(i.e. Δ = ru – st = ±1) and a quadratic surd ξ = (a + √d)/c, where c divides d – a2 and produces η = (rξ + s)/(tξ + u).
Then ξ and η will have regular continued fraction expansions which ultimately agree. Similarly for the nearest integer and nearest square continued fraction expansions.
If ξ=(a + √d)/c, then η = (A + √d)/C, where
A = Δ(brt + a(ru + st) + cus), C = Δ(bt2 + 2atu + cu2) and b = (a2 – d)/c.
To prove 2, suppose p is a prime dividing A,B and C. Then
brt + a(ru + st) + cus ≡ 0 (mod p)
br2 + 2ars + cs2 ≡ 0 (mod p)
bt2 + 2atu + cu2≡ 0 (mod p).
= -Δ, so p divides a,b and c.This is a BCMATH translation of a BC program.
Last modified 29th October 2009
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