Can the Number of Solutions $xy(x-y-1)=n$ be Unbounded?

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'

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

Best Answer

Although this is not a mathematical answer I will put the results of my brute force search as an aswer as requested by jerr18. I didn't get anywhere with the thinking part.

Code

You can find the (non-optimal) C-code I wrote under my webpages. The biggest limitation with the program is that it uses 64-bit integers. Feel free to run, test, tweak and/or mutilate the code as you wish.

The program constructs first $n$ with a recursion and then $y$ with a recursion (this way I avoid considering values of $y$ that don't divide $n$). Finally it checks if the positive solution $x$ to the equation $$xy(x-y-1) = n$$ is an integer.

Results

Here are some values found using this program (in roughly 4 hours).

36 positive solutions

$$n = 2597882099904000 = 2^9 · 3^3 · 5^3 · 7 · 13 · 17 · 23 · 29 · 31 · 47$$

30 positive solutions

$$ n = 34747990981704000 = 2^6 · 3^4 · 5^3 · 7^2 · 11 · 13 · 17 · 19^2 · 29 · 43 $$

28 positive solutions

$$n = 105140926800 = 2^4 · 3^3 · 5^2 · 7^2 · 13 · 17 · 29 · 31 $$ $$n = 8659883232000 = 2^8 · 3^3 · 5^3 · 7 · 11 · 13 · 17 · 19 · 31 $$ $$n = 3783439308448800 = 2^5 · 3^4 · 5^2 · 7^3 · 11 · 13 · 19 · 31 · 43 · 47$$ $$n = 9928464968822400 = 2^7 · 3^4 · 5^2 · 7^2 · 11^2 · 13 · 17 · 23 · 31 · 41$$ $$n = 18680310941292000 = 2^5 · 3^4 · 5^3 · 7 · 11^2 · 13^2 · 17 · 19 · 29 · 43$$ $$n = 88550619849291600 = 2^4 · 3^5 · 5^2 · 7^2 · 11^2 · 13 · 17 · 19 · 23 · 37 · 43$$

Note that I did not check the results after my program handed them to me...

Edit: Just out of curiosity I tried also with $+1$ instead of $-1$. For example the equation $$ xy(x-y+1) = 388778796252000 = 2^5 · 3^3 · 5^3 · 7^2 · 11 · 17 · 19 · 23 · 29 · 31$$ has 38 positive solutions.