Functional equation $f(f(k+1)+3)=k$

Question

Find all functionas $f:\mathbb{Z}\to\mathbb{Z}$ such that $\forall k\in\mathbb{Z}$, $$f(f(k+1)+3)=k$$. Obvious solution is $f(k)=k-2$. I have found following: $k+3=f(f(k+1)+3)+3\implies f(k+3)=f(f(f(k+1)+3)+3)=f(k+1)+2\implies f(k)=\begin{cases} k+a,\quad 2\nmid k \\ k+b,\quad 2\mid k\end{cases}$ for some $a, b\in\mathbb{Z}$. And I can't continue after this. Can anyone help me, please?

Answer 1

What you have done is good. We can continue to search for constraints on $a$ and $b$.

First, $a$ must be even: Because, if $a$ was odd, then setting $k=0$ gives $$0=f(f(\underbrace{1}_\text{odd})+3) = f(1+a+3) = f(\underbrace{4+a}_\text{odd}) = 4+2a \implies a=-2,$$ contradiction.

Now that we know that $a$ is even, $$0=f(f(\underbrace{1}_\text{odd})+3) = f(1+a+3) = f(\underbrace{4+a}_\text{even}) = 4+a+b \implies b=-a-4.$$

In summary, $a$ is even and $$ f(k)=\begin{cases} k+a, & k \text{ odd} \\ k-a-4, & k \text{ even}\end{cases} $$

(and $a=-2$ gives the "obvious" solution $f(k)=k-2$).