Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1995-1996. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 23)

LlPTAI, K., An approximation problem concerning linear recurrences

14 Kálmán Liptai and \x\ < |Ä

n|. Using the equation of the plane (9) (a + ß)x - 2y + 2 = 0 we get the following equalities VDI = |(a — ß)x — z\ — I [ax — (

ßx + z)\ = \ax — (2 y — ax)\ = 2 — y | and (11) y/Dyaz = I (a - ß)y - (2 ay - a(a + ß)x)\ = \a + ß\ \ax - y\ . Thus, from (1), (5), (6), (8), (10) and (11) we obtain the inequality az - yI < \ß\ d x, y,z A

2 + 5 1 + a

2 +L> and so using |x| < |Ä

n| and (1), we get A

2 + 5 y a x < M 1 + a

2 + D ' l~(ß/a)

n 1 -(ß/a) < a From this, using the mentioned theorem of P. Kiss and its proof, we obtain x — R{, y = R{+

1 and by (9) z = 2y — [a + ß)x = V

n, for some z, if n is sufficiently large. Thus d

XiVj

Z < d

n. But by (6), d^ < d

n, only if k > n, so i > n. It can be seen that | R

t \ , l-ßt+i | ,... is an increasing sequence if t is sufficiently large, so |x| = \R{\ > \R

n\, which contradicts the assumption \x\ < IR-nI

• To complete the proof, we have to show that in the case \B\ > 1 there axe lattice points (x,y,z) for which d

X)Vt

Z < d

n and |x| < |Ä

n| for some n. Suppose \B\ > 1. If \ß\ > 1, then by (6), d

n —

> oo as n —> oo, so there are such lattice points for any sufficiently large n. If \ß\ — 1 the d

n is a constant and there are infinetely many n and points (x,y,z) which fulfill the assumptions. Suppose \ß\ < 1. Let y/

x be a convergent of the simple continued fraction expansion of the irrational a. Then, by the elementary properties of continued fraction expansions of irrational numbers and by (10), (11), we have the inequalities , 1 \ax -yI < V~Dx — z V~Dy — az = 2\ax - y\ < = \a + ß\ \ax -y\<\a + ß\ -. x