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... 9 Using the definitions and the last assumptions we can finish the proof by induction for n: fn+ 2 (^1 , • • • 5 xn) = xnfn+

1 {

xl ,••••> xn-l) + fn{

xl xn~2) ~

xn9n+1 (^1 5 • • • > xn-l) + 9ti (#1 5 • • • , xn-2 ) = xn

xl9n{

x2 , • • • •> xn-l) + xn9n1 {

x3, • • • i

xn1 ) ~\~

xl9n — l(

x2 •> • • • , ^n-2) + 9n-2(

x3, • • • ,Z

n_

2) = (zn/n(z2, • • • , ^

n-l) + fn-l{

x2,- • - ,

xn-2)) + (^n/n-l (^3, • • • 5 Zn-l) + ín-2 (^3 5 • • • , xn-2)) — x\ fn+

1 {

x2 1 ' • • 1 xn —I 1 xn) "i" fn(

x3 1 ••'•> xn — \ 1 xn) +xig

n

+i(x

2,...,x

n) + g

n(x

3,

... ,x

n) = g

n+

2{

xi,. • •, x

n). Lemma 6. If X{ = x

n+

2-i holds for every i (1 < i < n -f 1) then fn+2(^11 • • •

t

xn) — fn+2{

x2, • • • ,

xn +

1 ) is also vaUd for every integer n (n > —1). Proof. This is evident for — 1 < n < 2, because /1 = 0, /2 = 1, /3(^1 ) = xl = x

2 - fz(

x2 ) and /4(^1,^2) = X2

X\ + 1 = 2:32:2 + 1 = /4(2:2,2:3) (since xi = x

3). Let n > 2. Assume that if yi = y

n-i holds for every i (1 < í < n — 1) then /n(l/l , • • • , Vn2) = /n(2/2, • • • , 2/n-l) is also valid. Let = x

I

+i for every i (1 < i < n — 1). Then Vi — xi+i = 2:

n+2-( j+i) = 2;

n_

t+

1 = and so In{

x2 1 • • • 1 xn-l) = fn{

x3 1 • • • 1 xn)Using this equation, Lemma 5. and the relation X\ = a:

n

+i we obtain, that /n+2(^1, • • • 1 xn) = 9n + 2 {

xl,...,x

n) = X

lg

n+i (x

2 , . . . , X

n) + 9n{

x3 , • • • , In) = X n-)_ 1 f

n+1 ( x 2 , • • • 1 x n ) + /71(2:3, • • • » xn) ~

xn+l fn+1 (2:2 > • • • 5 xn) + fn(

x2 , • • • , 2:

n-l ) — fn + 2(2:2 5 • • • , 2:

n + l )