Az Egri Ho Si Minh Tanárképző Főiskola Tud. Közleményei. 1984. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 17)
II. TANULMÁNYOK A TERMÉSZETTUDOMÁNYOK KÖRÉBŐL - Kiss Péter: Feli egyenletek megoldása lineáris rekurzív sorozatok segítségével
has integer (x;y) solutions, then all solutions are obtained by (x;y) = = (± (Gf 2 n + aG 2n+ 1) ;±G 2n+ 1), n = 0, 1, 2, ..., using finitely many linear recurrences G(2a, —1, G 0, G^. These sequences satisfy the conditions 0 G 1 < 2a j/N or 0 s Gj < (2a 2 +1) f- N/(a 2+ l) according as N > 0 or N < 0. THEOREM 2. Let N 0) and a ( > 2) be integers. If the equation x 2 — (a 2 — 4)y 2 = 4N has integer (x ;y) solutions, then all solutions are obtained by (x;y) = (±H n; + G n), n = 0, 1, 2, ..., using finitely many linear recurrences G(a, 1, G 0, GjJ and their associate sequences H. These sequences satisfy the conditions 0 ^ Gj, < |/N or 0 g G x < a / —N/(a 2— 4) according as N > 0 or N < 0. THEOREM 3. Let N (^ 0) and a (> 0) be integers. If the equation x 2 — (a 2 + 4)y 2 = 4N has integer (x;y) solutions, then all solutions are obtained by (x;y) = = (±H 2 n; + G 2 n), n = 0, 1, 2, ..., using finitely many linear recurrences G(a, — 1, G 0, Gj) and their associate sequences H. These sequences satisfy the conditions 0 ^ G 0 < a /N or 0 ^ G 0 < (a 2+ 2) / — N/(a 2 + 4) according as N > 0 or N < 0. THEOREM 4. If the pairs (x;y) = ( + H 2 n; + G 2 n), n = 0, 1, 2, . . ., are solutions of the equation x 2—(a 2 + 4)y 2 = 4N, then the numbers (x;y) = = (±H 2n+ 1; + G 2n+ 1), n = 0, 1, 2, ..., are solutions of the equation x 2— (a 2 + 4)y 2 = — 4N. The proofs of the theorems are based on the following two results : 1. If (A, —1, G 0, G x) is a linear recurrence and k = G 0 2 -f AGoGj 2 — Gj 2, then G„ 2 + A.G n.G n+ 1-G 2 + 1 = (-l)".k for every integer n ^ 0. 2. Let G = G(A, B, G 0, G,) be a linear recurrence and let D = A 2 — 4B ^ 0. If H is the associate sequence of G, then H n 2 —D .G n 2 = 4bc-B n for every integer n ^ 0, where b and c are defined in (i). 824