20220830, 12:35  #23 
Feb 2017
Nowhere
6224_{10} Posts 
As to the numerators N and denominators D for sqrt(23), the only ones with obvious divisibility properties are the ones with N^2  23*D^2 = 1. If
Mod(F_{k}*x + L_{k}, x^2  23) = Mod(5*x + 24, x^2  23)^k then N_{4k} = L_{k} and D_{4k} = F_{k}. The F and L sequences have divisibility properties similar to those of the Fibonacci and Lucas numbers, respectively. The SCF for sqrt(n^2 + 1) has partial quotients n, 2n, 2n, ... [period has length 1]. so the numerators and denominators are the solutions to N^2  (n^2 + 1)*D^2 = ±1. We have D_{k} = F_{k}, N_{k} = L_{k}, where Mod(F_{k}*x + L_{k}, x^2  (n^2 + 1)) = Mod(n + x, x^2  (n^2 + 1))^k for nonnegative integer k. With the single exception n = 2, Mod(n + x, x^2  (n^2 + 1)) is the fundamental unit for Q(Mod(x, x^2  (n^2 + 1)) [i.e. Q(sqrt(n^2 + 1))]. So with the single exception n = 2, the numerators and denominators will have divisibility properties similar to those of the Lucas and Fibonacci numbers. With the SCF for sqrt(n^2 + 4), however, we see something similar to what happens with sqrt(5). [Note that 5 is the only positive integer which is simultaneously of the forms n^2 + 1 and n^2 + 4.] For n > 2, the sequence of partial quotients n, ... 2n will have period d > 1. Every dth numerator N and denominator D will satisfy N^{2}  (n^2 + 4)*D^{2} = ±1. But in this case, the fundamental unit is Mod((x+n)/2, x^2  (n^2 + 4)), and we have Mod(D_{dk}*x + N_{dk} , x^2  (n^2 + 4)) = Mod((F_{3k}*x + L_{3k})/2, x^2  (n^2 + 4)) similar to the case n = 2. I was unable to discern any divisibility properties of the numerators N and denominators D for which N^{2}  (n^2 + 4)*D^{2} > 1. Last fiddled with by Dr Sardonicus on 20220830 at 23:13 Reason: xinfig topsy 
20220901, 14:43  #24  
May 2003
FF_{16} Posts 
Quote:
The 23 was incidentally completely arbitrary, which is why I immediately put it to one side and looked at smaller discriminants instead. 

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