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

10 Béla Zay which completes the proof of the lemma. Proof of Lemma 1. It is well known [see in [6] p. 317] that in the representation of y/~D as a simple conntinued fraction, the sequence ai, 02,..., a

s-iis symmetric, that is a

l = a

s-i, for every i (1 < i < s — 1) and (17) a

s = 2a

0. If ^ is the n

t

h convergent of (a\

, 02,. ..) = (ai

, a

2,.. . , a

s) then /i_ 1 =1, ho = a

u h

n = a

n

+i/i

n_i + /i„_2, n > 1 and &-1 — 0, ^o = 1? ^n — an+l^n-l + &n-2? n > 1. Using this last definition and (10) by Lemma 6. we obtain, that k

s-2 = /

s(a

2,

... ,a

s_ i) = /

s(ai,...,a

s_

2) = It is known (and it is easy to see by induction for n) that (19) K

n = h

n.

u n> 0 and (20) H

n = a

0h

n_I + , n > 0. By (19), (12), (17), (18) and (20) h

s_i + k

s-2 = + /c

r_2 = a

5A'

s_i + K

s2 + =2a

0A'

s_i + 2k

s-2 = 2(a

0h

s„2 -f = 2i7

s. Using this equation we obtain (7) and (8) from (4) and (5) respectively. Thus the theorem is proved. To proofs of the Theorem 1., Theorem 2., Theorem 3. and Theorem 4. we use the Lemma 4. and the representation of \[T) as a simple continued fraction: Lemma 7. Let k be a rational integer. Then (21) y/(2k + l)

2 - 4 = (2 k,l,k - 1,2, k - 1,1,4/?) fork >2. (21) y/(2

k)

2 - 4 = (2k - 1,1, k - 2,1, 4k - 2) for k > (23) \A

2 — 1 = (Ar — 1,1, 2A: — 2) fork >2, (24) y/k

2 + 1 = (Jfc, 2Ä

7) for fc > 1.