Re: Pell's equations and beyond

[Kurt Foster <drsardonicus@earthlink.net>]

On Dec 18, 2023, at 12:31 PM, hermann@stamm-wilbrandt.de wrote:

<snip>
> 1) How do these commands from Karim's posting reveal x and y?
>
> ? quadunit(61)
> %15 = 17 + 5*w
<snip>

If D > 0 is square free and D == 1 (mod 4), then w = (1 + sqrt(D))/2.   
So if the coefficient of w in the quadunit u is odd, then u is not in  
the order Z[w], which I'll call R.

This can't happen if D == 1 (mod 8); it can only happen if D == 5 (mod  
8).  But sometimes even then (e.g. D = 37, 101, 141) the unit is in  
the order Z[sqrt(D)].

The reason it can't happen when D == 1 (mod 8) is because 2R splits  
into two distinct prime ideal factors P and P' , with the residue  
rings R/P and R/P' both copies of the finite field with 2 elements.

Since u isn't in either P or P', u-1 is in both P and P' so that u-1  
is in 2R, hence u is in Z[sqrt(D)].

If D is square free and D == 5 (mod 8) then 2R = P is a prime ideal,  
and the residue ring R/P is the finite field with 4 elements.  The  
multiplicative group of nonzero elements in R/P is cyclic of order 3.

So if D == 5 (mod 8) then either u is in Z[sqrt(D)] or it is not, in  
which case u^3 is the smallest power of u in Z[sqrt(D)].

Now u^3 has the same norm as u.  So if u is not in Z[sqrt(D)] and  
norm(u) = +1, then u^3 is in Z[sqrt(D)] and has norm +1.

But if u is not in Z[sqrt(D)] and norm(u) = -1, the smallest power of  
u in Z[sqrt(D)] with norm 1 is u^6.

If D is a prime and D == 1 (mod 4) then norm(u) = -1.  If D is  
composite (and squarefree), D == 1 (mod 4) and D  has any factors  
congruent to 3 mod 4, then norm(u) = +1.

If D is composite, and all its prime factors are congruent to 1 (mod  
4), I don't know how to tell in general whether norm(u) is +1 or -1  
other than by calculating u.

If D == 5 (mod 8), I don't know how to tell whether u is in Z[sqrt(D)]  
other than by calculating u.