Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae,. (1997) pp. 9 33 Some congruences concerning second order linear recurrences JAMES P. JONES PÉTER KISS Abstract. Let U n V n (n=0,1,,...) be sequences of integers satisfying a second order linear recurrence relation with initial terms U 0 =0, U 1 =1, V 0 =, V 1 =A. In this paper we investigate the congruence properties of the terms U nk V nk, where the moduli are powers of U n V n. Let U n V n (n = 0,1,,...) be second order linear recursive sequences of integers defined by U n = AU n 1 BU n (n > 1) V n = AV n 1 BV n (n > 1), where A B are nonzero rational integers the initial terms are U 0 = 0, U 1 = 1, V 0 =, V 1 = A. enote by α,β the roots of the characteristic equation x Ax + B = 0 suppose = A 4B 0 hence that α β. In this case, as it is well known, the terms of the sequences can be expressed as (1) U n = αn β n α β V n = α n + β n for any n 0. Many identities congruence properties are known for the sequences U n V n (see, e.g. [1], [4], [5] [6]). Some congruence properties are also known when the modulus is a power of a term of the sequences (see [], [3], [7] [8]). In [3] we derived some congruences where the moduli was Un 3, V n or V n 3. Among other congruences we proved that U nk kb n k 1 Un (mod U 3 n) Research supported by the Hungarian National Research Science Foundation, Operating Grant Number OTKA T 16975 0095.
30 James P. Jones Péter Kiss when k is odd a similar congruence for even k. In this paper we extend the results of [3]. We derive congruences in which the moduli are product of higher powers of U n V n. Theorem. Let U n V n be second order linear recurrences defined above let = A 4B be the discriminant of the characteristic equation. Then for positive integers n k we have 1. U nk kb k 1 n U n + k(k 1) B k 3 n U 3 n (mod U 5 n ), k odd,. U nk k B k n V n U n + k(k 4) 48 B k 4 n V n Un 3 (mod V n Un 5 ), k even, 3. V nk k( 1) k 1 B k 1 n V n + k(k 1) ( 1) k 3 B k 3 n Vn 3 (mod Vn 5 ), k odd, 4. V nk ( 1) k B k n + k 4 k ( 1) B k n V n (mod V 4 n ), k even, 5. U nk U n ( 1) k 1 B k 1 n + k 1 8 ( 1) k 3 B k 3 n U n V n (mod U n V 4 n ), k odd, 6. U nk k k ( 1) B k n U n V n + k(k 4) 48 ( 1) k 4 B k 4 n U n Vn 3 (mod U nvn 5 ), k even, 7. V nk B k 1 n V n + k 1 8 B k 3 n V n Un (mod V n Un 4 ), k odd, 8. V nk B k n + k 4 B k n U n (mod U 4 n ), k even. We note that the congruences of [3] follow as consequences of this theorem. For the proof of the Theorem we need some auxiliary results which are known (see e.g. [6]) but we show short proofs for them. In the followings we suppose that A > 0 hence that α = A + β = A, so that α β =, α + β = A, αβ = B hence by (1) () U n = αn β n Lemma 1. For any integer n 0 we have U 3n = 3U n B n + U 3 n. Proof. By (), using that αβ = B, we have to prove that α 3n β 3n = 3 αn β n ( α (αβ) n n β n ) 3 +,
Some congruences concerning second order linear recurrences 31 which follows from α 3n β 3n = 3(α n β n )α n β n + (α n β n ) 3. Lemma. For any non-negative integers m n we have U m+n = V n U m+n B n U m. Proof. Similarly as in the proof of Lemma 1, α m+n β m+n = (α n + β n ) αm+n β m+n (αβ) n αm β m is an identity which by (1) (), implies the lemma. Lemma 3. For any n 0 we have V n = B n + U n = V n B n U n = U n V n. Proof. The identities ( α α n +β n = (αβ) n n β n ) + αn β n = αn β n (α n + β n ) prove the lemma. Proof of the Theorem. We prove the first congruence of the Theorem by double induction on k. For k = 1 k = 3, by Lemma 1, the congruence is an identity. Suppose the congruence holds for k k +, where k 1 is odd. Then by Lemma 3 we have (3) U n(k+4) = U nk+4n = V n U nk+n B n U nk = (B n + U n)u n(k+) B n U nk (B n + U n )Q Bn R (mod U 5 n ), where (4) Q = (k + )B k+1 n U n + (k + )((k + ) 1) B k 1 n Un 3 (5) R = kb k 1 n U n + k(k 1) B k 3 n U 3 n.
3 James P. Jones Péter Kiss After some calculation (3), (4) (5) imply (6) U n(k+4) U n T + U 3 n S (mod U 5 n ), where T = ((k + ) k)b k+3 n = (k + 4)B (k+4) 1 S = (k + )B k+1 n + (k + )( (k + ) 1 ) k(k 1) B k+1 n = (k + 4)( (k + 4) 1 ) so by (6), U n(k+4) (k + 4)B (k+4) 1 U n + (k + 4)( (k + 4) 1 ) B k+1 n B (k+4) 3 n, B (k+4) 3 n U 3 n (mod U 5 n ). Hence the congruence holds also for k + 4 for any odd positive integer k. The other congruences in the Theorem can be proved similarly using Lemma 1,, 3 the identities U n = V n U n, V n = V n B n = B n + U n, U 3n = U n V n B n U n, V 3n = V 3 n 3Bn V n = B n V n + V n U n, U 4n = U n V 3 n B n U n V n, V 4n = V 4 n 4B n V n + B n. References [1]. Jarden, Recurring sequences, Riveon Lematematika, Jerusalem (Israel), 1973. [] J. P. Jones P. Kiss, Some identities congruences for a special family of second order recurrences, Acta Acad. Paed. Agriensis, Sect. Math. 3 (1995 96), 3 9.
Some congruences concerning second order linear recurrences 33 [3] J. P. Jones P. Kiss, Some new identities congruences for Lucas sequences, iscuss Math., to appear. [4]. H. Lehmer, On the multiple solutions of the Pell Equation, Annals of Math. 30 (198), 66 7. [5]. H. Lehmer, An extended theory of Lucas functions, Annals of Math. 31 (1930), 419 448. [6] E. Lucas, Theorie des functions numériques simplement périodiques, American Journal of Mathematics, vol. 1 (1878), 184 0, 89 31. English translation: Fibonacci Association, Santa Clara Univ., 1969. [7] S. Vajda, Fibonacci & Lucas numbers, the golden section, Ellis Horwood Limited Publ., New York-Toronto, 1989. [8] C. R. Wall, Some congruences involving generalized Fibonacci numbers, The Fibonacci Quarterly 17.1 (1979), 9 33. James P. Jones epartment of Mathematics Statistics University of Calgary Calgary, Alberta TN 1N4 Canada Péter Kiss Eszterházy Károly Teachers Training College epartment of Mathematics Leányka u. 4. 3301 Eger, Pf. 43. Hungary E-mail: kissp@gemini.ektf.hu