Find all odd functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(t+1)=f(t)+1$ and $f\left(\frac{1}{t}\right)=\frac{1}{t^2}f(t)$ Please help me to solve this example:

Find all functions $f: \mathbb{R} \to \mathbb{R}$, so that for every $t\in\mathbb{R}$ apply:
$$f(-t)=-f(t)\text,$$
$$f(t+1)=f(t)+1\text{, and}$$
$$f\left(\frac{1}{t}\right)=\frac{1}{t^2}f(t)\text {, for } t\ne 0$$

Thank you very much for your help and your attention.
 A: Note that $f(1-t)= f((-t)+1)=f(-t)+1=1-f(t)$.
Clearly , $f(0)=0, f(-1)=-1$. Choose $x\neq 0,-1$.
$$f(\frac{1}{x+1})=\frac{f(x+1)}{(x+1)^2}=\frac{f(x)+1}{(x+1)^2}$$
and
$$\begin{aligned}f\left(\frac{1}{x+1}\right) =f\left(1-\frac{x}{x+1}\right) & = 1-f\left(\frac{x}{x+1}\right)\\ & = 1-f\left(\frac{1}{1+\frac1x}\right) \\ & =1-\frac{f(\frac1x+1)}{(\frac1x+1)^2} =1-\frac{\frac{f(x)}{x^2}+1}{(1+\frac1x)^2}\end{aligned}$$
Hence comparing we get $f(x)=x$ for all such $x\neq 0,-1$.
A: Substitute $t=0$ in your first equation to find $f(0)$.
Use that, the second equation, and induction to find $f(t)$ for integers $t$.
Use that and the third equation to find $f(t)$ for reciprocals of the integers.
Use that and the second equation to find $f(t)$ for all rational numbers.
It looks like (I am not sure) the irrational numbers are then not defined by your equations. Given the value at one irrational number you could find values for many other, but not all, irrational numbers. See Vitali sets for clues how to handle that. Of course, if you know that $f$ is continuous, $f(t)$ will then be uniquely defined for all $t$.
