Is this divisibility problem correct? Let $n$ be a natural number and let $1 \le a_1<a_2<...<(a_k=n)$ be all of its divisors.
Find all $n$ such that $a_2^3+a_3^2-15=n$ .
It seems impossible to find all such numbers.

There are two posibilities for $a_2$ and $a_3$:
$$
a_2=p,\ a_3=p^2\quad\text{or}\quad a_2=p,\ a_3=q
$$
where $p$ and $q$ are prime divisors of $n$ and $p<q<p^2$.
Note: the restriction $q<p^2$ is not necessary. See barak manos comment below.
For the first one we have $p^3+p^4-15=n$. Since $p^2\mid n$ this implies that $p^2\mid15$, which is not possible.
For the second one we get $p^3+q^2-15=n$. Since $q\mid n$ it must be hat $q\mid p^3-15$ (and remenber that $p<q$.) Let's see the possible values of $p$.


*

*$p=2$. Then $q\mid7$ which is not possible.

*$p=3$. Then $q\mid12$ which is not possible.

*$p=5$. Then $q\mid110=2\cdot5\cdot11$, which again is not possible.

*$p=7$. Then $q\mid328=2^3\cdot41$, $q=41$ and $n=7^3+41^2-15=2009=7^2\cdot41$ is a solution.

*$p>7$. For a proof see user298259's answer.

