# 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'

• If one drops $-1$, which answer is yes? The unboundedness of the number of solutions? or the large number of solutions giving moderate rank curves? (or both?). Is there any advantage to studying the $-1$ case rather than (or in addition to) the case without? Would $xy(x-y-17)=n$ (say) be just as interesting? – Gerry Myerson Dec 27 '10 at 16:53
• The unboundedness is "yes". The construction uses multiples of a point to find integer solutions, so the rank depend on the original'' $n$. – jerr18 Dec 27 '10 at 17:27
• Re: advantages. No, the curve was chosen pseudorandomly. I gave a single curve to define the computational challenge better. – jerr18 Dec 27 '10 at 17:50
• Gerry Myerson the new edit hopefully addresses your comments. – jerr18 Dec 28 '10 at 8:01
• According to Magma the minimal Weierstrass model of the 26 pts curve is: [ 1, 0, 0, 195643523275200, 38276388199533724134935040000 ]. Positive x,y give $6$ for lower bound of the rank. "Mwrank" abort()s even with the -p switch. – jerr18 Dec 28 '10 at 14:01

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.

• I confirmed your 36 record, good catch. – jerr18 Jan 1 '11 at 16:40
• The 36 record curve is of rank at least 7. – jerr18 Jan 4 '11 at 11:49

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.

• If the $-1$ is dropped in the original question, one find that the $-nX$ is dropped in the above Weierstrass question and so the elliptic surface is constant. Here it is a non-isotrivial elliptic surface of rank $0$. – Chris Wuthrich Dec 29 '10 at 13:04
• Thank you. Are there other types of elliptic curves with easy to find integral points? (I lack to knowledge to understand your comment, never mind) – jerr18 Dec 29 '10 at 14:08
• If you dropped $-1$, you find that the elliptic curves you are considering are isomorphic for varying $n$. So finding it on one will give you the answers on all others. Not so for the family with the $-1$ as there the equation $E_n$ depends really on $n$. For any given elliptic curve $E/\mathbb{Q}$ it is "easy" to find the integral points, i.e. computer software (magma, sage,...) can do it "quickly". It is much harder do say anything in families of curves. – Chris Wuthrich Dec 29 '10 at 14:46
• <"can do it "quickly"> Hm, are you sure about this? The points in this question come from divisors only. mwrank failed to find a single point on a record curve.... – jerr18 Dec 29 '10 at 14:54
• Sure, that is why I put " " around it. By "easy" I mean that there is an implementation which in principle can do it for reasonably sized coefficients quite quickly. From such a curve, you could construct other examples like your $xy(x-y)=n$. Your question is much harder as it would require theoretical input, say if you wish to decide if such a bound existed or whether it is linked to the rank of the curve. – Chris Wuthrich Dec 29 '10 at 15:09

$n=938995200$ 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.

• $ab(a+b+9)=195643523275200$ has $48$ positive integer points. – Aaron Meyerowitz Dec 28 '10 at 8:32
• I wrote a small program that searches for numbers $n$ which are products of small primes and which have lots of solutions for the desired equation. So far (after about 20 min running time on my slow laptop) it has found out that $$xy(x−y−1) = 8659883232000 = 2^8*3^3*5^3*7*11*13*17*19*31$$ has 28 positive integer solutions. However, I don't think it will find a 100-solution number in a reasonable time without some (clever) optimization - currently it does the search by brute force. – Tapio Rajala Dec 29 '10 at 20:15
• I modified the program so that it searches for solutions in equations of the form $xy(x-y-k)=n$ where $0 < k < 200$. Here are some numbers I found which might help you figure out where to find the rich $n$s: $$xy(x-y-156)=4503303604800 = 2^6 · 3^6 · 5^2 · 7^2 · 11 · 13 · 19 · 29 \quad (50 \text { solutions})$$ $$xy(x-y-77)=894776186304000 = 2^9 · 3^6 · 5^3 · 7^3 · 11 · 13 · 17 · 23 \quad (40 \text { solutions})$$ $$xy(x-y-1)=61448587200 = 2^6 · 3^3 · 5^2 · 7^3 · 11 · 13 · 29 \quad (26 \text { solutions})$$ It seems that the 28-solution $n$ in my previous post was a lucky hit. – Tapio Rajala Dec 30 '10 at 8:38
• The number of solutions of $xy(x-y-a)=n$ is unbounded. ($x$,$y$) -> ($dx,$dy) is on $xy(x-y-a d)=n d^3$ Find many rational points, multiply by suitable $d$ to make them integer. The original question included $-1$ on purpose. – jerr18 Dec 30 '10 at 12:32
• I updated my answer to include a 28 point curve for $xy(x-y+1)=n$. That seems just as hard as the $xy(x-y-1)=n$ variant. My program completely bogs down at 41 and did not turn up anything great for 37. So computed records may await a better idea computer or program. – Aaron Meyerowitz Jan 1 '11 at 8:43