Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$, such that $$f(f(x)+y)(x-f(y)) = f(x)^2-f(y^2).$$
Problem
Source: IMOC 2023 A2
Tags: algebra
StarLex1
09.09.2023 17:34
deleted i see the wrong version of the problem
pco
09.09.2023 17:45
StarLex1 wrote: so all the sols is $x,-x,0$ No, $f(x)=-x$ $\forall x$ is not a solution.
StarLex1
09.09.2023 17:50
yup you are correct in the resub i made some mistake RHS should be x^2+y^2 that's why $f(x) = -x$ doesn't suit
CrazyInMath
09.09.2023 18:15
StarLex1 wrote: $P(0,0)= (f(f(0))(-f(0)) =0$ it should be $f(f(0))(-f(0))=f(0)^2-f(0)$
If $f(a)=f(b)$, $P(a,y)-P(b,y)$ gives $a=b$ or $f(x)=0$, so $f$ is injective if $f(x)\neq 0$
$P(f(x),x)\Longrightarrow f(f(x))^2=f(x^2)$.
$P(f(x),-x)\Longrightarrow f(f(f(x))-x)(f(x)-f(-x))=f(f(x))^2-f(x^2)=0$
Let $f(x)=f(-x)$, $P(x,-y)\Longrightarrow f(f(x)-y)(x-f(y))=f(x)^2-f(y^2)=f(f(x)+y)(x-f(y))$, so $f(x)+y=f(x)-y$, contradiction.
So there exists $c$ such that $f(c)=0$. $P(c,c)\Longrightarrow 0=f(c^2)=f(f(c))^2=f(0)^2$ so $f(0)=0$
As $f(f(f(x))-x)=0\Longrightarrow f(f(x))=x$, we know that $f(f(x))^2=x^2=f(x)^2$.
$P(0,x)\Longrightarrow f(x)^2=f(x^2)$, so $f(x)=x$ for all positive $x$.
For negative $y$ just take $x$ super big and we have $f(y)=y$.
Edit: Incorrect
alba_tross1867
10.08.2024 17:16
CrazyInMath wrote: StarLex1 wrote: $P(0,0)= (f(f(0))(-f(0)) =0$ it should be $f(f(0))(-f(0))=f(0)^2-f(0)$
If $f(a)=f(b)$, $P(a,y)-P(b,y)$ gives $a=b$ or $f(x)=0$, so $f$ is injective if $f(x)\neq 0$
$P(f(x),x)\Longrightarrow f(f(x))^2=f(x^2)$.
$P(f(x),-x)\Longrightarrow f(f(f(x))-x)(f(x)-f(-x))=f(f(x))^2-f(x^2)=0$
Let $f(x)=f(-x)$, $P(x,-y)\Longrightarrow f(f(x)-y)(x-f(y))=f(x)^2-f(y^2)=f(f(x)+y)(x-f(y))$, so $f(x)+y=f(x)-y$, contradiction.
So there exists $c$ such that $f(c)=0$. $P(c,c)\Longrightarrow 0=f(c^2)=f(f(c))^2=f(0)^2$ so $f(0)=0$
As $f(f(f(x))-x)=0\Longrightarrow f(f(x))=x$, we know that $f(f(x))^2=x^2=f(x)^2$.
$P(0,x)\Longrightarrow f(x)^2=f(x^2)$, so $f(x)=x$ for all positive $x$.
For negative $y$ just take $x$ super big and we have $f(y)=y$.
You haven't proved yet injectivity in point $0$, you should still find that if there is non-zero roots to $f$ then $f$ is $0$.
CrazyInMath
10.08.2024 18:51
alba_tross1867 wrote: CrazyInMath wrote: StarLex1 wrote: $P(0,0)= (f(f(0))(-f(0)) =0$ it should be $f(f(0))(-f(0))=f(0)^2-f(0)$
If $f(a)=f(b)$, $P(a,y)-P(b,y)$ gives $a=b$ or $f(x)=0$, so $f$ is injective if $f(x)\neq 0$
$P(f(x),x)\Longrightarrow f(f(x))^2=f(x^2)$.
$P(f(x),-x)\Longrightarrow f(f(f(x))-x)(f(x)-f(-x))=f(f(x))^2-f(x^2)=0$
Let $f(x)=f(-x)$, $P(x,-y)\Longrightarrow f(f(x)-y)(x-f(y))=f(x)^2-f(y^2)=f(f(x)+y)(x-f(y))$, so $f(x)+y=f(x)-y$, contradiction.
So there exists $c$ such that $f(c)=0$. $P(c,c)\Longrightarrow 0=f(c^2)=f(f(c))^2=f(0)^2$ so $f(0)=0$
As $f(f(f(x))-x)=0\Longrightarrow f(f(x))=x$, we know that $f(f(x))^2=x^2=f(x)^2$.
$P(0,x)\Longrightarrow f(x)^2=f(x^2)$, so $f(x)=x$ for all positive $x$.
For negative $y$ just take $x$ super big and we have $f(y)=y$.
You haven't proved yet injectivity in point $0$, you should still find that if there is non-zero roots to $f$ then $f$ is $0$. ok then my solution is wrong. I'm not gonna fix my solution though. (don't want to do another F.E.)