BdMO Online Forum

Let $n$ be a positive integer. Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such
that, for all reals $y$ and all nonzero reals $x$
\[x^nf(y)-y^nf(x)=f(\frac{y}{x})\]

My solution steps:

Let $P(x,y)$ be the assertion.$P(x,0)\Rightarrow f(0)=0$Now take $x\neq 0,y\neq 1,-1$.
$P\left (\displaystyle \frac {y}{x},y\right )\Longrightarrow f(x)=\left (\displaystyle \frac {y}{x}\right )^n\displaystyle \frac {1-x^{2n}}{1-y^{2n}}f(y)$
$\Longrightarrow f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\; or\; -a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$ where $a$ is a fixed real.Now assume $\exists c,d$ such that $f(c)=a\left (c^n-\displaystyle \frac {1}{c^n}\right ), f(d)=-a\left (d^n-\displaystyle \frac {1}{d^n}\right )$.Now take $P(c,d)$ to get a contradiction.So all the functions are:
$(i)f(0)=0, \; f(x)=a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$
$(ii)f(0)=0, \; f(x)=-a\left (x^n-\displaystyle \frac {1}{x^n}\right )\forall x\in \mathbb {R}\backslash 0$