General solutions for a functional equation

Question

I wonder if we can show the existence of solutions other than $f(x)=x$ for the functional equation $$f(x+1)=f(x)+1,\text{ }f(x^3)=(f(x))^3$$ where f is a real-valued function on the real line.

Answer 1

For $x=0,\pm 1$ we have $x^3=x$, so that the second equation implies $f(x)=f(x)^3$ and thus $f(x)=0,\pm 1$. In order for $f(1)=f(0)+1 = f(-1)+2$, we must have $f(-1)=-1$, $f(0)=0$, and $f(1)=1$. Thus by $f(x+1) = f(x)+1$ we must have $f(x)=x$ for all integers.

The second equation can be transformed to $f(x)=f(x^{1/3})^3$. Combined with the above observation, this implies that for any $x$ which is a cube root of an integer, $f(x)=x$. The using the first equation again, any $x$ which is a cube root of an integer mod 1 (i.e. $x=n^{1/3}-\lfloor n^{1/3}\rfloor$ for some integer $n$) must also have $f(x)=x$.

Now cube roots of integers mod 1 are dense in the interval $[0,1]$, and thus $f(x)=x$ on a dense set of $[0,1]$, and by the first equation the same holds on a dense set of $\mathbb{R}$. So if you require continuity for $f$, then $f(x)=x$ is the only solution.

If you don't require continuity, then there will be other solutions. This is analogous to how the Cauchy equation has uncountably many solutions if you don't impose some condition like continuity: by iterating the two equations defining $f$ (i.e. adding/subtracting 1 and taking cubes or cube roots), you can only reach a countable number of points on the real line. Without imposing continuity, all the other points are unconstrained. So there are uncountably many solutions, but giving a closed-form description of any of the non-trivial ones is difficult.