Infinite set of AP's that partition $\mathbb{N}$

Question

Inspired by this question.

Suppose we want to partition the set of positive integers $\mathbb{N}$ into distinct (possibly infinitely many) arithmetic progressions $AP_1, AP_2 \cdots$ such that no two of them have the same common difference, $\mathbb{N} = AP_1\cup AP_2\cdots $ and $AP_i\cap AP_j = \phi$ $ \forall j, \forall i\neq j$.

What i tried :

If $AP_i$ is the $i$th AP then it should be of the form $\{s_{i-1}, s_{i-1}+d_i,s_{i-1}+2d_i,\cdots\}$ where $s_{i-1}$ the smallest number in $\mathbb{N}\setminus AP_1\cup\cdots\cup AP_{i-1}$, $d_i \neq d_j \forall j<i$ and $gcd(d_i,d_j)\nmid s_{i-1} - s_{j-1} \forall j<i$ since otherwise there'd be common terms. So we're after a sequence of common differences that satisfy these conditions. But I'm not able to figure out much more. Any help is very much appreciated!

Answer 1

There are of course infinitely many ways to obtain such a partition into infinitely many APs. The most straightforward way is to let $AP_i$ run through all odd multiples of $2^{i-1}$

Answer 2

Just observe that you can write the odd numbers into union of arbitrary number of $AP's$ hence there will be infinitely many such partitions.

Use remainder by some power of $2$, like you can write the odd numbers as a union of two distinct $AP's$ $\{4k+1\} \cup \{4k+3\}$.