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.
 A: Use the fact that $f(n)=n$ in 3 to get the value for reciprocals of all integers.  Then use 2 to get reciprocals plus integers.  I believe you need to be told that $f$ is continuous.  Without that I don't think you can say anything about the value at irrationals.
A: From the given conditions you can prove that $f(x) = x$ for all $x$.  Hint:  Suppose that for some number $a$, $f(a) = b$.  The given conditions allow you to compute $f(-a)$, $f(a+1)$, and $f(1/a)$ (if $a \ne 0$).  Use these steps repeatedly to reach other values.  If you can find two different sequences of steps that lead to the value of $f$ at the same input, then you can set the results equal to each other and solve for $b$.
A: Well, I just saw Dan Vellerman's proof (hint actually), which is very elegant, and which I did not think of.
I will nevertheless add the proof that $f(x)=x$ on $\mathbb{Q}$ because the induction techniques used might be useful for some.
Let's prove that $f(x)=x$ on $\mathbb{Q}$.

*

*Notice that $f(0) = -f(-0) = -f(0)$. Therefore $f(0)=0$.


*Let's prove $f(x+n)=f(x)+n$ for all $n\in\mathbb{N}$. For $n=0$ this is obiously true. By induction $f(x+n+1) = f(x+1+n) = f(x+1)+n = f(x)+1+n = f(x) + n+1$. Therefore it is true for all $n\in\mathbb{N}$.


*Let us now prove $f\left( \frac{a}{b} \right) = \frac{a}{b}$ for all $a\in \mathbb{Z}$ and $b\in\mathbb{N}^+$. Suppose that $a\in\mathbb{N}$ and continue by induction on $a$.  If $|a|=0$ we reduce to the first case above. Otherwise:
$$
f\left( \frac{a}{b} \right) = \frac{f\left( \frac{b}{a}\right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) + \lfloor \tfrac{b}{a} \rfloor a}{a} \right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) }{a} + \left\lfloor \tfrac{b}{a} \right\rfloor \right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) }{a}\right) + \left\lfloor \tfrac{b}{a} \right\rfloor  }{\left(\frac{b}{a}\right)^2} = \frac{\frac{(b \mod a) }{a} + \left\lfloor \tfrac{b}{a} \right\rfloor  }{\left(\frac{b}{a}\right)^2} = \frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2} = \frac{a}{b}
$$
Note that the induction step was used in saying that $f\left(\frac{b \mod a}{a}\right) = \frac{b \mod a}{a}$ since $(b \mod a) < a$.
Finally if $a<0$ then $f\left(\frac{a}{b}\right) = f\left(-\frac{|a|}{b}\right) = -f\left(\frac{|a|}{b}\right) = -\frac{|a|}{b}= \frac{a}{b}$.
A: 
$f:\Bbb{R} \to \Bbb{R}(\text{odd}), f(x+1)=f(x)+1, f(\frac{1}{x})=\frac{f(x)}{x^2}.$

\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}
