Problem: Determine all composite positive integers $n$ for which
it is possible to arrange all divisors of $n$ that are greater than $1$
in a circle so that no two adjacent divisors are relatively prime.

Solution: The only bad ones are $n = pq$, products of two distinct primes.
Clearly they can't be so arranged, so we show all others work.
\begin{itemize}
\ii If $n$ is a power of a prime, the result is obvious; any arrangement works.

\ii If $n = p_1^{e_1} \dots p_k^{e_k}$ for some $k \ge 3$,
then first situate $p_1p_2$, $p_2p_3$, \dots, $p_kp_1$ on the circle.
Then we can arbitrarily place any multiples of $p_i$
between $p_{i-1} p_i$ and $p_i p_{i+1}$.
This finishes this case.

\ii Finally suppose $n = p^a q^b$.
If $a > 1$, say, we can repeat the argument by first placing
$pq$ and $p^2q$ and then placing multiples of $p$ in one arc
and multiples of $q$ in the other arc.
On the other hand the case $a=b=1$ is seen to be impossible.
\end{itemize}