The question that I am talking about is:
A natural number $n$ has property $(*)$, if $n=pqrs$, product of $4$ distinct primes, and, if we order its $16$ divisors, i.e. $$1=d_1<d_2<…<d_{16}=n,$$we have: $d_9-d_8=22$.
Are there infinitely many numbers $n$, that satisfy $(*)$ ?
In fact, by discussing the cases of $s$, compared with the other primes, the above problem is equivalent to finding solutions to one of the following equations: ($p<q<r<s$)
$1) s=pqr+22$, if $pqr<s$;
$2) s=pqr-22$, if $qr<s<pqr$;
$3) ps=22+qr$, if $qr/p<s<qr$;
$4) qr=22+ps$, if $s<qr/p$.
If any of $1)-4)$ has infinitely many solutions, then the original problem is solved.
My attempt to the first one, treated separately:
First of all, it is easy to notice that $p,q,r \neq 2,11.$
I tried to reduce the degrees of freedom, by setting, for example, $p=3,q=5$, in order to get some sort of arithmetic progressions, but I don't know if Dirichlet's theorems can help.
P.S. The above question is derived from $1995$ Irish Math Olympiad problem. At that time, the task was similar to finding the minimum solution, that is $n=3\cdot 5 \cdot 7 \cdot 19$, which is a solution to $3)$. My apologies for asking such a difficult question.
IMPORTANT EDIT: The problem is still open, although I have accepted an answer (or comment), for being the most relevant.
Best Answer
THIS IS NOT AN ANSWER, it is just a comment maybe not impertinent and it is put here for obvious reasons of space.
It is clear that $2$ must be discarded. For all pair of odd primes $p,q$ the diophantine equation $px-qy=22$ has infinitely many integer solutions in such a way that there are infinitely many primes for $x$ and $y$ (Dirichlet). The point is that for our purpose we need these primes appear simultaneously (which could deserve to be an interesting problem maybe). We look at some examples.
$$3x-5y=22\iff(x,y)=(5n+4,3n-2)\Rightarrow n=3 \space\text {gives }(r,s)=(19,7)\\5x-7y=22\iff(x,y)=(7n+3,5n-1)\Rightarrow n=4 \space\text {gives }(r,s)=(31,19)\\13x-17y=22\iff(x,y)=(17n+3,13n+1)\Rightarrow n=4 \space\text {gives }(r,s)=(19,7)$$ It must be said that it could be the case that only one of this class of equations is sufficient (to show which would ultimately be a refinement of Dirichlet's Theorem). Another thing is that we have not taken into account the required eighth and ninth position but have limited ourselves to finding four odd primes.