Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1993. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 21)

Aleksander Grelak: On The Equation (x2 - 1 )(y2 - 1) = z2

(3) x

2-l = dr, y

2 - ds\ (/%s) = l By (3) it follows that (4) (x

2-\)(y

2-\) = d

2rs. From (1) and (4) we have r -r

2, s = s

2 and consequently (5) x

2 -l = dr

2 =D(ur

x)

2 , y

2 1 -ds

2 = D(us

x)

2 what proves first part of our Lemma. Now, let < A

i,B

i > and < A

}

,B

} > denote arbitrary solutions of the equation A

2 ~DB

2 = 1 with squarefree D. Then we have A

2-\ = DB

2 and A

2-\ = DB

2. Hence (A

2 -I)(A

2 -l) = (DB

jB

J)\ Putting z = DB

xB

}

> x = A, and y = A

. we get second part of our Lemma and the proof is complete. Proof of the Theorem By well-known formulae from the theory of Pell's equation and our Lemma it follows that 1 (6) 2 1 U+v^J+U-^J {A,+yfDB

x)

} +{A

x-jDBy From (6) and (1) we obtain z = DB

1B

] and z = (A

x + JDB

x )' - [A, - -JDB, ) [A, + JDB,- (A, - y[DB

x y and the proof of our Theorem is complete. 93