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 (c) If N = 1 - 2k then M = {(»,»): x = (a - l)a

3m+

1 - (ß - l)3*

m+l 2(a — ß

) (d) If 1 < \N\ < 2k, N ^ 1 - 2k and N isn't a square of a natural number then M = 0 (empty set). Theorem 2. Let k (k > 3) be a natural number and D = (2k)

2 — 4. Let a and ß denote the zeros of f

2(x) = x

2 - 2kx + 1 with a > ß. Denote by M the set of positive (x, y) solutions of x

2 - Dy

2 — N

. (a) If N = 4/

2 and / (1 < I < y^f

1) is a natural number then ( a

m - 3

m } M = {(x,y):x = l(a

m+ß

m), y = I —m> 1 j . (b) If N = (21 - l)

2 and I is a natural number (l < / < \ + yjk

2 - l) then , , 1\ Q

2

m- ß

2

m (c) If 1 < |A

r| < and A isn't a square of a natural number then M = 0. Theorem 3. Let k (k > 2) be a natural number and D = k

2 - 1. Let a and 3 denote the zeros of fo(x) = x

2 — 2kx + 1 where a > ß. Denote by M the set of positive solutions of x

2 — Dy

2 = N

. (a) If N = I

2 and 1 < I < then { I I (a

n+

l - 3

n+

l) 1 M= <(x,y):x = -(a*+

l+ß

n+

1), y= [ J m> 1 . (b) If 1 < \N j < 2k — 1 and N isn't a square of a natural number then M = 0.