Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1998. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 25)

ZAY , B., An application of the continued fractions for ... in solving some types of Pell's equations

An application of the continued fractions for \FD... 7 The first 2s terms of sequences (H

n

) and (Ä'

n) can be got from the following well known relations (9) and (10) H-m — aTTi H

M-I + //

m_2 , H-L =1, H

0 = CLq , Km = a

m/t

m_i + Km1, A'_1 - 0, A'o = 1, for any m > 0. The following algorithm for representing the number y/D as a simple continued fraction is well known (see in [6], p. 319): We set a

0 — yfD |, L j bi = a

0, C\ — D — al and we find the numbers a

n_!

, b

n and c

n successively using the formulae fln-l 00 + Cn-1 Now consider the sequence 5 ^n — ^n-l^n-l ^n — li Cn — D-bl Cn1 (h ,c

2), (^3,C

3), (64, C

4),... and find the smallest index s for which b

s

+\ = bi and c

s

+i = c\

. Then the representation of \pD as a simple continued fraction is \f~D = (a

0,ai,a

2,...,a

s). We shall use two other results from [5] (pp 158-159). Lemma 2. If D is a positiv integer, not a perfect square, then H* — DK\ = (-l)

n1c

n

+i for all integer n > -1. Lemma 3. Let D be a positive integer not a perfect square, and let the convergents to the continued fraction expansion of y/D be H

n/K

n. Let N be an integer for which |TV

[ < D

. Then any positive solution x — u. y = t of x

2 — Dy

2 = N with (u

, t) = 1 satisfies u = H

n,t — K

n, for some positive integer n. Recalling that c

n = c

n+

s in the Lemma 2., we can formulate Lemma 4. which is a consequence of the first three lemmas. Lemma 4. Let D be a positive integer not a perfect square, and let Vd = (a

0,ai,

... ,a

s)