Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1998. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 25)
JONES, J. P. and Kiss, P., Representation of integers as terms of a linear recurrence with maximal index
32 James P. Jones and Péter Kiss (3.3.2) F r +1 < X < 2 F r y 1 < y < F r_ 2 and y < x. But not both. Proof. By Theorem 2.22, N is either a double or a single. Hence there are only two cases to consider. If N is a single, then (x,y) = (a,b) and condition (3.3.1) holds by Lemma 3.1. (i). If N is a double, then (x,y) = (a, 6) or (x,y) — (c, d). In the first case, by Lemma 3.1 (ii) (3.3.2) holds. In the sceond case, by Lemma 3.1 (ii) (3.3.1) holds. Lemma 3.4. Suppose R(N) — r. Then all solutions of N = H r(x.y ) in positive integers (x,y) satisfy the conditions x < 2F r and y < F r. Proof. By Lemma 3.3, either (3.3.1) holds or (3.3.2) holds. (3.3.1) implies x < F T < 2F r and y < F r. (3.3.2) imphes x < 2F r and y < F r-% < F r. Hence x < 2F r and y < F r. Lemma 3.5. If 0 < k, then for all positive integers a and b, 0 < II k(a,b) < H k+ l{a,b). Proof. Prom the definition it follows that //„(a, b ) is a strictly increasing sequence of positive integers. Theorem 3.6. There exist integers x and y such that (3.6) N = H n(x,y) and 1 < y < x < F n iffn = R(N). Furthermore x and y are unique. Proof. To prove the first part of the theorem suppose R(N) — n. Then by Lemma 3.2 there exist unique integers x and y suet that N = H n(x,y ) a^d 1 < y < % < F n, i.e. (3.6). To prove the second part suppose x and y are integers satisfying (3.6). Then n > 0. Let R(N) — r. Then n < r. Let k — r — n. By definition of R(N) there are positive integers a and b such that N = H r(a,b). By Lemma 1.3 (ii), since n — r — k, we have N — H r{a , b) = H n(H k(a, 6), H k+ 1 (a, b)) so that N = H n{II k(a, 6), II k+ l (a, b)). Thus x = H k(a,b) and y = H k+i(a,b) are particular solutions to the linear diophantine equation N = xF n-\ + yF n. Since (F n,F n +i) = 1, all solutions to the equation are given by x = H k(a,b) - tF n and y - H k+ l (a, 6) + iF n_i , where t is an integer. Since y < x, we have for some t the inequality H k +i (a, b) + tF n—1 < H k(a , b) - tF n. This implies t < (H k(a, b) — Hk+i (a, b))jF n+\ , so that
