Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:45:53Z http://mathoverflow.net/feeds/question/50479 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50479/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-z-be-unbounded-as-n-var Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? jerr18 2010-12-27T08:33:46Z 2011-01-07T19:25:18Z <p>Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies?</p> <p>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).</p> <p>If yes, will large number of solutions give moderate rank EC?</p> <p>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.</p> <p>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.</p> <p>What I find intriguing is that <strong>all integral points</strong> in this model come from factorization/divisors only.</p> <p><strike> Current record is n=<strong>179071200</strong> with 22 solutions with positive x,y. Due to Matthew Conroy.</p> <p>Current record is n=<strong>391287046550400</strong> with 26 solutions with positive x,y. Due to Aaron Meyerowitz</p> <p>Current record is n=<strong>8659883232000</strong> with 28 solutions with positive x,y. Found by Tapio Rajala. </strike></p> <p>Current record is n=<strong>2597882099904000</strong> with 36 solutions with positive x,y. Found by Tapio Rajala.</p> <p>EDIT: $ab(a+b+9)=195643523275200$ has 48 positive integer points. – Aaron Meyerowitz (<em>note this is a different curve and 7 &lt;= rank &lt;= 13</em>)</p> <p>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.</p> <p>Is it possible some relation in the primes or primes or divisors of certain form to produce records: Someone is trying in $\mathbb{Z}[t]$ <a href="http://mathoverflow.net/questions/51193/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-zt-be-unbounded-as-n" rel="nofollow">Can the number of solutions xy(x−y−1)=n for x,y,n∈Z[t] be unbounded as n varies?</a> ? Read an article I didn't quite understand about maximizing the Selmer rank by chosing the primes carefully.</p> <p>EDIT: The curve was chosen at random just to give a clear computational challenge.</p> <p>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).</p> <p>EDIT: Related search seems <strong>"thue mahler" equation'</strong></p> <p>Related: <a href="http://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves" rel="nofollow">unboundedness of number of integral points on elliptic curves?</a></p> <p>Crossposted on MATH.SE: <a href="http://math.stackexchange.com/questions/14932/can-the-number-of-solutions-xyx-y-1-n-for-x-y-n-in-z-be-unbounded-as-n" rel="nofollow">http://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></p> http://mathoverflow.net/questions/50479/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-z-be-unbounded-as-n-var/50485#50485 Answer by Aaron Meyerowitz for Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? Aaron Meyerowitz 2010-12-27T10:31:35Z 2011-01-01T08:30:11Z <p>$n=938995200$ 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). </p> <p>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. </p> <p><strong>later</strong> The following sounded plausible but does not turn out to work that well</p> <p>This suggests seeking $n$ from the <a href="http://oeis.org/A002182" rel="nofollow">highly composite numbers</a> (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 <a href="http://oeis.org/A002182/b002182.txt" rel="nofollow">longer list</a> linked there and only found two points for $j=1$. I did not look at other $j$. </p> <p><strong>Continued</strong> 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 <code>$1.7 \, 10^{28}$</code>(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.</p> <p>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 $|j|&lt;50$</p> <p><strong>even later THEN UPDATED</strong> Consider these four integers</p> <p><code>$$\begin{eqnarray} 2888071057872000=&amp;&amp;2^7\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot 31\\ 8659883232000 =&amp;&amp;2^8\cdot 3^3\cdot 5^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 31\\ 32607253879200=&amp;&amp;2^{5}\cdot3^{3}\cdot\cdot5^{2}\cdot7^{2}\cdot11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\\ 1248124550400=&amp;&amp;2^{8}\cdot 3^{3}\cdot 5^{2}\cdot 7^{2}\cdot 13\cdot 17\cdot 23\cdot 29 \end{eqnarray}$$</code></p> <p>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.</p> http://mathoverflow.net/questions/50479/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-z-be-unbounded-as-n-var/50652#50652 Answer by Chris Wuthrich for Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? Chris Wuthrich 2010-12-29T13:01:14Z 2010-12-30T10:21:57Z <p>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.</p> <p>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 $1$. </p> <p>(Edit:) Now, I have a reason to believe that the number <strong>is</strong> bounded. As pointed out by Felipe Voloch in <a href="http://mathoverflow.net/questions/50661/unboundedness-of-number-of-integral-points-on-elliptic-curves" rel="nofollow">this question</a>, the <a href="http://www.springerlink.com/content/xgbvqxm4383nwhqu" rel="nofollow">paper</a> by Abramovich shows that:</p> <p><em>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.</em> </p> <p>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.</p> <p>Maybe a descent via three-isogeny could help to give an upper bound on the rank.</p> http://mathoverflow.net/questions/50479/can-the-number-of-solutions-xyx-y-1n-for-x-y-n-in-z-be-unbounded-as-n-var/50863#50863 Answer by Tapio Rajala for Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z$ be unbounded as n varies? Tapio Rajala 2011-01-01T15:35:12Z 2011-01-01T18:07:41Z <p><em>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.</em></p> <h2>Code</h2> <p>You can find the (non-optimal) C-code I wrote under <a href="http://users.jyu.fi/~tamaraja/temp/solu.c" rel="nofollow">my webpages</a>. 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.</p> <p>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.</p> <h2>Results</h2> <p>Here are some values found using this program (in roughly 4 hours).</p> <h3>36 positive solutions</h3> <p>$$n = 2597882099904000 = 2^9 · 3^3 · 5^3 · 7 · 13 · 17 · 23 · 29 · 31 · 47$$</p> <h3>30 positive solutions</h3> <p>$$n = 34747990981704000 = 2^6 · 3^4 · 5^3 · 7^2 · 11 · 13 · 17 · 19^2 · 29 · 43$$</p> <h3>28 positive solutions</h3> <p>$$n = 105140926800 = 2^4 · 3^3 · 5^2 · 7^2 · 13 · 17 · 29 · 31$$ $$n = 8659883232000 = 2^8 · 3^3 · 5^3 · 7 · 11 · 13 · 17 · 19 · 31$$ $$n = 3783439308448800 = 2^5 · 3^4 · 5^2 · 7^3 · 11 · 13 · 19 · 31 · 43 · 47$$ $$n = 9928464968822400 = 2^7 · 3^4 · 5^2 · 7^2 · 11^2 · 13 · 17 · 23 · 31 · 41$$ $$n = 18680310941292000 = 2^5 · 3^4 · 5^3 · 7 · 11^2 · 13^2 · 17 · 19 · 29 · 43$$ $$n = 88550619849291600 = 2^4 · 3^5 · 5^2 · 7^2 · 11^2 · 13 · 17 · 19 · 23 · 37 · 43$$</p> <p><em>Note that I did not check the results after my program handed them to me...</em></p> <p><strong>Edit:</strong> 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.</p>