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#1 h Apr 24, 2025, 8:32 PM • 1 Y

Y by cubres

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

This post has been edited 1 time. Last edited by skellyrah, Apr 25, 2025, 7:42 PM

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#2 h Apr 24, 2025, 8:35 PM • 1 Y

Y by cubres

skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2 + (x+1)f(x)$

What is IR?

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#3 h Apr 24, 2025, 8:41 PM • 1 Y

Y by cubres

skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2 + (x+1)f(x)$

If $IR=\mathbb R$ then:
Let $P(x,y)$ be the assertion $xf(x+yf(xy))+f(f(y))=f(xf(y))^2+(x+1)f(x)$ .
$P(0,0)\Rightarrow f(f(0))=f(0)^2+f(0)$
$P(f(0),0)\Rightarrow f\left(f(0)^2\right)=0$
$P\left(0,f(0)^2\right)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow\boxed{f(x)=0}$ for all $x$ which fits.

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#4 h Apr 24, 2025, 9:24 PM

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Made a mistake sorry

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#5 h Apr 24, 2025, 9:24 PM

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skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

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#6 h Apr 25, 2025, 12:35 AM

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skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

What is IR? Just $\mathbb R$ , right? ive never seen the notation before

This post has been edited 2 times. Last edited by jasperE3, Apr 25, 2025, 12:49 AM

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#7 h Apr 25, 2025, 9:34 AM

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yeah its the set of real numbers

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#8 h Apr 25, 2025, 7:42 PM

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refresh its a nice fe
and difficult

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#9 h Apr 25, 2025, 7:56 PM

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jasperE3 wrote:

skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

What is IR? Just $\mathbb R$ , right? ive never seen the notation before

I think IR resembles a R with a vertical bar, which looks similar to the symbol $\mathbb R$

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#10 h Apr 28, 2025, 3:17 AM

Y by

jasperE3 wrote:

skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2 + (x+1)f(x)$

If $IR=\mathbb R$ then:
Let $P(x,y)$ be the assertion $xf(x+yf(xy))+f(f(y))=f(xf(y))^2+(x+1)f(x)$ .
$P(0,0)\Rightarrow f(f(0))=f(0)^2+f(0)$
$P(f(0),0)\Rightarrow f\left(f(0)^2\right)=0$
$P\left(0,f(0)^2\right)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow\boxed{f(x)=0}$ for all $x$ which fits.

Why f(f(0)^2)=0?

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#11 h Apr 28, 2025, 3:28 AM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

jasperE3 wrote:

skellyrah wrote:

find all functions IR->IR such that $xf(x+yf(xy)) + f(f(y)) = f(xf(y))^2 + (x+1)f(x)$

If $IR=\mathbb R$ then:
Let $P(x,y)$ be the assertion $xf(x+yf(xy))+f(f(y))=f(xf(y))^2+(x+1)f(x)$ .
$P(0,0)\Rightarrow f(f(0))=f(0)^2+f(0)$
$P(f(0),0)\Rightarrow f\left(f(0)^2\right)=0$
$P\left(0,f(0)^2\right)\Rightarrow f(0)=0$
$P(x,0)\Rightarrow\boxed{f(x)=0}$ for all $x$ which fits.

Why f(f(0)^2)=0?

Actually we don't need $P(0,0)$ .
We have:
$xf(x+yf(xy))+f(f(y))=f(xf(y))^2+(x+1)f(x).$ Substitute $y=0$ :
$xf(x)+f(f(0))=f(xf(0))^2+(x+1)f(x).$ Substitute $x=f(0)$ :
$f(0)f(f(0))+f(f(0))=f\left(f(0)^2\right)^2+(f(0)+1)f(f(0)).$ Cancel terms (since $f(0)f(f(0))+f(f(0))=(f(0)+1)f(f(0))$ ):
$0=f\left(f(0)^2\right)^2,$ and the conclusion follows.

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#12 h Apr 29, 2025, 6:25 PM

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@above

The problem was edited.

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#13 h Apr 29, 2025, 7:05 PM

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mkultra42 wrote:

@above

The problem was edited.

I know.
The problem got edited after I solved it, my solution is still valid for the original problem.

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