Two bijections between set of integers I have the following interesting question:
Let $b$ be a $\mathbb{Z} \rightarrow \mathbb{Z}$ bijection, where $\mathbb{Z}$ denotes the set of integers. Is it possible that there exist a bijection between $\mathbb{Z}$ and $k+b(k)$ (where $k\in \mathbb{Z}$)?
That is: is it possible that $k+b(k)$ are all distinct integers and for every integer $S$ there exist (exactly one) an integer $n$ for which $S=n+b(n)$?
 A: We construct a sequence of maps $b_n\colon A_n\to \Bbb Z$ such that

*

*$A_n\subsetneq A_{n+1}$

*$b_{n+1}|_{A_n}=b_n$

*$b_n$ is injective

*$c_n:=b_n+\operatorname{id}$ is injective

*if $n>6m$ then $m\in A_n$, $m\in b_n[A_n]$, $m\in c_n[A_n]$
Then $b(n):=b_{6n}(n)$ will have the desired properties.
We start with $A_0=\{0\}$ and $b_0(0)=0$.
To obtain $A_{n+1}$, we pick suitable $x\in\Bbb Z\setminus A_n$, $y\in \Bbb Z\setminus b_n[A_n]$, $z\in \Bbb Z\setminus c_n[A_n]$ and set $A_{n+1}=A_n\cup\{x\}$ and $b_{n+1}(x)=y$. We just need to take care that $x+y=z$.

*

*If we first pick a (valid) $x$, the conditions on $y$ and $z$ exclude only finitely many values, hence we can certainly find a valid $y$ (and then $z$)

*If we first pick a (valid) $y$, the conditions on $x$ and $z$ exclude only finitely many values, hence we can certainly find a valid $x$ (and then $z$)

*If we first pick a (valid) $z$, the conditions on $x$ and $y$ exclude only finitely many values, hence we can certainly find a valid $x$ (and then $y$)

So by in turns picking the absolutely smallest "missing" $x$ or $y$ or $z$, we mange to comply with the last bullet point above.
