Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in\mathbb Z$ be unbounded as $n$ varies?

Question

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? ? 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?

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

Answer 1

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.

Answer 2

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.

Answer 3

$n=938995200$ also has $22$ solutions. It might be nicer to put $a=y$ and $b=x-y-1$ so $x=a+b+1$ and one has $ab(a+b+1)=n$. For $n=391287046550400$ there are $26$ solutions with $a,b>0$ which grows to $78$ if one allows negative values and shrinks to $13$ if $[a,b]$ and $[b,a]$ are considered the same (and must be positive).

It would seem reasonable that if one chose a number $n$ with "lots" of factors relative to the size of $n$ then any of the curves $ab(a+b+j)=n$ with a "small" j would have a fair number of points and at least some of them would have a large number of points (so one could start with $n$ and look for the most fruitful $j$). That could be made more precise (at least with regard to expectation), maybe not by me though.

later The following sounded plausible but does not turn out to work that well

This suggests seeking $n$ from the highly composite numbers (more divisors than any smaller number.) In fact $391287046550400$ is on that list! (Although I did not know that when I found it) However I tried $n=106858629141264000$ from further down the longer list linked there and only found two points for $j=1$. I did not look at other $j$.

Continued I found $391287046550400$ by looking for products $ab(a+b+1)=n$ with all prime factors under 30 (and no prime over 7 repeated in $n$), and looking for $n$ which turned up frequently. Then I decided to look at the hcn and found that $n$ on the list. However it appears that up to about $1.7 \, 10^{28}$(which is something like the first 260 such ) the appropriate curve has 26 positive points in that one case, 14 in another, and 12 and 10 just a handful of times.

Among those $n$ values, the curve $ab(a+b-7)=481880599200$ has $28$ positive integer points and the curve $ab(a+b+9)=195643523275200$ has $48$ positive integer points but those are the only ones $ab(a+b+j)=n$ which better $ab(a+b+1)=391287046550400$ with a smaller $n$ and $|j|<50$

even later THEN UPDATED Consider these four integers

$$\begin{eqnarray} 2888071057872000=&&2^7\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\\ 8659883232000 =&&2^8\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 31\\ 32607253879200=&&2^{5}\cdot3^{3}\cdot\cdot5^{2}\cdot7^{2}\cdot11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\\ 1248124550400=&&2^{8}\cdot 3^{3}\cdot 5^{2}\cdot 7^{2}\cdot 13\cdot 17\cdot 23\cdot 29 \end{eqnarray}$$

The first has $32,768$ factors which makes it a hcn since every smaller integer has fewer. The second is the excellent value of $n$ found by tapio which makes $ab(a+b+1)=n$ have 28 positive solutions. The third is also a hcn and the fourth makes $ab(a+b-1)=n$ have 28 positive solutions (or if you prefer, $xy*(x-y+1)=n$ an equation which seems as hard as the chosen one). That is where I would look for similar examples, n a hcn maybe modified by putting in or taking out a couple of large primes and fiddling with the exponent of the smallest primes. Brute calculations will not prove anything of course.