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Source: Evan Chen

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#1 h Apr 27, 2021, 2:21 AM • 3 Y

Y by HWenslawski, oVlad, nathantareep

Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$ .

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#2 h Apr 27, 2021, 2:35 AM • 4 Y

Y by HamstPan38825, MathLuis, Mango247, Mango247

For context, this is a made-up functional equation mentioned at the end of section 6 of https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf, where it's shown that $f$ injective would let you finish. However, $f$ is not necessarily injective, e.g. if $f$ is constant. I make no promise that there is a nice solution to this equation.

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gghx

#3 h Apr 28, 2021, 1:27 AM • 1 Y

Y by Mango247

Can't solve the entire thing. But add the additional condition $f(f(0))=-f(0)$ and it is solvable, and relatively hard too. Try it!

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#4 h Sep 20, 2021, 10:22 PM

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What is the solution with $f(f(0))=-f(0)$ ?

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gghx

#7 h Sep 21, 2021, 12:13 AM

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$P(x,y): f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$ .
$P(0,0): f(f(0))+f(0)=f(1)^2-1$

Case 1: $f(1)=1$ .
Note $f(f(x)+1)=f(x+f(1))=f(x+1)$ .
$P(f(x),y),P(f(y),x): f(xf(y))=f(yf(x))$ . Taking $x=0,f(0)=f(yf(0))$ .
If $f(0)\ne 0$ , $f$ is constant. Easy to find solutions.
If $f(0)=0$ , $P(x,0): f(x+1)=f(x)+1$ .
Original equation becomes $f(x+f(y))+f(xy)=f(x)f(y)+f(x)+f(y)$ .
Putting $y\rightarrow y+1$ , $f(x+f(y))+f(xy+x)=f(x)f(y)+2f(x)+f(y)$ .
Hence, $f(xy+x)=f(x)f(y)+f(x)$ which is a well known FE. (Cauchy+multiplicative).
Only solution $f(x)=x$ (or $0$ which dosent fit).

Case 2: $f(1)=-1$ .
$f(f(x)+1)=f(x+f(1))=f(x-1)$ .
$P(x,1): f(x-1)+f(x)=f(x+1)f(2)-1$ or $f(f(x)+1)+f(x)=f(x+1)f(2)-1$ .
This means that if $f(a)=f(b)$ then $f(a+1)=f(b+1)$ .
If $f(a)=f(b)$ ,
$P(x,a),P(x,b): f(x)=f(cx)$ for some $c$ .
$P(x,yc): f(y+1)=f(yc+1)=f(y+1/c)$ .
This means that $f(x)=f(x+d)$ for some $d$ . $P(x,y),P(x,y+d)$ gives $f$ constant.

Else $f$ is injective and it is easy to finish.

I can't remember if this is what I thought of back in April but it should be correct.

Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.

This post has been edited 2 times. Last edited by gghx, Sep 21, 2021, 12:24 AM

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#8 h Nov 3, 2022, 3:51 PM

Y by

gghx wrote:

$P(x,y): f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$ .
$P(0,0): f(f(0))+f(0)=f(1)^2-1$

Case 1: $f(1)=1$ .
Note $f(f(x)+1)=f(x+f(1))=f(x+1)$ .
$P(f(x),y),P(f(y),x): f(xf(y))=f(yf(x))$ . Taking $x=0,f(0)=f(yf(0))$ .
If $f(0)\ne 0$ , $f$ is constant. Easy to find solutions.
If $f(0)=0$ , $P(x,0): f(x+1)=f(x)+1$ .
Original equation becomes $f(x+f(y))+f(xy)=f(x)f(y)+f(x)+f(y)$ .
Putting $y\rightarrow y+1$ , $f(x+f(y))+f(xy+x)=f(x)f(y)+2f(x)+f(y)$ .
Hence, $f(xy+x)=f(x)f(y)+f(x)$ which is a well known FE. (Cauchy+multiplicative).
Only solution $f(x)=x$ (or $0$ which dosent fit).

Case 2: $f(1)=-1$ .
$f(f(x)+1)=f(x+f(1))=f(x-1)$ .
$P(x,1): f(x-1)+f(x)=f(x+1)f(2)-1$ or $f(f(x)+1)+f(x)=f(x+1)f(2)-1$ .
This means that if $f(a)=f(b)$ then $f(a+1)=f(b+1)$ .
If $f(a)=f(b)$ ,
$P(x,a),P(x,b): f(x)=f(cx)$ for some $c$ .
$P(x,yc): f(y+1)=f(yc+1)=f(y+1/c)$ .
This means that $f(x)=f(x+d)$ for some $d$ . $P(x,y),P(x,y+d)$ gives $f$ constant.

Else $f$ is injective and it is easy to finish.

I can't remember if this is what I thought of back in April but it should be correct.

Furthermore, it is interesting to see if anyone can remove the condition / use a weaker one. I probably tried this back then and failed.

Can you help me how to prove that there are only 2 cases for $f(1)$ ?

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#9 h Nov 3, 2022, 4:13 PM

Y by

Dennis1208 wrote:

Can you help me how to prove that there are only 2 cases for $f(1)$ ?

gghx is assuming $f(f(0)) = -f(0)$

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#10 h Apr 27, 2025, 8:15 PM

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jasperE3 wrote:

Find all functions $f:\mathbb R\to\mathbb R$ such that $f(x+f(y))+f(xy)=f(x+1)f(y+1)-1\forall x,y\in\mathbb R$ .

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

If $f(1)=f(2)=0$ we have:
$P(1,1)\Rightarrow 0=-1$ , contradiction.
So $f(1)\ne0$ or $f(2)\ne0$ .

Claim: $f$ is injective (barring constant solutions)
Suppose $f(a)=f(b)$ for some $a\ne b$ .
$P(0,a)\Rightarrow f(f(a))+f(0)=f(1)f(a+1)-1$
$P(0,b)\Rightarrow f(f(a))+f(0)=f(1)f(b+1)-1$
Comparing, we get $f(1)(f(a+1)-f(b+1))=0$ .
$P(1,a)\Rightarrow f(1+f(a))+f(a)=f(2)f(a+1)-1$
$P(1,b)\Rightarrow f(1+f(a))+f(a)=f(2)f(b+1)-1$
Comparing, we get $f(2)(f(a+1)-f(b+1))=0$ .
So $f(a+1)=f(b+1)$ , whether $f(1)\ne0$ or $f(2)\ne0$ .
$P(x,a)\Rightarrow f(x+f(a))+f(xa)=f(x+1)f(a+1)-1$
$P(x,b)\Rightarrow f(x+f(a))+f(xb)=f(x+1)f(a+1)-1$
Comparing, we get $f(xa)=f(xb)$ . If one of $a,b$ is zero, then the other isn't, so $f$ is constant. The only constant solutions are $\boxed{f(x)=\frac{\sqrt5-1}2}$ and $\boxed{f(x)=\frac{-\sqrt5-1}2}$ , so assume from now on that $ab\ne0$ , we get $f(x)=f\left(\frac abx\right)$ .
$P\left(y,\frac abx\right)\Rightarrow f(y+f(x))+f(xy)=f\left(\frac abx+1\right)f(y+1)-1$
$P(y,x)\Rightarrow f(y+f(x))+f(xy)=f(x+1)f(y+1)-1$
Comparing, we get $f(y+1)\left(f\left(\frac abx+1\right)-f(x+1)\right)=0$ . Setting $y\in\{0,1\}$ such that $f(y+1)\ne0$ , this becomes $f\left(\frac abx+1\right)=f(x+1)$ . Then:
$f(x)=f\left(\frac abx-\frac ab+1\right)=f\left(x-1+\frac ba\right).$ Let $p=\frac ba-1\ne0$ , this is just $f(x)=f(x+p)$ .
$P(x,y+p)\Rightarrow f(x+f(y))+f(xy+px)=f(x+1)f(y+1)-1$
$P(x,y)\Rightarrow f(x+f(y))+f(xy)=f(x+1)f(y+1)-1$
Comparing, we get $f(xy+px)=f(xy)$ . Setting $y=0$ , we get $f(px)=f(0)$ , so $f$ is a constant function whose solutions we already encountered.

Swapping $x,y$ we get $f(x+f(y))=f(y+f(x))$ , so using injectivity and setting $y=0$ , we get $f(x)=x+f(0)$ , testing gives $\boxed{f(x)=x}$ .

This post has been edited 1 time. Last edited by jasperE3, Apr 27, 2025, 8:34 PM

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