Here is an excerpt of a problem:
Let $A$ be nonempty set, and let $f:A \rightarrow A$ be a function. Suppose that $f$ is bijective. For each $n \in \mathbb{N}$, let $f^n$ denote the function $A \rightarrow A$ given by
$$f^n = f \circ \cdots \circ f$$
where $f$ iterated $n$ times.
The function $f^n$ is the $n$-fold iteration of $f$. We now extend the definition of $f^n$ to all $n \in \mathbb{Z}$. Let $f^0 = i_A$ where $i_A: A \rightarrow A$ is the identity mapping. Because $f$ is bijective, it follows that $f^n$ is bijective. Hence $f^n$ has an inverse. For each $n \in \mathbb{N}$, let $f^{-n} = (f^n)^{-1}$. It can be verified that $f^a \circ f^b = f^{a+b}$ and $(f^a)^b = f^{ab}$ for all $a,b \in \mathbb{Z}$.
(1) Let $x,y,z \in A$. Prove that the following holds.
$x = f^n(x)$ for some $n \in \mathbb{Z}$
The reason why I doubt my solution is because it seems to simple, yet I have no logical reason to fear that my answer is correct. Since the quantifier in the problem is existential, I merely need to exhibit an $n \in \mathbb{Z}$ such that the equation holds. Merely setting $n = 0$ will satisfy the equation, and the proof is complete.
Is there a problem with this approach? Is there a way to build an $n$ that is equal to a specific $n_0$, as opposed to $n$ having a concrete value like 0?
Consider the case $A=\mathbb{Z}$ and $f(m)=m+1$. It follows that $f^{n}(m)=m+n$ for all $n\in\mathbb{Z}$. The equation $m=f^{n}(m)=m+n$ implies $n=0$. What does this imply?
Addendum: If $A$ is finite, the situation is different. Try playing around with $A=\mathbb{Z}/k\mathbb{Z}$ (integers modulo some $k$) and $f(m)=m+1$.