Uniqueness of the -primitive decomposition for fixed square-free part

Complete proof. For a positive integer $u=\prod_p p^{e_p}$, let

$$\operatorname{sf}(u)=\prod_{e_p\text{ odd}}p$$

denote its square-free part. For every positive integer $c$,

$$\operatorname{sf}(c^2u)=\operatorname{sf}(u).$$

Proposition 3.2 of the source establishes that every $n\in\mathbb N(\Delta)$ admits a decomposition $n=c^2m$, where $c\ge1$ and $m$ is $\Delta$-primitive. Equivalently, repeatedly remove a nontrivial square factor whenever its quotient remains in $\mathbb N(\Delta)$. The positive quotients strictly decrease, so the process terminates at a primitive quotient.

Fix a square-free $s$ with $|P(s)|=1$, and suppose $n\in\mathbb N(\Delta)$ has square-free part $s$. If

$$n=c_1^2m_1=c_2^2m_2$$

are primitive decompositions, square-free-part invariance gives

$$\operatorname{sf}(m_1)=\operatorname{sf}(n)=s=\operatorname{sf}(m_2).$$

Thus $m_1,m_2\in P(s)$. Since $P(s)$ is a singleton, $m_1=m_2$; consequently $c_1^2=c_2^2$, and positivity gives $c_1=c_2$. The decomposition therefore exists and is unique. The nonsquare assumption is unnecessary.

More precisely, writing $n=su^2$, its primitive decompositions are in bijection with

$$\{d\mid u:sd^2\in P(s)\}$$

via $m=sd^2$ and $c=u/d$. Conversely, if $P(s)$ contains distinct $m_i=sd_i^2$, then

$$N=s\,\operatorname{lcm}(d_1,d_2)^2$$

has two distinct primitive decompositions, and every square multiple of $N$ does as well. For $s>1$, all these examples are nonsquares.