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Source: CJMO 2022/5 (https://aops.com/community/c594864h2791269p24548889)

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#1 h Mar 20, 2022, 7:12 AM • 1 Y

Y by kimyager

Find all functions $f:\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$ , $f(f(xy)+y)=(x+1)f(y).$
Proposed by novus677

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#2 h Mar 20, 2022, 7:12 AM • 1 Y

Y by StarLex1

Note that the zero and the identity functions both work. Let $f$ be a solution different from these two.

Taking $y=0$ in the given implies $f(f(0))=(x+1)f(0)$ , so $f(0)=0$ . Now, we present the following Claim:

Claim: $f(x)=0$ only when $x=0$ .
Proof: Assume otherwise, and let $f(k)=0$ for some $k \neq 0$ . Then $x=\dfrac{k}{y}$ in the given implies $f(y)=0$ for all non-zero $y$ , so $f$ is identically zero, a contradiction $\blacksquare$

Now, take $x \rightarrow x/y$ in the given, so $f(f(x)+y)=(\frac{x}{y}+1)f(y)$ for all $x$ and for all nonzero $y$ .

Then, $y=-f(x)$ in the above relation implies $f(-f(x))(x-f(x))=0,$ for all $x \neq 0$ .

However, if we pick a $x$ such that $f(x) \neq 0, f(x) \neq x$ (it must exist since $f$ is neither the zero nor the identity function), then using the Claim we have

$f(-f(x)) \neq 0$ and $x-f(x) \neq 0,$

a contradiction.

To sum up, the only solutions are $f \equiv 0$ and $f \equiv \rm id$ .

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#3 h Mar 20, 2022, 7:39 AM • 1 Y

Y by StarLex1

Let $P(x,y)$ denote the assertion.
$P(-1,y)$ gives $f(f(-y)+y)=0$ .
$P(f(-y)+y,1)$ gives
$\begin{align*} f(f(f(-y)+y)+1)&=(f(-y)+y+1)f(1) \\ f(1)&=(f(-y)+y+1)f(1) \\ f(1)(f(-y)+y)&=0. \end{align*}$ Case 1: $f(1)\neq 0$ .
Then, $f(-x)+x=0$ for all $x$ , so $f(x)=x$ for all $x$ , which is indeed a solution.
Case 2: $f(1)=0$ .
$P\left(\tfrac{1}{x},x\right)$ for $x\neq 0$ gives
$\begin{align*} f(f(1)+x)&=\left(\frac{1}{x}+1\right)f(x) \\ f(x)&=\left(\frac{1}{x}+1\right)f(x) \\ \frac{f(x)}{x}&=0, \end{align*}$ so $f$ vanishes on all nonzero reals. Now, suppose $f(0)=c\neq 0$ . Then, $P(0,0)$ gives
$\begin{align*} f(f(0)+0)&=f(0) \\ f(c)&=f(0) \\ 0&=f(0), \end{align*}$ contradiction. So $f(0)=0$ as well, which means in this case, $f$ is the zero function.
Therefore, the only solutions are $f(x)=0$ and $f(x)=x$ for all $x$ . $\blacksquare$

This post has been edited 1 time. Last edited by Pleaseletmewin, Mar 20, 2022, 7:39 AM

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#4 h Mar 20, 2022, 11:33 AM

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Notice that the only constant function satisfying $f$ is the zero function henceforth assume $f$ is non-constant and there exists some number such that $f(\text{that number}) \neq 0$ . Let $P(x,y)$ denote the given assertion

$P(-1,0) \implies f(f(0))=0$ .
$P(0,0) \implies f(0)=0$ .

Fixing $y$ such that $f(y) \neq 0$ gives that $f$ is surjective.
Claim: $f$ is injective at $0$ .
Proof. Assume there exists $a \neq 0$ such that $f(a)=0$ then $P(x/a , a)$ gives
$f(f(x)+a) = 0$ but since $f$ is surjective it means that $f$ is constant which violates our assumption. Now, taking $y \neq 0$ to be a free variable choose $a$ such that $f(a)=-y$ and plugging $P(a/y , y)$ gives
$0 = f(-y + y) = (a/y + 1)f(y) \implies a=-y$ so $f(-y)=-y$ for all nonzero $y \implies f(x)=x$ for all real $x$ (because $0$ is also satisfied by the relation). To conclude, the two functions satisfying the original equation are $f(x)=0$ and $f(x)=x$ .

This post has been edited 1 time. Last edited by MrOreoJuice, Mar 20, 2022, 11:34 AM

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#5 h Mar 20, 2022, 11:37 AM

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Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that for all real numbers $x$ and $y$ , $f(f(xy)+y)=(x+1)f(y)$ The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x)=x}$ .

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

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

$P(-1,0): f(f(0))=0$ .

So $f(0)=0$ .

Now we take cases based on whether there exists a $k\ne 0$ with $f(k)=0$ .

Case 1: There exists such a $k$ .
$P\left(\frac{k}{x},x\right): f(x)=\left(\frac{k}{x}+1\right)f(x)$ . Now $\frac{k}{x}+1\ne 1$ , so $f(x)=0\implies f\equiv 0$ is the only solution here.

Case 2: There does not exist such a $k$ .
$P(-1,-x): f(f(x)-x)=0\implies f(x)-x=0\implies f(x)=x$ .

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CT17

#6 h Mar 20, 2022, 12:08 PM

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Misplaced?

Let $P(x,y)$ denote the assertion that $f(f(xy) + y) = (x+1)f(y)$ .

First, note that $f(x)\equiv 0$ is a solution because $0 = (x+1)0$ for all $x$ and $f(x)\equiv x$ is a solution because $xy + y = (x+1)y$ for all $x$ and $y$ . Hence, we can assume from now on that there exist (not necessarily distinct) $u$ and $v$ such that $f(u)\neq 0$ and $f(v)\neq v$ . Let $c = f(v) - v$ .

$P(x,u)\implies f(f(xu) + u) = (x+1)f(u)$ . As $f(u)\neq 0$ the RHS can take any real value, so $f$ is surjective.

$P(-1,-v)\implies f(f(v) - v) = 0f(v)\implies f(c) = 0$ .

$P(x,c)\implies f(f(cx) + c) = 0$ . Since $c\neq 0$ and $f$ is surjective, $f(cx) + c$ can take any real value. However, this implies that $f(x) = 0$ for all $x$ , contradicting the fact that $f(u)\neq 0$ . Hence, the only such functions are $f(x)\equiv 0$ and $f(x)\equiv x$ .

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#7 h Mar 20, 2022, 12:34 PM

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The answer is $f \equiv 0$ and $f(x)=x$ only, which both work, so we show that no other solutions exist. Suppose $f \not \equiv 0$ , and let $P(x,y)$ denote the assertion.
Pick some $x$ with $f(m) \neq 0$ . By letting $x$ vary in $P(x,m)$ , it follows that $f$ is surjective.
Now suppose $f(n)=0$ for some $n$ . Then $P(x,n)$ gives $f(f(xn)+n)=0$ . If $n \neq 0$ , then by varying $x$ again and using surjectivity shows that $f \equiv 0$ , which is impossible because then it's not surjective. Thus $f(0)=0$ and $f$ is injective about $0$ .
To finish, from $P(-1,x)$ we have $f(f(-x)+x)=0 \implies f(-x)=-x \implies f(x)=x$ , as desired. $\blacksquare$

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#9 h Mar 20, 2022, 1:42 PM • 1 Y

Y by kimyager

meh.
We claim that the only solutions to the F.E. are $f(x)=x$ and $f(x)=0$ .
Let $P(x,y)$ the assertion of the functional Equation, clearly $f(x)=0$ works so here we will assume that there exists $c$ such that $f(c) \ne 0$
$P(x,0)$
$f(f(0))=(x+1)f(0) \implies f(0)=0$ Case 1: $f(1)=0$
We use our $c$ in a smart way, we got $c \ne 0$ so now $P \left(\frac{1}{c},c \right)$
$f(c)=\left(\frac{1}{c}+1 \right)f(c) \implies \frac{1}{c}+1=1 \implies \frac{1}{c}=0 \; \text{contradiction!!}$ Case 2: $f(1) \ne 0$
Assume that there exists a real $d \ne 0$ such that $f(d)=0$ , then by $P(d,1)$
$f(1)=(d+1)f(1) \implies d=0$ Hence $f$ is injective at $0$ and now by $P(-1,-x)$
$f(f(x)-x)=0 \implies f(x)=x$ Hence all the functions that work are $\boxed{f(x)=0,x}$ thus we are done

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#10 h Mar 20, 2022, 3:46 PM • 1 Y

Y by megarnie

Let $P(x,y)$ be the assertion $f(f(xy)+y)=(x+1)f(y)$ . If $f(0)\ne0$ then:
$P\left(\frac{f(f(0))+2021}{f(0)}-1,0\right)\Rightarrow 2021=0$
so $f(0)=0$ . Suppose that $f(k)=0$ for some $k\ne0$ .
$P\left(\frac kx,x\right)\Rightarrow f(x)=0$ for $x\ne0$ , and since $f(0)=0$ $\boxed{f(x)=0}$ for all $x$ .
Otherwise, $f(k)=0$ implies $k=0$ .
$P(-1,-x)\Rightarrow f(f(x)-x)=0\Rightarrow f(x)-x=0\Rightarrow\boxed{f(x)=x}$
which are both solutions.

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#11 h Mar 20, 2022, 5:48 PM

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CT17 wrote:

Misplaced?

Agreed. Swap this w/ P1.

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math31415926535

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math31415926535

#12 h Mar 20, 2022, 6:16 PM

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is it just me or kinda similar to damo p4

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#13 h Mar 20, 2022, 6:59 PM

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math31415926535 wrote:

is it just me or kinda similar to damo p4

Much easier but it does hav3 the same solutions

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#14 h Mar 21, 2022, 5:49 PM

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math31415926535 wrote:

is it just me or kinda similar to damo p4

LOL I did not expect that

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#15 h Apr 5, 2022, 6:57 PM

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Let $P(x,y)$ be our assertion then,
$P(0,0): f(f(0))=f(0)$
$P(x,0):f(f(0))=(x+1)f(0) \implies (x+1)f(0)=f(0) \implies f(0)=0$
Let $f(a)=f(b)$ for some $a,b \in \mathbb{R}$ ,
$P(a,1)=P(b,1) \implies (a-b)f(1)=0$
Case 1:- $a=b ; f(1) \neq 0$
Then the function is injective .
$P(-1,x):f(f(-x)+x)=0=f(0) \implies f(x)=x$
Case 2:- $a \neq b;f(1)=0$
$P(\frac{1}{x},x):f(f(1)+x)=(\frac{1}{x}+1)f(x) \implies \frac{1}{x}f(x)=0 \implies f(x)=0 \ \forall x \in \mathbb{R}-{0}$
Therefore we have two solutions, $\boxed{f(x)=0;f(x)=x}$

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#16 h Apr 7, 2022, 1:49 PM

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

$P(x, 0): f(f(0))=(x+1)f(0).$
$x=-1; \ f(f(0))=0.$
$\therefore \ (x+1)f(0)=0.$
$x \neq -1; \ f(0)=0.$

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

$\text{if } f(1)=0:$
$P\Big(\dfrac {1} {x}, x\Big): f(f(1)+x)=\Big(\dfrac {1}{x}+1\Big)f(x).$
$\therefore \ \dfrac {1}{x} f(x)=0.$
$x \neq 0; \ \color{blue}{f \equiv 0.}$

$\text{if } f(1) \neq 0:$
$\text{let } f(a)=f(b).$
$\Rightarrow P(a, 1) \text{ vs } P(b, 1):$
$(LHS): \text{same.}$
$(RHS): (a+1)f(1)=(b+1)f(1).$
$f(1) \neq 0; \ a=b, f: \text{injective}.$

$f(y) \neq 0;$
$P\Big(\dfrac{a}{f(y)}-1, y\Big): f(\text{Some complicated expressions})=a.$
$\Rightarrow f: \text{ BIJECTIVE.}$

$\text{let } f^{-1}: \text{ inverse function of } f.$
$P\Big(\dfrac{f^{-1}(-y)}{y}, y\Big): f(f(f^{-1}(-y))+y)=\Big(\dfrac{f^{-1}(-y)}{y}+1\Big)f(y).$
It seems a bit complicated, but after some cleanups, we can say that:
$\Big(\dfrac{f^{-1}(-y)}{y}+1\Big)f(y)=0. (\because f(f^{-1}(-y))=-y.)$
$f(y) \neq 0.$
$\therefore \dfrac{f^{-1}(-y)}{y}+1=0.$
$\Rightarrow f^{-1}(-y)=-y, -y=f(-y).$

$\therefore f(y)=y \text{ for } \forall y \neq 0.$
$f(0)=0.$
$\therefore \color{blue} {f(x)=x.}$

$\color{blue} f \equiv 0, f(x)=x._{\small \blacksquare}$

This post has been edited 1 time. Last edited by WhatistheName, Apr 7, 2022, 2:03 PM
Reason: Typos

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#17 h May 17, 2022, 6:07 AM • 1 Y

Y by Mango247

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ZETA_in_olympiad

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#18 h May 28, 2022, 2:33 PM

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Let $P(x,y)$ denote the given assertion.
We have $f(0)=0$ from $P(0,0)$ and $P(x,0).$

A. $\exists a\neq 0: f(a)=0:$
Then $P(ab^{-1},b)$ yields that $f$ is identically vanishing, fits.
B. $f(a)=0\iff a=0:$
Then $P(-1,-b)$ yields that $f$ is the identity, fits.

We are done after exhausting cases.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, May 28, 2022, 2:56 PM
Reason: Fits

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Taco12

#19 h Jun 26, 2022, 7:06 PM • 2 Y

Y by RedFireTruck, Mango247

We claim the only solutions are $f(x)=0$ and $f(x)=x$ .

$P(0,0) \rightarrow f(f(0))=f(0)$
$P(-1,0) \rightarrow f(f(0))=0$

Thus, we have $f(0)=0$ .

We have 2 cases on whether or not there is a $n\ne 0$ with $f(n)=0$ .

Case 1: There does exist $n$ .
$P\left(\frac{n}{x},x\right) \rightarrow f(x)=\left(\frac{n}{x}+1\right)f(x)$ . We have $f(x)=0$ .

Case 2: There does not exist $n$
$P(-1,-x)\rightarrow f(x)-x=0\rightarrow f(x)=x$ .

Both of these solutions work, so we are done. $\blacksquare$

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CyclicISLscelesTrapezoid

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#20 h Sep 4, 2022, 5:59 AM • 1 Y

Y by Mango247

We claim that the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x$ , which work.

Let $P(x,y)$ denote the assertion that $f(f(xy)+y)=(x+1)f(y)$ . By $P(x,0)$ , we have $f(f(0))=(x+1)f(0)$ , so $f(0)=0$ . If there exists $a \ne 0$ such that $f(a)=0$ , then for nonzero $x$ , $P(\tfrac{a}{x},x)$ gives $f(f(a)+x)=(\tfrac{a}{x}+1)f(x)=f(x)$ , and since $\tfrac{a}{x}+1 \ne 1$ , we have $f(x)=0$ . This gives the solution $f(x) \equiv 0$ . Otherwise, $f(a)=0$ implies that $a=0$ . Then, $P(-1,-x)$ gives $f(f(x)-x)=0$ , so $f(x)=x$ .

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#21 h Nov 8, 2023, 4:43 PM

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We claim that the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x$ , which obviously work.
Let $P(x,y)$ denote the assertion,
$P(x,1) \rightarrow$ $f(1)=0$ or $f$ is injective
CASE $1$ : f is injective
then $P(0,0) \rightarrow$ $f(0)=0$
then $P(-1,y) \rightarrow$ combined with injectivity gives $f(y)=y$
$CASE-2: f(1)=0$ $P(x,1) \rightarrow$ $f(f(x)+1)=0$
then $0=f(f(f(xy)+y)+1) = f((x+1)f(y)+1)$
However the thing wrapped in function in RHS is clearly surjective,
$\rightarrow f(x)=0$
$\blacksquare$

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#22 h Feb 5, 2024, 11:36 PM • 2 Y

Y by megarnie, OronSH

Solved with GrantStar.

The answer is $f(x) = x$ and $f(x) = 0$ , obviously working.

Letting $x=-1$ shows that there exists a root of $f$ .

Letting $x = \frac{r}{y}$ , where $r$ is some root of $f$ , gives us
$0 = r \cdot f(y)$ for all nonzero $y$ .

So, we have two cases:

If the only root is $0$ , take $x=-1$ , giving us
$f(f(-y)+y) = 0,$ so $f(-y)+y = 0$ , forcing $f(y) = y$ .
If there exists a nonzero root $r$ , then $f(y) = 0$ for all nonzero $y$ . Plugging in $y=0$ gives us
$f(f(0)) = (x+1) f(0),$ so $f(0) = 0$ , giving us the solution $f(x) = 0$ .

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#23 h Mar 8, 2025, 4:22 AM • 2 Y

Y by peace09, imagien_bad

The answer is $f(x) \equiv x$ or $f(x) \equiv 0$ . These both clearly work, so henceforth assume $f$ is nonzero.

Claim: $f(0) = 0$

Proof. Observe $\begin{align*} P(x, 0) \implies f(f(0)) = (x+1)f(0)\end{align*}$ so $f(0) = 0$ . $\square$

Claim: $f$ is injective

Proof. For contradiction, assume not. Take $f(a) = f(b)$ where $a \neq b$ . From $\begin{align*} P(a, 1) \implies f(f(a)+1) = (a+1)f(1) ~~~ (\clubsuit) \\ P(b, 1) \implies f(f(b)+1) = (b+1)f(1) ~~~ (\spadesuit) \end{align*}$ Equating $(\clubsuit)$ and $(\spadesuit)$ , we require either $f(1) = 0$ or $a = b$ . If $f(1) = 0$ , then we have $\begin{align*} P \left( x, \tfrac{1}{x} \right) \implies f \left( f(1) + \tfrac{1}{x} \right) = f(\tfrac{1}{x}) = (x+1)f(\tfrac{1}{x})\end{align*}$ clearly implying $f$ must be zero. $\square$

To finish, observe that $\begin{align*} P(-1, -y) \implies f(f(y)-y) = f(0) = 0\end{align*}$ This implies that $f(y) - y = 0$ . We can check for the pointwise trap, and we are done. $\blacksquare$

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#24 h May 21, 2025, 12:21 AM

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Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for all positive real numbers $x$ and $y$ :
$f(f(xy)+y)=(x+1)f(y).$

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#25 h Jun 4, 2025, 3:39 AM

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The solutions to this functional equation are $\boxed{f(x)\equiv 0, f(x)\equiv x}$ , which obviously works. Let $P(x,y)$ be the usual assertion.
$P(-1,0)\implies f(f(0))=0 ,$
$P(0,0)\implies f(f(0))=f(0)\implies f(0)=0$
Now we separate the problem into two cases.

Case 1: There exists a $c\in \mathbb{R}$ not equal to $0$ such that $f(c)=0$ .
$P(\frac{c}{y},y)\implies f(f(c)+y)=(\frac{c}{y}+1)f(y)\implies f(y)=0$ or $\frac{c}{y} =0$ .
Clearly, $\frac{c}{y}=0$ is impossible since $c\neq 0.$ Therefore in this case, the only possible solution for $f$ is $\boxed{f(x)\equiv0}.$

Case 2: There doesn't exist a $c\in \mathbb{R}$ not equal to $0$ such that $f(c)=0$ .
$P(x,1)\implies f(f(x)+1)=(x+1)f(1)$
This clearly shows that $f$ is bijective since $f(1)\neq 0$ .
$P(-1,y)\implies f(f(-y)+y)=0=f(0)\implies f(-y)+y=0\implies f(y)=y$ .
Therefore in this case, the only possible solution for $f$ is $\boxed{f(x)\equiv x}.$

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ezpotd

#26 h Jun 4, 2025, 4:41 AM

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The solutions are $f(x) = x, f(x) = 0$ , both of which clearly work.

Assume $f$ is nonzero for at least one input. Taking $y$ such that $f(y)$ is nonzero, $f$ is bijective. Set $x = 0$ , then $f(f(0) + y)= f(y) \rightarrow y = y + f(0)$ , so $f(0) = 0$ . Then set $x = -1$ , so $f(f(-y) + y) = 0$ , so $-y = f(-y)$ , so $f$ is the identity as desired.

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#27 h Jun 10, 2025, 7:52 AM

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solving FE because I got nothing left (btw I hate FE)

solution

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