On Sun, May 11, 2014 at 1:59 PM, Bill Gosper <billgosper@gmail.com> wrote:
All I found was a slow but amusing Mma solution. E.g. changing 5 to 14, In[838]:= Intersection @@ Transpose[ Reap[NestWhile[1/(2*Floor[#] + 1 - #) &,1/3, (If[NumberQ[#[[1]]], Sow[#]]; True) &[{1, #}/Sqrt[(#^2 - 1)/14]] &, 1, 9999]][[2, 1]]]
Out[838]= {-((65 I)/12), 65/12}
Mystery: Why doesn't this find 337/120 for 7? --rwg NeilB just privately sent a big improvement to the search technique, plus some interesting results. I'll leave it up to him to post it here.
As Rich points out, this is exactly the congruent number problem as formulated in http://oeis.org/A003273 (as opposed to the triangle area formulation in http://en.wikipedia.org/wiki/Congruent_number. I completely forgot reading of the equivalence in Koblitz until eating fish just now. (Rosemary was insufficient.)) --rwg Back in Sep 2009 Bernie Cosell posted here the link http://www.eurekalert.org/pub_releases/2009-09/aiom-att091809.php apparently claiming the tabulation of the first trillion congruent numbers, but I can't determine whether or not they assumed Birch & Swinnerton-Dyer, and if not what they did instead. And whether they did a full trillion squarefree.
---------- Forwarded message ---------- From: Neil Bickford <techie314@gmail.com> Date: Thu, May 8, 2014 at 10:01 PM Subject: Question for tomorrow To: Bill Gosper <billgosper@gmail.com>
I came across this problem in a recreational mathematics book allegedly by H.E. Licks (apparently the I.J. Matrix of the late 19th century), who attributes it to Fibonacci- specifically, that Fibonacci solved it in a mathematical competition of the time:
Find a rational number n such that: n^2+5 is a square of a rational, and n^2-5 is a square of a rational.
I've been working on this problem for some time, and while it's almost trivial to find a solution by brute force, I haven't been able to find an elegant solution. So far, the best observation I've been able to make is that it's Pellish.
Any thoughts?
--Neil