Let $R$ be a commutative ring with $1$ and suppose $q^2\mid p^2,$ for $p,q \in R$. Unless $R$ is a UFD, I don't believe I can conclude that $q\mid p,$ but I would like to know a concrete counterexample.
Does anyone know an example of such a ring, where there exist elements $p,q$ such that $q^2$ divides $p^2$ but $q$ does not divide $p$?
On
Just a remark: you need much less than a UFD to conclude that $p^2|q^2 \implies p \mid q$.
Namely, if $A$ is an integrally closed domain then this holds.
(If $q^2 = a p^2$ for some $a,p,q \in A$ with $p \neq 0$, then $(q/p)^2 = a$, so --- by integral closeness of $A$ --- we see $\alpha := q/p \in A$, so $q = p \alpha$ and hence $p | q$ in $A$.)
Note that UFD's are integrally closed, and so the case of a UFD follows from this one.
Note also that Martin Brandenburg's example in comments, namely $\mathbb C[x,y,z]/(zx^2 - y^2)$, corresponds to a surface in $\mathbb A^3$ whose singular locus is an entire curve (the line $x = y = 0$), which is the basic geometric cause of an affine ring being non-integrally closed. (Integrally closed implies that the singular locus has codimension at least two.)
In $\mathbb{Z}/4$ take $q=0$ and $p=2$.
For an example of an integral domain, take $\mathbb{Z}[x,y]/(x y^2-4)$ and $p=y$, $q=2$.