A CRITERION FOR POLYNOMIALS TO BE CONGRUENT TO THE PRODUCT OF LINEAR POLYNOMIALS (mod ) ZHI-HONG SUN Deartment of Mathematics, Huaiyin Teachers College, Huaian 223001, Jiangsu, P. R. China e-mail: 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 1 + + 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 1 + + 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 1 + + 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 1 + + a m (a 1,..., a m Z) is irreducible in Z[x], then the 1
set S of rimes such that x m + a 1 x m 1 + + 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 1 + + 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 1 + + 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 1 + + 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),
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 2 + + 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 1. 3. Alication to cubic congruences. 3 k=0
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 3 1 + 9a 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 )/6 2 1 3a 2 4 )(a 2 1 3a 2 ) 1 ( 2 (mod ).
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,. 84-86. 2. Z.H. Sun, Linear recursive sequences and owers of matrices, Fibonacci Quart. 39 (2001), 339-351. 3. Z.H. Sun, Cubic and quartic congruences modulo a rime, J. Number Theory 102 (2003), 41-89. 5
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