A CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod p) ZHI-HONG SUN

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1 A CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod ) ZHI-HONG SUN Deartment of Mathematics, Huaiyin Teachers College, Huaian , Jiangsu, P. R. China hyzhsun@ublic.hy.js.cn (Submitted November 2004-Final Revision February 2005) Abstract. Let {u n } be defined by u 1 m = = u 1 = 0, u 0 = 1 and u n +a 1 u n a m u n m = 0 (m > 2, n > 1). In this aer we show that the congruence x m + a 1 x m 1 + +a m 0 (mod ) has m distinct solutions if and only if u m u 2 0 (mod ) and u 1 1 (mod ), where is a rime such that > m and -a m. 1. Introduction. In [2] the author extended Lucas series to general linear recurring sequences by defining {u n (a 1,..., a m )} as follows: u 1 m = = u 1 = 0, u 0 = 1, u n + a 1 u n a m u n m = 0 (n = 1, 2, 3,... ), (1) where m 2 and a 1,..., a m are comlex numbers. Let Z be the set of integers. In this aer we establish the following result. Theorem 1. Let m 2, m, a 1,..., a m Z, u n = u n (a 1,..., a m ), and let be a rime such that > m and a m. Then the congruence x m + a 1 x m a m 0 (mod ) has m distinct solutions if and only if u m u 2 0 (mod ) and u 1 1 (mod ). (2) The famous Chebotarev density theorem imlies that (see for examle [4]) if the olynomial x m + a 1 x m a m (a 1,..., a m Z) is irreducible in Z[x], then the 1

2 set S of rimes such that x m + a 1 x m a m 0 (mod ) has m solutions has a ositive density d(s), that is, d(s) = Thus, by Theorem 1 we have lim x + { : x, S} { : x, is a rime} > 0. Corollary 1. Let m 2, a 1,..., a m Z and u n = u n (a 1,..., a m ). If x m + a 1 x m a m is irreducible in Z[x], then there are infinitely many rimes satisfying (2). 2. Proof of Theorem 1. Let f(x) = x m + a 1 x m a m. If f(x) 0 (mod ) has m distinct solutions b 1,..., b m, then we have f(x) (x b 1 ) (x b m ) (mod ) and b i b j (mod ) for i j (see [1, Theorem 108]). Suose (x b 1 ) (x b m ) = x m + A 1 x m A m. Then m i=1 (a i A i )x m i 0 (mod ) for any integer x. Since > m, by [1, Theorem 107] or Lagrange s theorem we must have a i A i (mod ) for i = 1, 2,..., m. By the definition of {u n }, it is evident that u n u n (A 1,..., A m ) (mod ) for all n 1 m. Since a m we see that b 1 b m. Hence, alying [2, Theorem 2.3] and Fermat s little theorem we obtain u n+ 1 u n+ 1 (A 1,..., A m ) = m i=1 b n+ 1+m 1 i m (b i b j ) j=1 j i m i=1 = u n (A 1,..., A m ) u n (mod ) (n 1 m). Note that u 1 m = = u 1 = 0 and u 0 = 1. So (2) holds. Conversely, suose (2) is true. Let Then we see that c k = a 0 = 1, g(x) = 0 i m 0 j 1 m i+j= 1 k a i u j = 1 m j=0 b n+m 1 i m (b i b j ) j=1 j i 1 u j x 1 m j and f(x)g(x) = c k x k. max{0,m k} i min{m, 1 k} 2 k=0 a i u 1 k i (0 k 1),

3 where max{a, b} and min{a, b} denote the maximum and minimum elements in the set {a, b} resectively. Clearly we have c 1 = a 0 u 0 = 1 and c 0 = a m u 1 m (u 1 + a 1 u a m u 1 m ) u 1 = u 1 1 (mod ). For k {1, 2,..., 2} we claim that c k = max{0,m k} i m a i u 1 k i. (3) If 1 k m, then clearly (3) holds. If 1 1 k < m, for k i m we have 1 m 1 k i 1 and so u 1 k i = 0. Thus, m i= k a iu 1 k i = 0 and hence (3) is also true. If m k 2, from (1) and (3) we see that c k = m i=0 a iu 1 k i = 0. If 1 k m 1, by (1), (3) and the fact that u m u 2 0 (mod ) we get c k = = m k i m 0 i m k 1 a i u 1 k i = 0 i m a i u 1 k i a i u 1 k i 0 (mod ). Therefore c k 0 (mod ) for k = 1, 2,..., 2. Now, utting the above together we obtain ( m f(x)g(x) = a i x i=0 m i)( 1 m j=0 0 i m k 1 a i u 1 k i u j x 1 m j) 1 = c k x k x 1 1 (mod ). (4) Since x 1 1 (x 1)(x 2) (x +1) (mod ) by Lagrange s theorem (see [1, Theorem 112]), we see that f(x) is congruent to the roduct of distinct linear olynomials (mod ). This comletes the roof of Theorem Alication to cubic congruences. 3 k=0

4 Theorem 2. Let a 1, a 2, a 3 Z, u n = u n (a 1, a 2, a, a = (a 2 1 3a 2 ) 3, b = 2a a 1 a 2 27a 3, and let > 3 be a rime such that aba 3 (b 2 4a). Then the following statements are equivalent: (i) x 3 + a 1 x 2 + a 2 x + a 3 0 (mod ) has three solutions, (ii) u 1+n u n (mod ) for all n 2, (iii) u 3 u 2 0 (mod ) and u 1 1 (mod ), (iv) u 2 0 (mod ), (v) U ( ( )/3 0 (mod ), (vi) s +1 a 2 1 2a 2 (mod ), (vii) V ( ( )/3 2(a 2 1 3a 2 ) 1 ( 2 (mod ), (viii) if ( a ) = 1, then U ( ( )/6 ; if ( a ) = 1, then V ( ( )/6, where ( n m ) is the Legendre symbol, and {U n}, {V n }, {s n } are given by U 0 = 0, U 1 = 1, U n+1 = bu n au n 1 (n 1), V 0 = 2, V 1 = b, V n+1 = bv n av n 1 (n 1), s 0 = 3, s 1 = a 1, s 2 = a 2 1 2a 2, s n+3 + a 1 s n+2 + a 2 s n+1 + a 3 s n = 0 (n 0). Proof. From the definition of u n we see that (ii) is equivalent to (iii). As b 2 4a and b2 4a 27 is the discriminant of x 3 +a 1 x 2 +a 2 x+a 3, the congruence x 3 +a 1 x 2 +a 2 x+ a 3 0 (mod ) has no multile solutions. By Theorem 1, (i) and (iii) are equivalent. According to [3, Theorem 4.3], (i) is equivalent to (iv). By [3, Theorem 3.2(i)], (iv) and (v) are equivalent. From [3, Theorem 4.1] we know that (i) is equivalent to (vi). By [3, Lemma 3.1], (vi) is equivalent to (vii). It is well known that (see [5]) Thus we have V ( ( )/3 = V 2 ( ( 3 U 2n = U n V n, V 2n = V 2 n 2a n and V 2 n (b 2 4a)U 2 n = 4a n. ( ( 3 ))/6 2a ) a 6 V( ( 2 2 )/ a 2 4 )(a 2 1 3a 2 ) 1 ( 2 (mod ).

5 Therefore (vii) is equivalent to V 2 ( ( )/6 2 (1 + ( a 2 1 3a 2 ))(a 2 1 3a 2 ) 1 ( 2 (mod ). As V 2 n (b 2 4a)U 2 n = 4a n, the above congruence is equivalent to (b 2 4a)U 2 ( ( )/6 2 (1 ( a 2 1 3a 2 ))(a 2 1 3a 2 ) 1 ( 2 (mod ). Thus, (vii) and (viii) are equivalent and the theorem is roved. Remark 1. Let a 1, a 2, a 3 Z be such that x 3 + a 1 x 2 + a 2 x + a 3 is irreducible in Z[x]. From Theorem 2 and Chebotarev density theorem we know that there are infinitely many rimes satisfying (i)-(viii) in Theorem 2. Let be a rime such that > 3 and a 2 1 3a 2. From [3, Theorems 4.1 and 4.2] and [3, Lemma 3.1] we know that and x 3 + a 1 x 2 + a 2 x + a 3 0 (mod ) has no solutions s +1 a 2 (mod ) V ( ( )/3 (a 2 1 3a 2 ) 1 ( 2 (mod ) x 3 + a 1 x 2 + a 2 x + a 3 0 (mod ) has one and only one solution s +1 a 2, a 2 1 2a 2 (mod ) V ( ( )/3 (a 2 1 3a 2 ) 1 ( 2, 2(a 2 1 3a 2 ) 1 ( 2 (mod ). By Chebotarev density theorem, there are also infinitely many rimes satisfying one of the above conditions in terms of {s n } or {V n }. References 1. G.H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers, 5th edition, Oxford Univ. Press, Oxford, 1981, Z.H. Sun, Linear recursive sequences and owers of matrices, Fibonacci Quart. 39 (2001), Z.H. Sun, Cubic and quartic congruences modulo a rime, J. Number Theory 102 (2003),

6 4. D. Terr, Chebotarev Density Theorem, htt://mathworld.wolfram.com/chebotarevdensitytheorem. html. 5. H.C. Williams, Édouard Lucas and Primality Testing, Canadian Mathematical Society Series of Monograhs and Advanced Texts (V ol.22), Wiley, New York, 1998,. 74. AMS Classification Numbers: 11A07,11B39,11B50,11T06. 6