+f(x)+f(y)$ . Using surjectivity, $f(xy+x+y)=xy+x+y$ , and since $xy+x+y$ can span through all real numbers $\implies f(x)\equiv x$ .(This contradicts $f(y_0)=y_0$ )
Thus the only solution is $f(x)\equiv x$ .

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#5 h May 10, 2022, 8:45 AM • 1 Y

Y by megahertz13

Define $P(x,y)$ as usual.
Note that $P(-1,-1):f(-1)=-1$ . Let $f(0)=c$ .We have:
$P(-1,0):f(-c-1)=c-1$ so $P(0,-c-1):-1=2c-1$ and $f(0)=0$ .
Now by $P(x,0)$ we have $f(f(x))=f(x)$ .On the other hand , if $P(t)=0$ , by $P(t,t)$ we have $t=0$ . Now consider a nonzero $y$ :
$P(-\frac{y}{f(y)} , y) :f(f(-\frac{y}{f(y)}))=f(y) - \frac{y^2}{f(y)} + f(-\frac{y}{f(y)})$ so multiplying by $f(y)$ , we have $f(y)^2=y^2$ . The rest is easy

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#6 h May 15, 2022, 3:23 PM • 1 Y

Y by megahertz13

Sol as everyone

.

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#7 h Aug 17, 2022, 5:56 AM • 1 Y

Y by Mango247

FALSE SOLUTION

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pco

#8 h Aug 17, 2022, 6:20 AM

Y by

strong_boy wrote:

$P(x,A) : f(f(x)+A)= -xA+f(x)$ then $f :$ injective

Uhh ?
And what if $A=0$ ? (and in fact it is)

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#9 h Oct 15, 2022, 7:42 PM • 1 Y

Y by Mango247

Let $P(x,y)$ denote the assertion $f(xf(y)+f(x)+y)=xy+f(x)+f(y)$ . By $P(0,-f(0))$ , $f(k)=0$ for some $k$ . Then $P(k,k)$ gives $k=0.$ So $f(f(x))=f(x)$ and $P(-\tfrac{x}{f(x)},x)$ implies $f(x)\equiv \pm x$ . Let $f(v)=-v$ then $P(f(x),v)$ shows $v=0.$

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#10 h Dec 25, 2022, 2:42 PM

Y by

the only answer f(x)=x

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cursed_tangent1434

658 posts

cursed_tangent1434

#11 h Jan 15, 2024, 2:13 AM

Y by

We claim that $f(x)=x$ for all $x\in \mathbb{R}$ is the only solution.

First, note that, $P(0,-f(0))$ gives
$f(-f(0))=0$ Now, we have the following key claim.
Claim : $f((x)=0$ if and only if $x=0$ .
Proof : Consider $\alpha \in \mathbb{R}$ such that $f(\alpha)=0$ . Then, $P(\alpha,\alpha)$ implies
$f(\alpha f(\alpha)+f(\alpha) + \alpha)=\alpha^2 + f(\alpha) +f(\alpha) \implies f(\alpha)=\alpha^2 \implies \alpha =0$ Further, we know that indeed exists $\alpha \in \mathbb{R}$ such that $f(\alpha)=0$ and thus, we must have that $f(0)=0$ and $f(x)=0$ if and only if $x=0$ as desired.

Now note that, $P(x,0)$ gives
$f(f(x))=f(x)$ for all $x\in \mathbb{R}$ . Now, we look at $P(\frac{-y}{f(y)},)y$ for $y \neq 0$ (note that $\frac{-y}{f(y)} \in \mathbb{R}$ since $f(y) \neq 0$ for all $y\neq 0$ by the above claim).
This yields,
$\begin{align*} f\left(\frac{-y}{f(y)}f(y)+f\left(\frac{-y}{f(y)}\right)+y\right) &= \frac{-y}{f(y)}y + f\left(\frac{-y}{f(y)}\right) + f(y)\\ f\left(f\left(\frac{-y}{f(y)}\right)\right) &= f\left(\frac{-y}{f(y)}\right)+f(y)-\frac{y^2}{f(y)}\\ f\left(\frac{-y}{f(y)}\right) &= f\left(\frac{-y}{f(y)}\right) + f(y) - \frac{y^2}{f(y)}\\ f(y) &= \frac{y^2}{f(y)}\\ f(y)^2 &= y^2 \end{align*}$ Thus, $f(x)=x$ or $f(x)=-x$ for each $x \in \mathbb{R}$ .

Now, we consider $x\neq y$ such that $f(x)=x$ and $f(y)=-y$ . Then, $P(x,y)$ yields,
$f(-xy+x+y)=xy+x-y$ Now, we know that $f(-xy+x+y)=xy-x-y$ (which implies $x=0$ ) or $f(-xy+x+y)=-xy+x+y$ (which implies $y=0$ or $x=1$ ). Thus, $f(x)=x$ for all $x$ , $f(x)=-x$ for all $x$ , or $f(x)=-x$ for all $x\neq 1$ and $f(1)=1$ . Now, we can check that $f(x)=x$ for all $x$ works and $f(x)=-x$ for all $x$ does not. For the remaining case, plugging in $P(2,2)$ yields a contradiction. Thus, the only possible solution is
$f(x) = x \text{ for all } x\in \mathbb{R}$

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#12 h Mar 24, 2024, 9:04 PM

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My solution is just a bit different than the others. It's redacted this way bc I took it from a project and was too lazy to change it jeje:

If $x = y = -1$ (so we can cancel $xf(y) + f(x)$ ):
$f(-f(-1) + f(-1) - 1) = 1 + 2f(-1) \implies f(-1) = -1.$ If $y = 0$ (so we can cancel some stuff):
$f(xf(0) + f(x)) = f(x) + f(0).$ Then, if $x = -1$ :
$f(-f(0) - 1) = f(0) - 1.$ This might look useless at first, but it was very natural to arrive at it, so you might have it on a small corner in your paper.

Now, the right hand side of the equation is symmetric, and so
$f(xf(y) + f(x) + y) = f(yf(x) + f(y) + x).$ If $x=0$ in this equation:
$f(y + f(0)) = f(yf(0) + f(y)).$ If $y = -1-f(0)$ (so we can cancel $y + f(0)$ ):
$f(-1) = f(-f(0) - f(0)^2 + f(-1-f(0))),$ but $f(-1-f(0)) = -1 + f(0)$ (surprise surprise! It wasn't that useless), so $-1 = f(-f(0)^2 - 1)$ .

We now have another value $t$ such that $f(t) = -1$ . If $t \neq -1$ , letting $x = -1$ and $y = t$ :
$f(-f(t) + f(-1) + t) = -t + f(-1) + f(t) \implies f(t) = -t-2 \implies t = -1,$ but this is absurd. Therefore $t = -1$ and so $f(0)^2 - 1 = -1$ , or $f(0) = 0$ . If $y = 0$ , $f(f(x)) = f(x)$ .

Now, how can I cancell $xf(y) + y$ to get only $f(f(x))$ ? We need $x = \dfrac{-y}{f(y)}$ , but there's an issue if $f(y) = 0$ . Suppose there exists a number $t \neq 0$ with $f(t) = 0$ . If $x = y = t$ :
$f(tf(t) + f(t) + t) = 0 = t^2 + f(t) + f(t) = t^2 \implies t = 0,$ but this is absurd. So we can take $y \neq 0$ and $x = \dfrac{-y}{f(y)}$ , from where:
$f\left(f\left( \dfrac{-y}{f(y)}\right)\right) = \dfrac{-y^2}{f(y)} + f\left(\dfrac{-y}{f(y)}\right) + f(y),$ And as $f(f(x)) = f(x)$ ,
$\dfrac{y^2}{f(y)} = f(y) \implies f(y)^2 = y^2 \implies f(y) = \pm y.$ It's easy to check $f(x) = x$ is a solution, but $f(x) = -x$ is not. Can they both happen? If there exists some number $b$ such that $f(b) = -b$ , letting $x = y = b$ :
$f(-b^2) = b^2 - 2b.$ Now, $f(-b^2) = -b^2 \, b^2$ . In the first case we get $b = 0 \, 1$ , and on the second case $b = 0$ . But if $b = 1$ , $f(1) = -1$ , and therefore $1 = -1$ , which is absurd.

This means $f(x) = x$ for all reals $x$ is the only solution.

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#14 h Sep 4, 2024, 3:59 PM

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$P(0,-f(0)) : f(-f(0))=0 \rightarrow \exists a$ such that $f(a)=0$
$P(a,a) : f(a)=a^2 \rightarrow a=0 \rightarrow f(0)=0$
$P(x,0): f(f(x))=f(x)$ and $P(- \frac{y}{f(y)},y) : f(f(- \frac{y}{f(y)})) = - \frac{y^2}{f(y)} + f(- \frac{y}{f(y)} ) + f(y)$
so $\frac{y^2}{f(y)} = f(y) \rightarrow f(y)^2=y^2 \rightarrow f(y)=\pm y$
$f(x)=-x$ doesn't work in $P(x,y)$ and only possible answer is :
$f(x)=x$

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Nartku

#16 h Oct 6, 2024, 6:59 AM

Y by

This solution feels weird, please tell me if something is wrong

This post has been edited 1 time. Last edited by Nartku, Oct 6, 2024, 7:00 AM

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#17 h Oct 6, 2024, 7:47 AM

Y by

Nartku wrote:

This solution feels weird, please tell me if something is wrong

That's how i solve every problem on FE anyway

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#18 h Oct 7, 2024, 6:54 AM • 1 Y

Y by RetroFuel

Solved with Infinitum, Retrofuel, Apothem, FastTortoise and others from Discord. (Main progress by me and Retrofuel)
Claim 1: $f(0) = 0$ .
Proof: $P(0, -f(0), P(-f(0), -f(0))$ .

Claim 2: $f(-1) = -1$ .
Proof: $P(-1, -1)$ .

Claim 3: $f(f(x)) = f(x)$ .
Proof: $P(x, 0)$ .

Claim 3: $f(x)^2 = x^2 \forall x \ne 0$ .
Proof: $P \left(\frac{-x}{f(x)}, x \right)$ .

Claim 4: $f(x) = x \forall x$ .
Proof: Assume otherwise, then $\exists b : f(b) = -b$ , then $P(b, b)$ works.

Since $f(x) = x$ works, we're done.

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#19 h Apr 11, 2025, 4:57 PM

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Let $P(x,y)$ be the assertion $f(xf(y)+f(x)+y)=xy+f(x)+f(y)$ . Suppose $f(k)=0$ for some $k$ .
$P(k,k)\Rightarrow k=0$
Applying this:
$P(0,-f(0))\Rightarrow f(-f(0))=0\Rightarrow f(0)=0$
$P(x,0)\Rightarrow f(f(x))=f(x)$
Let $x\ne0$ , then $f(x)\ne0$ so:
$P\left(-\frac x{f(x)},x\right)\Rightarrow f(x)^2=x^2$ so $f(x)\in\{x,-x\}$ for each $x$ whether $x\ne0$ or $x=0$ .
Suppose there is some $a\ne0$ with $f(a)=-a$ , and let $x\ne0$ .
$P(f(x),a)\Rightarrow f(-af(x)+f(x)+a)=af(x)+f(x)-a$
If $f(-af(x)+f(x)+a)=af(x)-f(x)-a$ then this simplifies to $f(x)=0$ , impossible.
If $f(-af(x)+f(x)+a)=-af(x)+f(x)+a$ then this simplifies to $f(x)=1$ for all $x\ne0$ , which doesn't satisfy say $P(1,1)$ .
So we must have $\boxed{f(x)=x}$ for all $x$ .

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#21 h May 17, 2025, 7:26 PM

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clearly f is surjective. also as you can see in above solutions $f(f(x))=f(x)$ Now let $P(f(x),f(y)) \implies f(f(x)f(y)+f(x)+f(y))=f(x)f(y)+f(x)+f(y) (*)$
claim: for every real number $z$ there exist $x,y$ such that $f(x)f(y)+f(x)+f(y)=z$ .
proof: its easily concluded from the fact that f is surjective function.
so according to $(*)$ for every real number $z$ we have $f(z)=z$
done.

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#22 h May 19, 2025, 8:20 PM

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arzhang2001 wrote:

clearly f is surjective.

why?

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