Prove that six conditions on natural number $n$ are all logically equivalent. In studying a course on propositional logic, I ran into the following problem.
Prove that the following conditions on the natural number $n$ are all (logically) equivalent. Use as few conditional proofs as possible.

*

*$n$ is divisible by 3

*$n^2$ is divisible by 3

*$n^2$ is divisible by 9

*$n^3$ is divisible by 3

*$n^3$ is divisible by 9

*$n^3$ is divisible by 27

We need to prove that each statement has an "if and only if" (ie a material biconditional) relationship with each of the other statements.
Using as few conditional proofs as possible means that we should prove one "cycle", for example:
$$1 \implies 2 \implies 3 \implies 4 \implies 5 \implies 6 \implies 1$$
My question is how to do this.
Here is my attempt.
$6 \implies 5$
if $n^3$ divisible by 27 then $n^3=27m=3\times9m$, $m \in \mathbb{N}$. Therefore $n^3$ is divisible by 9.
$5 \implies 4$
if $n^3$ divisible by 9 then $n^3=3^2m, m \in \mathbb{N}$, therefore n^3 divisible by 3.
$4 \implies 3$
$3 \implies 2$
$n^2$ divisible by 9 then $n^2=3^2m, m \in \mathbb{N}$, therefore $n^2$ divisible by 3.
$2 \implies 1$
$n^2$ divisible by 3, then $n^2=3m, m \in \mathbb{N}$
Assume $m$ is not divisible by 3. Then $n=\sqrt{3m}$, which is not a perfect square, so $n \notin \mathbb{N}$, but this contradicts our premise that $n \in \mathbb{N}$. Therefore, $m$ is divisible by 3, so $n^2=3^2k, k \in \mathbb{N}$, and $n=3k$, divisible by 3.
$1 \implies 6$
$n$ divisible by 3 then $n=3m, m \in \mathbb{N}$. $n^3 = 27m$. Therefore, $n^3$ is divisible by 27.
Is this proof correct (with exception of the missing step $4 \implies 3$)?
What is the proof of the missing step?


$1 \implies 6$ $n$ divisible by 3 then $n=3m, m \in \mathbb{N}$.
$n^3 = 27m$. Therefore, $n^3$ is divisible by 27.

You forgot to cube the $m;$ fortunately, this doesn't affect what follows.

$2 \implies 1$ $n^2$ divisible by 3, then
$n^2=3m, m \in \mathbb{N}.$ Assume $m$ is not divisible by 3.
Then $n=\sqrt{3m}$,
which is not a perfect square, so $n \notin \mathbb{N}$, but this
contradicts our premise that $n \in \mathbb{N}$. Therefore, $m$ is
divisible by 3, so $n^2=3^2k, k \in \mathbb{N}$, and $n=3k$, divisible
by 3.

The final $k$ needs to be square-rooted, in which case it's no longer clear that $n$ is divisible by $3.$
If a prime number $p$ divides a perfect square $n^2,$ then, for some integer $q,$ $$n^2=p^2q^2.$$



*$n^2$ is divisible by 9

*$n^3$ is divisible by 3

$4 \implies 3$

If a prime number $p$ divides a perfect cube $n^3,$ then, for some integer $q,$ $$n^3=p^3q^3.$$
