Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in\mathbb Z$ 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 $7 \le \text{rank} \le 13$*)


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 $x,y,n\in\mathbb Z[t]$ be unbounded as $n$ varies?](https://mathoverflow.net/questions/51193/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-zt-be-unbounded-as-n) ? 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 $n=d\_1 d\_2 \cdots d\_k$ 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?](https://mathoverflow.net/questions/50661/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>



Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in\mathbb Z$ be unbounded as $n$ varies?
*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
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You can find the (non-optimal) C-code I wrote under [my webpages](http://users.jyu.fi/~tamaraja/temp/solu.c). 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
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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.


