Can I borrow some ECM of a friendly CPU?

[Dr Sardonicus]

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 non-negative 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 d-th numerator N and denominator D will satisfy N² - (n^2 + 4)*D² = ±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 |N² - (n^2 + 4)*D²| > 1.

[fatphil]

Originally Posted by Dr Sardonicus

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.

Thanks for this, I'm a bit rusty, and it will take several re-reads to sink in. I wish I had the mastery of Pari/GP I used to, so I could play around with these 2-dimensional objects - I'd completely forgotten the Mod(..., x^2-d) trick despite it being my go-to tool a decade back!
The 23 was incidentally completely arbitrary, which is why I immediately put it to one side and looked at smaller discriminants instead.