Functional Square Root

[Leo.W]

On the topic of function composition. Is the successor function the only function f(x) such that f(f(x)) = f(x)+1?

The equation shows f is one Abel function of itself, and thus also the superfunction of itself.

If you assume that f has a singlevalued inverse function, then by the definition f must satisfy:
\[f^t(x)=f^{-1}(f(x)+t)\Leftrightarrow f^{t+1}(x)=f(x)+t\]
by taking t=-1, we simply get
\[f^0(x)=x=f(x)-1\]
this shows f(x)=x+1 is the only solution.

If not, we can still infer that for any natural number n, \[f^{n+1}(x)=f(x)+n\]
ps. This can be extended by the definition of iteration as
\[f^n(x)=f^{m+(n-m)}(x)=f^m(f^{n-m}(x))=f^m(f(x)+n-m-1), n>m\]
Consider n=3\[f(f(x))+1=f^2(f(x))=f^3(x)=f(f^2(x))=f(f(x)+1)\]
Let f(x)=t, \[f(t)+1=f(t+1)\]
Then this gives f(t+1)-f(t)=1, a recurrence, pretty easy to solve and get f(t)=t+C for some C, and easily prove C=1
So, this proves again that f is the only solution, if it's provided that f's domain and range are the same, which is the whole plane.