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China Northern MO 2009 p4 CNMO
parkjungmin   0
30 minutes ago
Source: China Northern MO 2009 p4 CNMO
China Northern MO 2009 p4 CNMO

The problem is too difficult.
Is there anyone who can help me?
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FE with a lot of terms
MrHeccMcHecc   1
N Apr 6, 2025 by jasperE3
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ $$f(x)f(y)+f(x+y)=xf(y)+yf(x)+f(xy)+x+y+1$$
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MrHeccMcHecc
Apr 6, 2025
jasperE3
Apr 6, 2025
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FE with a lot of terms
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MrHeccMcHecc
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MrHeccMcHecc
#1 h  Apr 6, 2025, 1:44 PM • 1 Y
Y by PikaPika999
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ $$f(x)f(y)+f(x+y)=xf(y)+yf(x)+f(xy)+x+y+1$$
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jasperE3
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#2 h  Apr 6, 2025, 3:37 PM • 2 Y
Y by PikaPika999, MrHeccMcHecc
MrHeccMcHecc wrote:
Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ $$f(x)f(y)+f(x+y)=xf(y)+yf(x)+f(xy)+x+y+1$$

Let $P(x,y)$ be the assertion $f(x)f(y)+f(x+y)=xf(y)+yf(x)+f(xy)+x+y+1$. Note that $f(x)=x+1$ isn't a solution, so:
$P(x,0)\Rightarrow f(0)=-1$
$P(2,2)\Rightarrow f(2)\in\{5,-1\}$
$P(1,1)\Rightarrow f(1)\in\{-1,1,2,4\}$
$P(x,1)\Rightarrow f(x+1)=(2-f(1))f(x)+(f(1)+1)x+2$

Case 1: $f(1)=-1$
We have $f(x+1)=3f(x)+2$. Comparing $P(x,y)$ with $P(x,y+1)$, we get:
$$f(xy+x)=(2y+1)f(x)+3f(xy)+2y+3$$and setting $y=\frac1x$, we have $f(x)=-1$ for $x\ne0$. Since $f(0)=-1$ our solution here is $\boxed{f(x)=-1}$ which fits.

Case 2: $f(1)=1$
We have $f(x+1)=f(x)+2x+2$. Comparing $P(x,y)$ with $P(x,y+1)$, we get:
$$(1+2y)f(x)+f(xy)-2xy+2y+1=f(xy+x)$$and setting $y=\frac1x$, we have $f(x)=x^2+x-1$ for $x\ne0$. Since $f(0)=-1$ our solution here is $\boxed{f(x)=x^2+x-1}$, which fits.

Case 3: $f(1)=2$
$P(x,1)$ immediately give $\boxed{f(x)=3x-1}$ which fits.

Case 4: $f(1)=4$
We have $f(x+1)=-2f(x)+5x+2$. Comparing $P(x,y)$ with $P(x,y+1)$, we get:
$$(2y+1)f(x)-2f(xy)+2y-2=5xy+f(xy+x)$$and setting $y=\frac1x$, we have $\left(\frac2x+3\right)f(x)=-\frac2x+5x+17$ which gives a contradiction when $x=-\frac23$. So no solutions in this case.
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