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. ---------- 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