XIUMEI LI AND MIN SHA

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1 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS arxiv: v [math.nt] 18 Apr 017 XIUMEI LI AND MIN SHA Abstract. In this paper we initiate a study on Gauss factorials of polynomials over finite fields, which are the analoues of Gauss factorials of positive inteers. 1. Introduction A well-known result in number theory, called Wilson s theorem, says that for any prime number p, we have (1.1 (p 1! 1 (mod p. If we replace p by any composite inteer in (1.1, it is not correct any more. Gauss proved a composite analoue of Wilson s theorem: for any inteer n > 1, 1 (mod n for n =,4,p k, or p k, j (1. 1 j n 1 cd(j,n=1 1 (mod n otherwise, where p is an odd prime and k is a positive inteer. In [3], Cosrave and Dilcher called the product (1.3 j 1 j N cd(j,n=1 a Gauss factorial, where N and n are positive inteers. We refer to [6] for a very ood survey on Gauss factorials of inteers, and [4, 5, 7, 8] for recent work. In particular, when n 3 is odd and N = (n 1/, the multiplicative order of the Gauss factorial (1.3 modulo n has been determined in [3, Theorem ]. Throuhout the paper, the letter p always denotes a prime, and q = p s for some inteer s 1. Let F q be the finite field of q elements. We denote by F q [X] the polynomial rin over F q. In the sequel, for simplicity let A = F q [X]. It is known that A has many properties in common with the inteers Z, and thus many number theoretic problems about Z have their analoues for A. 1

2 XIUMEI LI AND MIN SHA For any polynomial f(x A of deree def 1, we define the Gauss factorial (denoted by G(f of f as follows: G(f =. A 0 de <de f cd(,f=1 That is, G(f is the product of all non-zero polynomials of deree less than def and coprime to f. By convention, if f F q, we define G(f = 1. Note that if f is irreducible, the definition of G(f only depends on the deree of f. Moreover, iven inteer n 1 and polynomial f A, we define G(n,f = A 0 de n cd(,f=1 which is also called a Gauss factorial. In particular, G(f = G(def 1,f. In this paper, we initiate the study of Gauss factorials of polynomials over finite fields by eneralizin (1. and the main result in [3]. Certainly, there are many other thins remainin to be explored. We want to remark that one can similarly define factorials of polynomials over finite fields, and consider eneralizin related classical results (some of them are listed in [].. Preliminaries In this section, for the convenience of the reader we recall some basic results about polynomials over finite fields, which can be found in [11, Chapter 1 and Chapter 3]. For f A, set f = q def if f 0, and f = 0 otherwise. For a non-constant polynomial f A, write its prime factorization as (.1 f = ap e 1 1 P et t, where a F q, inteer e j 1 (1 j t, and each P j (1 j t is a monic irreducible polynomial. Here, a monic irreducible polynomial in A is said to be a prime polynomial, and so in (.1 each P j (1 j t is a prime divisor of f. Given a non-zero polynomial f A, denote by A/fA the residue class rin of A modulo f and by (A/fA its unit roup. Lemma.1. Let f A be a non-constant polynomial with prime factorization as in (.1. Then, we have (A/fA = (A/P e 1 1 A (A/P et t A.,

3 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 3 Lemma.. Let P A be an irreducible polynomial, and e a positive inteer. Then, we have (A/P e A = (A/PA (A/P e A (1, where (A/PA is a cyclic roup of order P 1, and (A/P e A (1 is a p-roup of order P e 1. For any non-zero polynomial f A, let Φ(f = (A/fA, which is the so-called Euler totient function of A. Lemma.3. For any non-zero polynomial f A, we have Φ(f = f P f ( 1 1/ P, where the product runs throuh all the prime divisors of f. Let P A be an irreducible polynomial. When q is odd, iven a non-zero polynomial A with cd(,p = 1, as usual we define the Leendre symbol ( P = ±1 such that ( P P 1 (mod P. Let f A be a non-constant polynomial with the prime factorization as in (.1. Given a( non-zero polynomial A with cd(,f = 1, the Kronecker symbol = ±1 is defined as f ( = f t ( ej. j=1 Asusual, fortwonon-zeropolynomials 1, Awithcd( 1,f = 1, we have ( ( ( 1 1 =. f f f Lemma.4 (The reciprocity law. Let f, A be relatively prime non-zero polynomials. Assume that q is odd, and both f and are monic. Then, we have ( f P j = ( 1 q 1 def de ( f. The followin lemma is essentially a special case of a result due to Artin [1, Section 18, Equation (10].

4 4 XIUMEI LI AND MIN SHA Lemma.5. Assume that q is odd. Let P A be a prime polynomial of odd deree. Denote by h( P the class number of the quadratic function field F q (X, P. Then, we have ( h( P =. P monic 0 de <de P Proof. Followin Artin [1] the quadratic function field F q (X, P is imainary (see also [11, Proposition 14.6] and the discussions therein. So, usin [1, Section 18, Equation (10] and [1, Section 17, Equation (4] directly (alternatively, applyin aruments similar to those in the proof of [11, Theorem 16.8], we obtain h( P = monic 0 de <de P ( P Here one should note that since P can be viewed as a monic polynomial with respect to X, we can apply the results in [1]. For each summation term in the above formula, usin the reciprocity law (Lemma.4 and noticin that dep is odd, we deduce that ( P = ( 1 (q 1de which implies the desired result. ( P =. ( P, Finally, we reproduce a useful result due to Miller [9, Section 1], which can yield a simple proof of (1.. Lemma.6. Let A be an abelian roup. Then, the product a A a is the identity element if A either has no element of order two or has more than one element of order two, otherwise the product is equal to the element of order two. 3. Conruence formulas In this section, our main objective is to eneralize (1. to G(f by followin the approach in [9]. One can see that we have two quite different cases dependin on the characteristic of A (that is, p. We first remark that it is known that for any irreducible polynomial P A, we have (for instance see [11, Chapter 1, Corollary ] G(P 1 (mod P,

5 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 5 which is an analoue of (1.1. Besides, it is easy to see that if non-zero f A is reducible and square-free, then we have 0 (mod f. A 0 de <de f Theorem 3.1. Assume that p is odd. Then, for any polynomial f A of deree def 1, we have 1 (mod f if f has only one prime divisor, G(f 1 (mod f otherwise. Proof. Note that p is odd. Then, for any irreducible polynomial P A and inteer e 1, by Lemma. the roup (A/P e A has only one element of order two (that is, 1. So, applyin Lemma.1, it is easy to see that the abelian roup (A/fA has only one element of order two if and only if f has only one prime divisor. Then, the desired result now follows from Lemma.6. Theorem 3.. Assume that p =. For any polynomial f A of deree def 1 with the prime factorization as in (.1, we have f/p 1 +1 (mod f if q =,dep 1 = 1, e 1 3, and all the G(f other exponents e j = 1 if they exist, 1 (mod f otherwise. Proof. We first remark that 1 is not an element of order two, because p =. We also note that for any irreducible polynomial P A, the order of (A/PA is an odd number, and thus the roup (A/PA has no element of order two. Let P A be an irreducible polynomial, and e a positive inteer. If h is an element of order two of (A/P e A, then P e (h + 1. So, it is easy to see that when e is even, the set of elements of order two of (A/P e A is { P e/ +1 : A,0 de < e dep }, whose cardinality is q 1 edep 1. Similarly, if e is odd and reater than 1, then the set of elements of order two of (A/P e A is { P (e+1/ +1 : A,0 de < e 1 } dep, whose cardinality is q 1 (e 1deP 1. Thus, (A/P e A has only one element of order two if and only if dep = 1, e 3, and q =.

6 6 XIUMEI LI AND MIN SHA Hence, the desired result follows by usin Lemma.1 and Lemma.6, and noticin that f/p is indeed an element of order two of (A/fA if e 1. Furthermore, we can et a conruence identity for G(n,f when n def. This can also be viewed as an analoue of (1.. Theorem 3.3. For any polynomial f A of deree def 1, if the inteer n satisfies n def, we have 1 (mod f if p is odd and f has only one prime divisor, G(n,f 1 (mod f otherwise. Proof. Fix an arbitrary inteer m def. For any polynomial A with de < def and cd(,f = 1, it is easy to see that the set of polynomials in A of deree m and conruent to modulo f is { +fr : r A,der = m def whose cardinality is (q 1q m def. Thus, we obtain n G(n,f G(f m=de f G(f qn+1 def }, G(f (q 1qm de f (mod f. WethenconcludetheproofbyusinTheorem3.1andTheorem Multiplicative orders As mentioned before, when the inteer n 3 is odd and N = (n 1/, Cosrave and Dilcher have determined the multiplicative order of the Gauss factorial (1.3 modulo n in [3, Theorem ]. That is, they only considered half of the positive inteers less than n. In this section, assumethatq isodd, thenouraimistoetsomesimilar results concernin a special divisor of G(f. For this divisor, we only consider half of the polynomials of deree less than def and coprime to f. For any non-zero A, we denote by sn( the leadin coefficient of, which is called the sin of. Let S be a subset of F q such that S = (q 1/ and for any α F q if α S then α S. Obviously, the set S has (q 1/ choices. Define δ(s = a S a. Notice that (4.1 δ(s = ( 1 q 1 a F q a = ( 1 q+1.

7 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 7 So, if q 3 (mod 4, we have δ(s = 1, and then δ(s = ±1. Notethatforanynon-zero A, sn( S ifandonlyifsn( S. So, for any polynomial f A of deree def 1, we easily have where G(f = ( 1 Φ(f/ G(f,S, G(f,S = A,sn( S 0 de <de f cd(,f=1 Thus, by Theorem 3.1 we obtain ( 1 Φ(f/ (mod f if f has only one prime divisor, G(f,S ( 1 Φ(f/ (mod f otherwise. This implies that the multiplicative order of G(f,S modulo f only can possibly be 1, or 4. Here, we want to determine the multiplicative order of G(f, S modulo f, which is denoted by ord f G(f,S. The main result is as follows. Theorem 4.1. Assume that q is odd, and let f A be a polynomial havin the prime factorization as in (.1. Then (1 ord f G(f,S = 4 when t = 1, and either q 1 (mod 4 or dep 1 is even. ( ord f G(f,S = when (a t = 1, q 3 (mod 4, dep 1 is odd, δ(s = 1, and e (h( P (mod, or (b t = 1, q 3 (mod 4, dep 1 is odd, δ(s = 1, and e (h( P (mod, or (c t =, q 1 (mod 4, and P 1 is not a quadratic residue modulo P, or (d t =, q 3 (mod 4, both dep 1 and dep are even, and P 1 is not a quadratic residue modulo P, or (e t =, q 3 (mod 4, and either dep 1 or dep is odd; (3 ord f G(f,S = 1 in all other cases. From Theorem 4.1 one can see that for many cases ord f G(f,S is independent of the choice of S. Especially some results are related to the class numbers of function fields. Actually, we do more in the paper: the value of G(f,S modulo f is either explicitly iven or easily computable. Before provin Theorem 4.1, we illustrate an example to confirm some results in Theorem 4.1 by usin the computer alebra system PARI/GP [1]. Choose q = 3 and P = X 3 +X +, then by Lemma.

8 8 XIUMEI LI AND MIN SHA.5 we have h( P = 7. If furthermore we choose S = {1}, we have δ(s = 1 and G(P,S = 1, and thus ord P G(P,S =, which is compatible with Theorem 4.1 (; otherwise if we choose S = { 1}, we et ord P G(P,S = 1, which is also consistent with Theorem 4.1 (3. In the followin, we divide the proof of Theorem 4.1 into several cases One prime divisor. We continue the eneral discussion about G(f,S. Suppose that f has the prime factorization as in (.1. Then, by Lemma.3 we et t Φ(f = q (e i 1deP i (q dep i 1. Note that q is odd, so i=1 ( 1 1 Φ(f = ( 1 1 t i=1 (qde P i 1. If t, that is, f has at least two distinct prime divisors, then we have ( 1 1 Φ(f = 1, and thus G(f,S 1 (mod f. Now assume that t = 1, that is, f has only one prime divisor P 1. Then it is easy to see that 1 if q 1 (mod 4, or dep 1 is even, G(f,S 1 otherwise. Thus, for any f A of positive deree and with the prime factorization as in (.1, we have (4. 1 (mod f if t = 1, and either q 1 (mod 4 or dep 1 is even, G(f,S 1 (mod f otherwise. The above equation immediately ives the followin partial result: Theorem 4.. Assume that q is odd. If f has only one prime divisor P, and either q 1 (mod 4 or P has even deree, then ord f G(f,S = 4. Otherwise, ord f G(f,S = 1 or. For further deductions, we need to et a new expression of G(f,S. Fix an arbitrary polynomial f A of deree def 1. Let i n be the

9 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 9 cardinality of the set { A : monic,de = n,cd(,f = 1}. Note that Denote de f 1 n=0 ( M(f = i n = Φ(f q 1. monic 0 de <de f, cd(,f=1 (q 1/, which is also a polynomial in A. Now, we deduce that (4.3 G(f,S = a S ( = a S ( def 1 n=0 a in Φ(f ( q 1 a = δ(s Φ(f q 1 M(f. monic de =n, cd(,f=1 monic 0 de <de f, cd(,f=1 So, our problem is reduced to studyin M(f modulo f. By definition we also rewrite G(f = 0 de <de f cd(,f=1 = = de f 1 n=0 de f 1 n=0 ( de =n cd(,f=1 ( a ( i n a F q = ( 1 Φ(f q 1 M(f. q 1 monic de =n, cd(,f=1 q 1 Now assume that f has the prime factorization as in (.1. So, usin Theorem 3.1 we obtain 1 (mod f if t = 1 and dep 1 is even, (4.4 M(f 1 (mod f otherwise. We first handle M(f in the case when f is irreducible. Lemma 4.3. Assume that q 3 (mod 4. If f A is an irreducible polynomial of odd deree with the form f = ap, where a F q and

10 10 XIUMEI LI AND MIN SHA P A is a prime polynomial, then M(f ( 1 1 (h( P 1 (mod P. Proof. We apply similar aruments as in[10]. By definition, we directly have M(f = M(P. So, it is equivalent to determine the value of M(P modulo P (or equivalently modulo f. Put d = dep. We first make some preparations. Let N (resp. R be the number of monic polynomials in A of deree less than d which are quadratic non-residues (resp. quadratic residues modulo P. So, N +R = 1+q + +q d 1, which, toether with Lemma.5, implies that N = 1 ( 1+q + +q d 1 h( P. Note that q 3 (mod 4 and d is odd, so we have 1+q + +q d 1 1 (mod 4. Thus, we et (4.5 N 1 ( h( P 1 (mod. Given a non-zero polynomial which is a quadratic residue modulo P, b is also a quadratic residue for any square element b F q (equivalently, b is a quadratic residue modulo P, and b = (q 1/. b F q,( b P=1 Besides, note that for any non-zero polynomial A, we can write = c 0 for some c F q and monic polynomial 0 A. Then, noticin q 3 (mod 4, we et (q 1/ = ± (q 1/ 0 = (± 0 (q 1/. Basedontheaboveobservationsandthefactthat 1isnotaquadratic residue modulo P and is the unique element of order two in (A/PA, we deduce that (4.6 1 = ( P = 1 0 de <d = sn( = ±1, ( P = 1 0 de <d (q 1/, where we also use the simple fact that theinverse of a quadratic residue is also a quadratic residue (modulo P.

11 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 11 Usin (4.6, we obtain ( M(P = monic, ( P = 1 0 de <d = ( 1 N monic, ( P = 1 0 de <d = ( 1 N ( 1 N sn( = ±1, ( P = 1 0 de <d (mod P. monic, ( P = 1 0 de <d (q 1/ (q 1/ Now, the desired result follows from (4.5. (q 1/ monic, ( P = 1 0 de <d ( (q 1/ In the followin, we extend the result in Lemma 4.3 to the case when f is a power of some prime polynomial up to a constant. Actually, we can obtain a more eneral form. Lemma 4.4. If f A is a polynomial of the form f = ap e, where a F q, P A is a prime polynomial and e is a positive inteer, then M(f ( 1 (e 1(q 1 dep M(P (mod P. Proof. Put d = dep. Note that for any non-zero polynomial A of deree de < def = de, by the Euclidean division we can write = 1 P + for some 1, A, where 1 and satisfy de 1 < d(e 1 and = 0 or 0 de < d. Then by definition, we have ( (q 1/, M(f = Q 0 Q 1 where Q 0 = monic 0 de <d and Q 1 = Now for Q 1, we deduce that ( Q 1 A 0 de <d A 1 monic 0 de <d 0 de 1 <d(e 1, q d(e 1 1 q 1 G(P qd(e 1 1 q 1 (mod P ( 1 d(e 1 (mod P, (mod P ( 1 P +.

12 1 XIUMEI LI AND MIN SHA where we have applied Theorem 3.1. Therefore (q 1/ M(f = (Q 0 Q 1 ( 1 d(e 1(q 1 M(P (mod P, which completes the proof. We also need a simple but useful result, which is an analoue of [3, Lemma 1]. One can prove it in a straihtforward manner; we therefore omit its proof. Lemma 4.5. Suppose that q is odd. Let f, A be two non-constant polynomials. Given an inteer e 1, assume that f 1 (mod e. Then, f ±1 (mod e if and only if f ±1 (mod, with the sins correspondin to each other. Now we are ready to et a partial result about the value of G(f,S modulo f. We use the product δ(s defined just before (4.1. Theorem 4.6. Assume that q 3 (mod 4. If f A is a polynomial of the form f = ap e, where a F q, P A is a prime polynomial of odd deree and e is a positive inteer, then G(f,S ( 1 e+1 (h( P 3 δ(s (mod f. Proof. By (4.3, we first note that G(f,S = δ(s Φ(f q 1 M(f = δ(sm(f, where we use the fact that δ(s = ±1 since q 3 (mod 4. From Lemma 4.4, we have M(f ( 1 (e 1(q 1 dep M(P ( 1 e 1 M(P (mod P, which, toether with Lemma 4.3, ives (4.7 M(f ( 1 e 1+1 (h( P 1 (mod P. Since M(f 1 (mod P e by (4.4, usin Lemma 4.5 we know that M(f ±1 (mod P e if and only if M(f ±1 (mod P, with the sins correspondin to each other. Thus, by (4.7 we obtain (4.8 M(f ( 1 e+1 (h( P 3 (mod P e. Now, the desired result follows.

13 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS Two or more prime divisors. In this section, we deal with the case when f has more than one prime divisors. We start with a key lemma. Lemma 4.7. If f A is a polynomial havin the prime factorization as in (.1, then e 1 1 ( 1 q 1 dep P Φ(P (mod P e 1 1 if t =, M(f ( P e 1 P Φ 1 Pe Pe t 1 t 1 t (mod P e 1 Pe t 1 t 1 if t 3. 1 Pe Proof. Put f = P e 1 1 Pe Pe t 1 t 1,d t = dep t. Note that for any nonzero polynomial A of deree de < def, by the Euclidean division we can write = 1 f + for some 1, A, where 1 and satisfy de 1 < d t e t and = 0 or 0 de < de f. Then by definition, we have ( (q 1/, (4.9 M(f = Q 0 Q 1 where Q 0 = monic 0 de <de f cd(,f=1 Now, we define Q 0 = and obtain monic 0 de <de f (, f=1 and Q 1 = and Q 1 = ( Q 1 A 0 de <de f cd(, f=1 A 1 monic 0 de <de f 0 de 1 <d te t, cd(,f=1 A 0 de <de f cd(, f=1 d q t e t 1 q 1 ( 1 1 monic 0 de 1 <d te t (mod f ( 1 f +. f +, G( f qd t e t 1 q 1 (mod f. For relatin Q j to Q j, we multiply all relevant multiples of P t back to Q 0 and Q 1. More precisely, on the riht-hand side of (4.9 we multiply

14 14 XIUMEI LI AND MIN SHA numerator and denominator by ( monic 0 de<de f+d t(e t 1 cd(, f=1 which is equal to ( q 1 ( P t monic 0 de<de f cd(, f=1 M( fp Φ( f t q 1 P t, A 1 monic 0 de <de f 0 de 1 <d t(e t 1, cd(, f=1 A 0 de <de f cd(, f=1 ( P t q d t (e t 1 1 q 1 M( fp Φ( f t G( f q d t (e t 1 1 (mod f. Hence, from the above discussions, we deduce that q 1 (Q 0 Q 1 M(f M( fp Φ( f t G( f qd t(e t 1 1 M( f G( f qd t e t 1 M( fp Φ( f t G( f qd t(e t 1 1 G( f 1 qdt(et 1 (q dt 1 P Φ( f t By Theorem 3.1, this completes the proof. q 1 (mod f. q 1 ( 1 f + P t (mod f (mod f (mod f Now, we first address thecasewhen f hasexactly two prime divisors. Theorem 4.8. Assume that q is odd, and f A is a polynomial havin the prime factorization as in (.1 with t =. Then, if one of the followin two conditions holds: (1 q 1 (mod 4, ( q 3 (mod 4 and both dep 1 and dep are even, we have ( P1 G(f,S (mod f; P

15 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 15 otherwise, we have G(f,S ±1 (mod f. Proof. First, since q is odd and t =, by (4.1 and (4.3 we have (4.10 G(f,S = δ(s Φ(f q 1 M(f = M(f. Put d j = dep j,j = 1,. By Lemma 4.7, we have M(f ( 1 (q 1d ( 1 (q 1d e 1 P Φ(P 1 ( P (mod P 1 P 1 1 ( ( 1 (q 1d P P 1 (mod P 1 (mod P 1. From (4.4, we know that M(f 1 (mod P e 1 1. So, applyin Lemma 4.5 we et (4.11 M(f ( 1 (q 1d By symmetry, we have (4.1 M(f ( 1 (q 1d 1 ( P P 1 ( P1 P (mod P e 1 1. (mod P e. Now, assume that either q 1 (mod 4, or q 3 (mod 4 and both d 1 and d are even. Then, by the reciprocity law (Lemma.4 we obtain ( P ( P1 =, P 1 P and thus M(f ( P1 P (mod P e 1 1 and (mod P e. Usin the Chinese Remainder Theorem, we et ( P1 (4.13 M(f (mod f. P So, by (4.10 we have G(f,S ( P1 P (mod f. In all other cases, one can similarly see that the product of the rihthand sides of (4.11 and (4.1 is equal to 1. Hence, applyin aain the Chinese Remainder Theorem, we can conclude the proof.

16 16 XIUMEI LI AND MIN SHA Finally, the case when f has more than two prime divisors is much easier. Theorem 4.9. Assume that q is odd. If f A is a polynomial havin the prime factorization as in (.1 with t 3, then we have Proof. By Lemma 4.7, we have M(f P Φ ( P e 1 G(f,S 1 (mod f. 1 Pe Pe t 1 t 1 t (mod P 1 ( ( Pt Φ P e Pe t 1 t 1 (mod P 1 P 1 1 (mod P 1. From (4.4, we know that M(f 1 (mod P e 1 1. So, applyin Lemma 4.5 we et M(f 1 (mod P e 1 1. By symmetry, we have M(f 1 (mod P e j j for j t. So, by the Chinese Remainder Theorem we obtain (4.14 M(f 1 (mod f. Noticin G(f,S = δ(s Φ(f q 1 M(f = M(f, we complete the proof. Acknowledements The authors are very rateful to the referee for careful readin and useful comments. The research of the first author was supported by National Natural Science Foundation of China Grant No , and the second author was supported by the Australian Research Council Grant DP References [1] E. Artin, Quadratische Körper im Gebiete der höheren Konruenzen I, II, Math. Zeit., 19 (194, [] M. Bharava, The factorial function and eneralizations, Amer. Math. Monthly, 107 (000, [3] J. B. Cosrave and K. Dilcher, Extensions of the Gauss Wilson theorem, Inteers, 8 (008, A39, available at [4] J. B. Cosrave and K. Dilcher, Mod p 3 analoues of theorems of Gauss and Jacobi on binomial coefficients, Acta Arith., 14 (010,

17 GAUSS FACTORIALS OF POLYNOMIALS OVER FINITE FIELDS 17 [5] J. B. Cosrave and K. Dilcher, The multiplicative orders of certain Gauss factorials, Int. J. Number Theory, 7 (011, [6] J. B. Cosrave and K. Dilcher, An introduction to Gauss factorials, Amer. Math. Monthly, 118 (011, [7] J. B. Cosrave and K. Dilcher, The Gauss Wilson theorem for quarter intervals, Acta Math. Hunar., 14 (014, [8] J. B. Cosrave and Karl Dilcher, A role for eneralized Fermat numbers, Math. Comp., DOI: [9] G. A. Miller, A new proof of the eneralized Wilson s theorem, Ann. Math., 4 (1903, [10] L. J. Mordell, The conruence (p 1/! ±1( (mod p, Amer. Math. Monthly, 68 (1961, [11] M. Rosen, Number theory in function fields, Spriner-Verla, New York, 00. [1] The PARI Group, PARI/GP version.7.5, Bordeaux, 015, School of Mathematical Sciences, Qufu Normal University, Qufu Shandon, 73165, China address: lxiumei013@hotmail.com School of Mathematics and Statistics, University of New South Wales, Sydney, NSW 05, Australia address: shamin010@mail.com