Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?
x,y are integral points on an Elliptic Curve and are easy to find using enumeration of divisors of n (assuming n can be factored).
If yes, will large number of solutions give moderate rank EC?
If one drops $-1$ i.e. $xy(x-y)=n$ the number of solutions can be unbounded via multiples of rational point(s) and then multiplying by a cube. (Explanation): Another unbounded case for varying $a , n$ is $xy(x-y-a)=n$. If $(x,y)$ is on the curve then $(d x,d y)$ is on $xy(x-y-a d)=n d^3$. Find many rational points and multiply by a suitable $d$. Not using the group law seems quite tricky for me. The constant $-1$ was included on purpose in the initial post.
I would be interested in this computational experiment: find $n$ that gives a lot of solutions, say $100$ (I can't do it), check which points are linearly independent and this is a lower bound on the rank.
What I find intriguing is that all integral points in this model come from factorization/divisors only.

Current record is n=179071200 with 22 solutions with positive x,y. Due to Matthew Conroy.
Current record is n=391287046550400 with 26 solutions with positive x,y. Due to  Aaron Meyerowitz
Current record is n=8659883232000 with 28 solutions with positive x,y. Found by  Tapio Rajala.

Current record is n=2597882099904000 with 36 solutions with positive x,y. Found by  Tapio Rajala.
EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (note this is a different curve and 7 <= rank <= 13)
A variation: $(x^2-x-17)^2 - y^2 = n$ appears to be eligible for the same question. The quartic model is a difference of two squares and checking if the first square is of the form $x^2-x-17$ is easy.
Is it possible some relation in the primes or primes or divisors of certain form to produce records: Someone is trying in $\mathbb{Z}[t]$ Can the number of solutions xy(x−y−1)=n for x,y,n∈Z[t] be unbounded as n varies? ? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.
EDIT: The curve was chosen at random just to give a clear computational challenge.
EDIT: On second thought, can a symbolic approach work? Set $n=d_1 d_2 ... d_k$ where d_i are variables. Pick, well, ?some 100? ($d_i$, $y_i$) for ($x$,$y$) (or a product of $d_i$ for $x$). The result is a nonlinear system (last time I tried this I failed to make it work in practice).
EDIT: Related search seems "thue mahler" equation'
Related: unboundedness of number of integral points on elliptic curves?
Crossposted on MATH.SE: https://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n
 A: With the transformation $X = -n/x$ and $Y= ny/x$, the curve becomes isomorphic to the Weierstrass model
$$ E_n\colon \  \ Y^2 - X\ Y - n\ Y = X^3.$$
The points in question are exactly the integral points in $E_n(\mathbb{Q})$ such that $X$ divides $n$. I do not see why the number of these points should be bounded independently of $n$; so my guess is that there is no bound and that it is going to be difficult to show this.
The curve $E_n$ has always two rational 3-torsion points $(0,0)$ and $(0,n)$. Unless $n$ is of the form $k\cdot (\tfrac{k-1}{2})^2$ for some integer $k\not\equiv 2\pmod{4}$, these are all the torsion points in $E_n(\mathbb{Q})$, otherwise there are 6 torsion points defined over $\mathbb{Q}$. Hence, if $n$ is not of the above form, then any integral point with $X$ dividing $n$ will be of infinite order and hence the rank will be at least $1$.
(Edit:)
Now, I have a reason to believe that the number is bounded. As pointed out by Felipe Voloch in
this question, the paper by Abramovich shows that:
if the conjecture by Lang and Vojta about rational poitns on varieties of general type holds, then the number of solutions is bounded as $n$ varies.
One has just to note that the equation $E_n$ is in fact minimal and that the curve $E_n$ is semistable for all $n$. For all primes $p$ dividing $n$, the curve has split multiplicative reduction with $3\cdot \text{ord}_p(n)$ components. For all primes $p$ dividing $27n+1$, the reduction can be shown to be multiplicative, as well.
Maybe a descent via three-isogeny could help to give an upper bound on the rank.
