Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$ 
Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions:

*

*$f(-x)=-f(x)$

*$f(x+1)=f(x)+1$

*$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$

I assume $f(x)=x$ satisfies the conditions. I could prove $f(0)=0 $ and $f(n)=n$ for all $n\in{\mathbb{N}}$, then I didn't know how to go any further.


$f:\Bbb{R} \to \Bbb{R}(\text{odd}), f(x+1)=f(x)+1, f(\frac{1}{x})=\frac{f(x)}{x^2}.$

\begin{align}
& f\left(1+\dfrac{1}{x}\right)=f\left(\dfrac{1}{x}\right)+1 = \dfrac{f(x)}{x^2}+1. \tag{1}\label{1} \\
& f\left(\dfrac{x+1}{x}\right)=\dfrac{f\left(\dfrac{x}{x+1}\right)}{\left(\dfrac{x}{x+1}\right)^2}\\
& = \dfrac{(x+1)^2}{x^2}f\left( 1-\dfrac{1}{x+1} \right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-f\left(\dfrac{1}{x+1}\right)\right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-\dfrac{f(x+1)}{(x+1)^2}\right) \\
& = \dfrac{(x+1)^2}{x^2}\left(1-\dfrac{f(x)+1}{(x+1)^2}\right). \tag{2}\label{2} \\
\ \\
\therefore \; & (\ref{1}) = (\ref{2}) \Rightarrow \dfrac{f(x)}{x^2}+1=\dfrac{(x+1)^2}{x^2}\left( 1-\dfrac{f(x)+1}{(x+1)^2} \right). \\
\Rightarrow \; & f(x)+x^2=(x+1)^2-f(x)-1, 2f(x)=2x, \boxed{f(x)=x}.
\end{align}
