Kurt Foster on Tue, 19 Dec 2023 01:27:57 +0100 |
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Re: Pell's equations and beyond |
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.