J. Number Theory 130(2010), no. 4, SOME CURIOUS CONGRUENCES MODULO PRIMES

Transcription

1 J. Number Theory 30(200, no. 4, SOME CURIOUS CONGRUENCES MODULO PRIMES L-Lu Zhao and Zh-We Sun Department of Mathematcs, Nanjng Unversty Nanjng 20093, People s Republc of Chna zhaollu@gmal.com, zwsun@nju.edu.cn Abstract. Let n be a postve odd nteger and let p > n + be a prme. We manly derve the followng congruence: 0< < < n <p ( 3 ( n 0 (mod p.. Introducton Smple congruences modulo prme powers are of nterest n number theory. Here are some examples of such congruences: (a (Wolstenholme p = / 0 (mod p2 for any prme p > 3. (b (Z. W. Sun [S02, (.3] For each prme p > 3 we have 0<<p/2 3 0<<p/6 ( (mod p. (c (Z. W. Sun [S07, Theorem.2] If p s a prme and a, n N = {0,, 2,... }, then n/p a! 0 (mod p a ( n ( ( p n/p a a (mod p. (d (Z. W. Sun and R. Tauraso [ST, Corollary.] For any prme p and a Z + = {, 2, 3,... } we have p a =0 ( 2 ( p a 3 (mod p 2, Key words and phrases. Congruences modulo a prme, p-adc ntegers, polynomals. 200 Mathematcs Subject Classfcaton. Prmary A07; Secondary A4, T06. The second author s responsble for communcatons, and supported by the Natonal Natural Scence Foundaton (grant of Chna.

2 2 LI-LU ZHAO AND ZHI-WEI SUN where ( 3 s the Legendre symbol. Let p > 3 be a prme. In 2008, durng hs study of p =0 ( 2 modulo powers of p wth R. Tauraso, the second author conjectured that.e., 0<<j<<p 0<<j<<p,2 (mod 6 ( ( 3 j j 0<<j<<p 4,5 (mod 6 0 (mod p, (. j (mod p. (.2 In ths paper we confrm the above conjecture of Sun by establshng the followng general theorem. Theorem.. Let n Z + and let p > n + be a prme. ( If n s odd, then 0< < < n <p,2 (mod 6 ( If n s even, then 0< < < n <p 0 (mod 3 n ( n 2 0< < < n <p 4,5 (mod 6 0< < < n <p 2,3,4 (mod 6 n (mod p. (.3 n (mod p. (.4 We deduce Theorem. from our followng result. Theorem.2. Let n Z + and let p > n + be a prme. Set F n (x = 0< < < n <p x n Z p [x], (.5 where Z p denotes the ntegral rng of the p-adc feld Q p. Then we have F n ( x ( n F n (x (mod p, (.6.e., all the coeffcents of F n ( x ( n F n (x are congruent to 0 modulo p. In the next secton we use Theorem.2 to prove Theorem.. Secton 3 s devoted to our proof of Theorem.2.

3 SOME CURIOUS CONGRUENCES MODULO PRIMES 3 2. Theorem.2 mples Theorem. Proof of Theorem. va Theorem.2. (.3 holds trvally when p = 3 and n =. Below we assume that p > 3. Let ω be a prmtve cubc root of unty n an extenson feld over Q p. Then, n the rng Z p [ω] we have the congruence F n ( ω 2 = F n ( + ω ( n F n ( ω (mod p. (2. For r Z we set Clearly S r = 0< < < n <p r (mod 6 n. F n ( ω =S 0 ωs + ω 2 S 2 S 3 + ωs 4 ω 2 S 5 =S 0 S 3 ω(s S 4 + ( ω(s 2 S 5 =S 0 S 3 S 2 + S 5 ω(s + S 2 S 4 S 5. Smlarly, F n ( ω 2 = S 0 S 3 S 2 + S 5 ω 2 (S + S 2 S 4 S 5. Thus F n ( ω + F n ( ω 2 =2(S 0 S 2 S 3 + S 5 + S + S 2 S 4 S 5 =2S 0 + S S 2 2S 3 S 4 + S 5 and F n ( ω F n ( ω 2 = (ω 2 ω(s + S 2 S 4 S 5. Note that (ω (ω 2 = 3 s relatvely prme to p. Therefore, by (2., f 2 n then S + S 2 S 4 S 5 0 (mod p; (2.2 f 2 n then 2S 0 + S S 2 2S 3 S 4 + S 5 0 (mod p. (2.3 To conclude the proof we only need to show that (2.3 s equvalent to S 0 S 3 2(S 2 + S 3 + S 4 (mod p. (2.4

4 4 LI-LU ZHAO AND ZHI-WEI SUN Recall that p x p (x j j= p = ( x (mod p (cf. Proposton 4.. of [IR, p. 40]. Comparng the coeffcents of x p n we get that 0 (mod p. (2.5 n 0< < < n <p So 5 r=0 S r 0 (mod p, whch mples the equvalence of (2.3 and (2.4. We are done. 3. Proof of Theorem.2 Proof of Theorem.2. We use nducton on n. Observe that p ( p ( x = + ( x p = p =0 ( p ( x = x p ( x p. For =,..., p clearly ( p ( p = ( ( p (mod p. Thus F (x p p ( p ( x = xp ( x p p = (mod p and hence F ( x F (x (mod p as desred. Ths proves (.6 for n =. For the nducton step we need to do some preparaton. For P (x = m a x Z p [x], =0 we defne ts formal dervatve by d dx P (x = 0< m a x.

5 SOME CURIOUS CONGRUENCES MODULO PRIMES 5 If m p and d dx P (x 0 (mod p, then a 0 (mod p for all =,..., m, and hence P (x a 0 = P (0 (mod p. Now assume that < n < p and F n ( x ( n 2 F n (x (mod p. Then d dx F n(x = and hence = 0< < < n <p 0< 2 < < n <p = F n (x x(x x x 2 n = < 2 < < n <p x2 2 n x 0< 2 < < n <p d ( Fn ( x ( n F n (x dx ( Fn ( x = ( x( x ( x + ( n ( Fn (x x(x x = ( n F n (x F n ( x x(x 0< 2 < < n <p 2 n 0< 2 < < n <p 2 n ( x + ( n x 2 n 2 n 2 = 0< 2 < < n <p Combnng ths wth the nducton hypothess and (2.5, we obtan x(x d dx (F n( x ( n F n (x 0 (mod p. x 2 n. For the fnte feld F p = Z/pZ, t s well nown that F p [x] s a prncpal deal doman. So we have and hence d dx (F n( x ( n F n (x 0 (mod p F n ( x ( n F n (x F n ( + ( n F n (0 = 0< < < n <p n 0 (mod p wth the help of (2.5. done. Ths concludes the nducton step and we are

6 6 LI-LU ZHAO AND ZHI-WEI SUN References [IR] K. Ireland and M. Rosen, A Classcal Introducton to Modern Number Theory, 2nd ed., Grad. Texts n Math. 84, Sprnger, New Yor, 990. [S02] Z. W. Sun, On the sum ( n r (mod m and related congruences, Israel J. Math. 28 (2002, [S07] Z. W. Sun, Combnatoral congruences and Strlng numbers, Acta Arth. 26 (2007, [ST] Z. W. Sun and R. Tauraso, On some new congruences for bnomal coeffcents, preprnt, arxv: