Functional equation $f(x+1)-f(x)=nf(x)$ I want to know how to solve this problem on functions. It seems that it may probably involve discrete differentiating.

Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ satisfying
$$f(x+1)-f(x)=nf(x)\text,$$
where $\mathbb{R}$ is the set of real number.

Thank you!

By applying the recursion formula $k$ times we find that for any $k\in\mathbb{Z}$ we have
$$f(x+k)=(n+1)^k f(x)$$
Now define any function $f$ on $[0,1)$. For any other $x\in \mathbb{R}$ we can write
$$x = \lfloor x\rfloor + \left<x\right>$$
where $\left<x\right>\in [0,1)$ is the fractional part of $x$. From the formula above (with $k=\lfloor x \rfloor$ and taking $\left<x\right>$ for $x$) we have
$$f(x)=(n+1)^{\lfloor x\rfloor} f(\left<x\right>)$$
Thus by specifying $f(x)$ on $[0,1)$ the recursion formula extends $f$ uniquely to all $x$ and consequently there are infinitely many solutions.
