Can anyone pleas help me?! I got stuck at a place where I got $\frac{f(x)}{f(-x)}=\frac{f(x)-f(y)}{f(y-x)}$. After that I can't find a way to finish the problem. EDIT: Has anyone proven $f$ odd?

Clearly if $f$ is constant then $f=0$ so there exists $d$ such that $f(d) \ne 0$ when $f$ is not constant. Let $P(x,y)$ the assertion of the F.E. (Here is the case $f$ non-constant) $P(x,0)$ for $x \ne 0$ $$x^nf(0)=0 \implies f(0)=0$$Assume that there exists $c \ne 0$ such that $f(c)=0$ then by $P(c,d)$ $$c^nf(d)=0 \implies c=0 \; \text{contradiction!!}$$Hence $f$ is injective at $0$ and hence $f(x) \ne 0$ for all $x \ne 0$. Let $a,b$ such that $f(a)=f(b)$, then by $P(x,a)-P(x,b)$ for any $x \ne 0$ $$f(x+a)=f(x+b)$$And now on this equation plug $x=-b$ hence $f(a-b)=0$ but since $f$ is injective at $0$ we have $a=b$ hence $f$ injective. Edit: i dont count wih much time rn, add my name if u got a proof using my injectivity

rafaello wrote: Let $n$ be a positive integer. Find all functions $f\colon \mathbb{R}\rightarrow \mathbb{R}$ that satisfy the equation $$ (f(x))^n f(x+y) = (f(x))^{n+1} + x^n f(y) $$for all $x ,y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(x)^nf(x+y)=f(x)^{n+1}+x^nf(y)$ If $f(u)=0$ for some $u\ne 0$, then $P(u,x)$ $\implies$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$, which indeed fits. So let us consider from now that $f(x)\ne 0$ $\forall x\ne 0$ $P(1,0)$ $\implies$ $f(0)=0$ $P(x,-x)$ $\implies$ $f(x)^{n+1}+x^nf(-x)=0$ $P(-x,x)$ $\implies$ $f(-x)^{n+1}+(-x)^nf(x)=0$ First line gives $f(-x)=-x^{-n}f(x)^{n+1}$ $\forall x\ne 0$ Plugging in second, we get $f(x)^{n^2+2n}=x^{n^2+2n}$ $\forall x\ne 0$ If $n$ is odd, this implies $f(x)=x$ $\forall x$, which indeed fits. If $n$ is even, this implies $f(x)=\pm x$ $\forall x$ If $\exists u,v\ne 0$ such that $f(u)=u$ and $f(v)=-v$, $P(u,v)$ $\implies$ $f(u+v)=u-v$, impossible since $f(u+v)$ is either $u+v$, either $-u-v$ and $uv\ne 0$ So : either $f(x)=-x$ $\forall x$, which indeed fits either $f(x)=x$ $\forall x$, which indeed fits And so $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ And $\boxed{\text{S3 : }f(x)=-x\quad\forall x}$ only when $n$ is even.

Case $1 : f$ is constant. Let $f(x) = a$ then we have $a^{n+1} = a^{n+1} + ax^n \implies a = 0$ so $f(x) = 0$ which fits the main equation. Case $2 : f$ isn't constant. $P(x,0) : (f(x))^nf(x) = (f(x))^{n+1} + x^nf(0) \implies f(0) = 0$. $P(x,-x) : (f(x))^{n+1} + x^nf(-x) = 0 \implies f(-x) = -\frac{(f(x))^{n+1}}{x^n}$. $P(-x,x) : (f(-x))^{n+1} + (-x)^nf(x) = 0 \implies f(x)^{n^2+2n} = x^{n^2+2n}$. so for even $n$ we have $f(x) = \pm x$ and for odd $n$ we have $f(x) = x$. Let's assume $a,b$ exists such that $f(a) = a$ and $f(b) = -b$. $P(a,b) : f(a+b) = a-b$ which gives contradioction cause $f(a+b) = \pm a+b$. Answers $: f(x) = x$ and $f(x) = -x$ for even $n$.