FE based on (x+1)(y+1)

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FE based on (x+1)(y+1)

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Source: 2023 CK Summer MSG I-A

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#1 h Aug 14, 2023, 5:11 AM • 1 Y

Y by farhad.fritl

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(xf(y)+f(x+y)+1)=(y+1)f(x+1)$ holds for all $x,y\in\mathbb{R}$ .

Proposed by owoovo.shih and CrazyInMath

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#2 h Aug 14, 2023, 5:57 AM

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The answer is $\boxed{f\equiv 0}$ and $\boxed{f(x) = x}$ . These work. Now we prove they are the only solutions.

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

$P(0,x): f(f(x) + 1) = (x+1)f(1)$ .

Case 1: $f(1) = 0$ .
Then $f(f(x) + 1) = 0$ for all $x$ .

$P(x-1,1): f(f(x) + 1) = 2f(x) \implies f(x) = 0$ for all reals $x$ .

Case 2: $f(1)\ne 0$ .
Then comparing $P(0,a)$ and $P(0,b)$ with $f(a) = f(b)$ gives that $a = b$ , so $f$ is injective.

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

$P(x,0): f( f(x) + 1) = f(x+1)$ , so $f(x) + 1 = x + 1 \implies f(x) = x$ .

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Taco12

#3 h Aug 16, 2023, 2:09 AM

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

Let $P(x,y)$ be the assertion.

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

Thus, there exists $n \in \mathbb{R}$ such that $f(n)=0$ .

$P(0,n) \rightarrow f(1)=(n+1)f(1)$ , so either $f(1)=0$ or $n=0 \rightarrow f(0)=0$ .

Case 1: $f(1)=0$
$P(0,x) \rightarrow f(f(x)+1)=0$
$P(x-1, 1) \rightarrow f(f(x)+1)=2f(x)$

Thus, $f(x)=0$ is the solution in this case.

Case 2: $f(0)=0$
Let $a,b$ be reals such that $f(a)=f(b)$ .
$P(0,a) \rightarrow f(f(a)+1)=(a+1)f(1)$
$P(0,b) \rightarrow f(f(b)+1)=(b+1)f(1)$

Since $f(a)=f(b)$ , the left hand sides are equal, implying $a=b$ , so $f$ is injective.

Now,

$P(x,0) \rightarrow f(f(x)+1)=f(x+1) \rightarrow f(x)=x$ , which is the solution in this case.

Exhausting all cases, we are done.

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ismayilzadei1387

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ismayilzadei1387

#4 h Nov 24, 2023, 4:04 PM • 1 Y

Y by FriIzi

$f(xf(y)+f(x+y)+1)=(y+1)f(x+1)$
$P(-1,x)$
$f(f(x-1)-f(x)+1)=(x+1)f(0)$
$P(0,0)$
$f(f(0)+1)=f(1)$
$P(x,1)$
$f(xf(1)+f(x+1)+1)=(2f(x+1))$
$P(x,0)$
$f(xf(0)+f(x)+1)=f(x+1)$
$P(0,y)$
$f(f(x)+1)=(y+1)f(1)$
$f(0) \ne 0$
$f$ surjective
assume there is a $z$ s.t $f(z)=0$
$P(0,z)$
$f(1)=(z+1)f(1)$
if $z=0$ then
$f(0)=0$
$P(x,0)$
$f(xf(0)+f(x)+1)=f(x+1)$
and $f$ injectivity
$f(f(x)+1)=f(x+1)$
$f(x)=x$
else $f(1)=0$
$P(-1,1)$
$f(f(0)+1)=2f(0)=f(1)$
$\implies$
$f(0)=0$
$\implies$
$f(f(x)+1)=0=f(x+1)$
$f(x)=0$
from 2 cases we get $f(0)=0$ which is contradicit with $f$ surjectivity iff $f$ constant case.

This post has been edited 2 times. Last edited by ismayilzadei1387, Nov 24, 2023, 4:11 PM

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#5 h Apr 10, 2025, 7:55 PM

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CrazyInMath wrote:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(xf(y)+f(x+y)+1)=(y+1)f(x+1)$ holds for all $x,y\in\mathbb{R}$ .

Proposed by owoovo.shih and CrazyInMath

Let $P(x,y)$ be the assertion $f(xf(y)+f(x+y)+1)=(y+1)f(x+1)$ . Testing, the only linear solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$ . Otherwise, there is some $j$ with $k:=(j+1)f(-1)+f(j)\ne-2$ .
$P(j+1,-1)\Rightarrow f(k+1)=0$
$P(x-1,k+1)\Rightarrow f(f(x+k)+1)=(k+2)f(x)$
$P(0,x+k)\Rightarrow f(f(x+k)+1)=(x+k+1)f(1)$
Comparing, we get that $f$ is linear, hence no more solutions.

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