FE f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)

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FE f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)

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steven_zhang123

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steven_zhang123

#1 h Mar 29, 2025, 11:27 PM

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Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ , we have $f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)$ .

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pco

#2 h Mar 30, 2025, 6:53 AM • 1 Y

Y by jasperE3

Sorry, wrong proof.

Thanks @jasperE3 for noticing it. :oops:

:oops:

This post has been edited 1 time. Last edited by pco, Mar 30, 2025, 7:34 AM
Reason: Wrong proof

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#3 h Mar 30, 2025, 7:18 AM • 1 Y

Y by pco

pco wrote:

steven_zhang123 wrote:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ , we have $f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)$ .

Let $P(x,y)$ be the assertion $f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)$

$P(0,0)$ $\implies$ $f(0)=1$
$P(x,1)$ $\implies$ $f(x+1)=f(x)$ and so $f(x)=1$ $\forall x\in\mathbb Z$

But then $P(2,1)$ is $2=4$ , contradiction

And so $\boxed{\text{No such function}}$

$P(x,1)\Rightarrow f(x+1)=f(x)(f(1)-1)-x^2+x+1$ isn't it?
and how would $P(2,1)$ be $2=4$ ?

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pco

#5 h Mar 30, 2025, 9:13 AM • 1 Y

Y by MihaiT

steven_zhang123 wrote:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ , we have $f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)$ .

Let $P(x,y)$ be the assertion $f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)$
Let $c=f(1)$

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

$P(x,1)$ $\implies$ $f(x+1)=(c-1)f(x)-x^2+x+1$ and so :
$P(1,1)$ $\implies$ $f(2)=c^2-c+1$
$P(2,1)$ $\implies$ $f(3)=c^3-2c^2+2c-2$
$P(3,1)$ $\implies$ $f(4)=c^4-3c^3+4c^2-4c-3$
$P(2,2)$ $\implies$ $(c^2-c+1)^2+1=2(c^4-3c^3+4c^2-4c-3)+8$
Which is $c^4-4c^3+5c^2-6c=0$ , and so $c(c-3)(c^2-c+2)=0$ and so $c\in\{0,3\}$

1) If $c=0$
$P(x,1)$ becomes $f(x+1)=-f(x)-x^2+x+1$ and so $f(x+2)=-(-f(x)-x^2+x+1)-(x+1)^2+(x+1)+1$
Which is $f(x+2)=f(x)-2x$ (and so $f(2)=1$ )

$P(2,\frac 12)$ $\implies$ $f(1)=\frac 32$ , contradiction

2) If $c=3$
$P(x,1)$ becomes $f(x+1)=2f(x)-x^2+x+1$
$P(x+1,y)$ $\implies$ $(2f(x)-x^2+x+1)f(y)+1=(2f(x+y)-(x+y)^2+(x+y)+1)+f(xy+y)+(x+1)y(x+y-1)$
Subtracting this from twice $P(x,y)$ , we get :
$f(xy+y)=2f(xy)-(x^2-x-1)f(y)+x^2y+xy^2-2xy+x^2-x-2$
Considering $x,y\ne 0$ and moving there $x\to\frac xy$ , we get :

$f(x+y)=2f(x)-(\frac{x^2}{y^2}-\frac xy-1)f(y)+\frac{x^2}y+xy-2x+\frac{x^2}{y^2}-\frac xy-2$

Swapping $x,y$ :
$f(x+y)=2f(y)-(\frac{y^2}{x^2}-\frac yx-1)f(x)+\frac{y^2}x+xy-2y+\frac{y^2}{x^2}-\frac yx-2$

Subtracting : $(\frac{y^2}{x^2}-\frac yx+1)f(x)+\frac{x^2}y-2x+\frac{x^2}{y^2}-\frac xy$ $=(\frac{x^2}{y^2}-\frac xy+1)f(y)+\frac{y^2}x-2y+\frac{y^2}{x^2}-\frac yx$
Setting there $y=1$ (and $f(y)=3$ ) : $(\frac 1{x^2}-\frac 1x+1)f(x)=x^2+1+\frac 1{x^2}$
And so $(x^2-x+1)f(x)=x^4+x^2+1=(x^2-x+1)(x^2+x+1)$

And so $\boxed{f(x)=x^2+x+1\quad\forall x}$ , which indeed fits
(in fact, we got $f(x)=x^2+x+1\forall x\ne 0$ but it is obviously still true when $x=0$ )

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#7 h Mar 30, 2025, 1:29 PM

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Nice work, pco! But I have a slight confusion: In the final step, you have only discussed the case where $f(x) \neq 0$ . So, how should the case where $f(x) = 0$ be considered? Thank you in advance.

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pco

#8 h Mar 30, 2025, 3:15 PM

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steven_zhang123 wrote:

Nice work, pco! But I have a slight confusion: In the final step, you have only discussed the case where $f(x) \neq 0$ . So, how should the case where $f(x) = 0$ be considered? Thank you in advance.

I dont understand what you mean.
I never considered the case $f(x)\ne 0$

Maybe you made a slight confusion : i considered in last steps that $x,y\ne 0$ (and not $f(x)\ne 0$ ) and I got $f(x)=x^2+x+1$ $\forall x\ne 0$ , which, has I wrote, is obviously still true when $x=0$

If you mean something else, dont hesitate to precise your interrogation.

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