arXiv:1807.00274v1 [math.CO] 1 Jul 2018 where scle the called is hogotti paper, this Throughout Introduction 1 and satisfying motn eainbetween relation important n sa d rm oe.Let power. prime odd an is is φ pligteMoisivrinfruat h bv qain efind we equation, above the to formula M¨obius inversion the Applying ( hnec primitive each then , n ,where ), xlctfcoiainof factorization Explicit ...Uiest fSine&Tcnlg,Mrhl113,India Murthal-131039, Technology, & Science of University D.C.R. ζ Keywords n integer and ahmtc ujc Classification Subject Mathematics nomials. over eaprimitive a be µ Let steMoisfnto.Mroe,if Moreover, M¨obius function. the is F q e q F sotie when obtained is n ≡ φ q ( hccooi oyoilover polynomial cyclotomic th eafiiefil fodcaatrsi containing characteristic odd of field finite a be n iiefils reuil atrzto;Ccooi pol Cyclotomic factorization; Irreducible fields, Finite : (mod 1 Φ n steElrTtetfnto.Nt htif that Note function. Totient Euler the is ) n ≥ ( x .I hsppr h xlctfcoiainof factorization explicit the paper, this In 1. n = ) [email protected] hro fuiyover unity of root th eateto Mathematics of Department k F n hro fuiyi loan also is unity of root th Φ Y k ,te Φ then ), q | n n k ( eoe nt edwith field finite a denotes x ( n x hccooi oyoiland polynomial cyclotomic th x n d = ) k eapstv nee uhta gcd( that such integer positive a be − ajtSingh Manjit sa d iio of divisor odd an is − Abstract = 1 gcd( 1) n µ Y j 1 j,n ( =1 ( x Y n k x n k | 2 n )=1 pisit h rdc of product the into splits ) ) n = Φ d ( k − x 20;1E0 12E20. 12E10; 12T05; : Y ( k x F | − n 1 ). q ( e F hntepolynomial the Then . ζ x vrafiiefield finite a over j q stelatpstv integer positive least the is n k ) h ereo Φ of degree The . − q 1. + 1) n hro fuiy An unity. of root th µ q ( k ) lmns where elements, , n q hro funity of root th elements x 2 n d k ,q n, − y- divides n 1 φ ( ( 1 = ) x that n is ) ) /e q monic irreducible polynomials over Fq of degree e. In particular, Φn(x) is irreducible over Fq if and only if e = φ(n) ([7, 10]). Factoring polynomials over finite fields is a classical topic of mathematics and play important roles in cryptography, coding theory and communication theory (see [1, 5, 7]). In recent years, the factorization of polynomials xn − 1 and Φn(x) have received extensive attention from researchers. A brief description of some of the past accomplishments regarding the factorization of xn 1 and Φ (x) over F is given below. − n q The factorization of Φ n (x) over F when q 1 (mod 4) can be found in 2 q ≡ [7, 14]. In 1996, Meyn [9] obtained the factorization of Φ2n (x) over Fq when q 3 (mod 4). Fitzgerald and Yucas [4] studied the explicit factorization of ≡ Φ n (x) over F , where r is an odd prime such that q 1 (mod r). They 2 r q ≡ ± also obtained explicit irreducible factors of Φ2n3(x). Wang and Wang [15] gave the explicit factorization of Φ2n5(x). Stein [12] obtained the factors of Φr(x) when q and r are distinct odd primes. Further, assuming the explicit factors of Φr(x) are known for arbitrary odd integer r > 1, Tuxanidy and Wang [13] obtained the irreducible factors of Φ2nr(x) over Fq . Singh and Batra [11] revisited the factorization of Φ2n (x) and introduced a recurrence relation to obtain the coefficients of all irreducible factors of Φ2n (x) over Fq when q 3 (mod 4). In 1993,≡ Blake et al. [2] explicitly determined all the irreducible factors 2n of x 1 over Fp , where p is a prime with p 3 (mod 4). Chen et al. [3] ± n l ≡ obtained the explicit factorization of x2 p 1 over F , where integer l 1 − q ≥ and p is an odd prime with p (q 1). In 2015, Mart´ınez et al. [8] generalized n the results discussed in [3] by| considering− the explicit factorization of x2 d 1 over F when d = pl1 pl2 plr , where integer l 1 and p is an odd prime− q 1 2 ··· r i ≥ i such that pi (q 1) for each i =1, 2, ,r. It is natural to ask what happens | − 2nd ··· for the case of factorizing x 1 over a finite field Fq when d divides q +1. − n In this paper, the factorization of x2 d 1 into the product of irreducible − factors over Fq is obtained when d is an odd divisor of q + 1. This factor- ization gives exact information regarding the type of irreducible factors, the number of irreducible factors and their degrees. The paper is organized as follows: In Sect. 2, we introduce some basic results concerning cyclotomic polynomials and irreducibility of decompos- able polynomials over finite fields. This section also contains two recursive sequences which are useful to obtain the coefficients of irreducible factors of 2nd x 1 over Fq . In Sect. 3, we find irreducible factors of the decomposable − d cyclotomic polynomials of the type Φ2n (x ) over Fq from irreducible factors of some decomposable cyclotomic polynomials of smaller orders. In Sect. 4, n we give the explicit factorization of x2 d 1 over F via factoring the cy- − q

2 d clotomic polynomial Φ2n (x ) into the product of irreducible factors over Fq in two different cases q 1 (mod 4) and q 3 (mod 4). In addition, two ≡ ≡ numerical examples are given to validate the results.

2 Preliminaries and auxiliary results

The paper follows the standard notation of finite fields and the multiplicative ∗ group of a finite field Fq , denoted by Fq . The following classical result presents necessary and sufficient conditions to verify the irreducibility of decomposable polynomials of the form f(xk) over finite fields.

Lemma 2.1. [7, Theorem 3.35] Let f1(x), f2(x), , fN (x) be all distinct monic irreducible polynomials over F of degree l and··· order e, and let t 2 q ≥ be an integer whose prime factors divide e but not (ql 1)/e. Also assume that 4 (ql 1) if 4 t. Then f (xt), f (xt), , f (xt) are− all distinct monic | − | 1 2 ··· N irreducible polynomials over Fq of degree lt and order et. Lemma 2.2. [7, Exercise 2.57] Suppose that p is an odd prime such that gcd(2p, q)=1. Then, in Fq[x], the following properties of cyclotomic poly- nomials hold: (i) Φ = Φ ( x) if n 3 and n is odd, 2n n − ≥ pk−1 (ii) Φnpk (x) = Φnp(x ) for any positive integers n, k

p 2r (iii) Φnp(x) = Φn(x ) if p divides n. In particular, Φ2k+r (x) = Φ2k (x ) for integers k 1 and r 0. ≥ ≥ p (iv) Φnp(x) = Φn(x )/Φn(x) if p ∤ n. For a fixed integer l 1, a reduced residue system modulo l can be found by deleting from 1, 2≥, ,l, a complete residue system modulo l, those integers that are not co-prime··· to l. Note that all reduced residue system modulo l contains φ(l) integers. Lemma 2.3. [6, Lemma 7.6] Let l be a positive integer and a be any integer with gcd(a, l)=1. Let r ,r , ,r be the positive integers l and rela- 1 2 ··· φ(l) ≤ tively prime to l. Then the least residues of the integers ar1, ar2, , arφ(l) modulo k are a permutation of the integers r ,r , ,r . ··· 1 2 ··· φ(l) We need the following conventions for further developments. For each integer l 1, ν2(l) denotes the maximum power of 2 dividing l. For any odd prime≥ power q, let s = ν (q 1) 1 and m = ν (q2 1), then 2 − ≥ 2 − m s = ν2(q +1) 1. Readily note that (i) m = s + 1 if and only if q 1 (mod− 4) and (ii) s≥= 1 if and only if q 3 (mod 4). ≡ ≡ 3 Lemma 2.4. Let q and d be odd integer such that gcd(q,d)=1. If β2k is a k ∗ primitive 2 th root of unity in F 2 . Then, for 2 k m 1, the complete q ≤ ≤ − factorization of cyclotomic polynomial Φ2k (x) over Fq2 is given by:

2k−2 d(2i−1) Φ k (x)= (x β k ). (2.1) 2 ± 2 Yi=1 Proof. Let d be an odd integer and integer k 2. For a fixed k, from Lemma 2.3, the least reduced residue system of the≥ integers d, 3d, , (2k 1)d ··· − modulo 2k is a permutation of the integers 1, 3, , 2k 1. Therefore, a k d 3d d(2 −1) ··· − collection of elements β2k , β2k , , β2k is a rearrangement of the elements k 3 2 −1 ··· ∗ d(2i−1) d(2j−1) β k , β k , , β k in some order in F 2 with β k = β k for 1 i< 2 2 ··· 2 q 2 6 2 ≤ j 2k−1 . For any 2 k m 1 and q is odd, we get 2k (q2 1). Since d≤(2i−1) k≤−1 ≤ k−1 ∗ | − β k for 1 i 2 are 2 distinct elements in F 2 , so we find that 2 ≤ ≤ q 2k−1 d(2i−1) Φ k (x)= (x β k ). 2 − 2 Yi=1 Using the fact k−2 k−1 d(2(2 +i)−1) d(2 )+d(2i−1) d(2i−1) k−2 β k = β k = β k for 1 i 2 , 2 2 − 2 ≤ ≤ the factorization of Φ2k (x) can be viewed as follow:

2k−2 2k−1 d(2i−1) d(2i−1) Φ k (x) = (x β k ) (x β k ) 2 − 2 − 2 Yi=1 i=2Yk−2+1 2k−2 d(2i−1) = (x β k ). ± 2 Yi=1

The following remark to Lemma 2.4 is easily deduced.

k Remark 2.1. If q 1 (mod 4) and α2k is a primitive 2 th root of unity in ∗ ≡ F . Then β k = α k for 2 k m 1= s. By Lemma 2.4 it follows that q 2 2 ≤ ≤ − 2k−2 d(2i−1) Φ k (x)= (x α k ). (2.2) 2 ± 2 Yi=1 Also if q 3 (mod 4), then 2k ∤ (q 1), but 2k (q +1) for 2 k m 1. ≡ − k | ≤ ≤ − In this case Φ2k (x) is the product of φ(2 )/2 trinomails of degree 2 over Fq (see [11, Lemma 2.6]).

4 The following three lemmas will be useful in the proof of Theorem 3.2 and Theorem 3.3.

∗ Lemma 2.5. For any positive integer ℓ relatively prime with q, let c Fq such that c = aℓ for some a F . Then ∈ ∈ q ℓ−1 xℓ c = (x abj), (2.3) − − Yj=0 where b is a primitive ℓth root of unity in some extension field of Fq . Further, if ℓ is an odd divisor of q +1, the explicit factorization of xℓ c over F is − q given by (ℓ−1)/2 xℓ c =(x a) (x2 a(bj + bqj)x + a2). (2.4) − − − Yj=1 Proof. Since gcd(ℓ, q) = 1, so there exists a least positive integer l such l ∗ that ℓ (q 1). Let b be a primitive ℓth root of unity in Fql . Let c Fq | − ℓ ∗ ℓ−1 ∈ such that c = a for some a Fq . Note that a, ab, , ab are ℓ distinct ∗ ∈j ℓ ℓ ℓ j ℓ ··· elements in F l such that (ab ) = a (b ) = a = c for all 0 j ℓ 1. q ≤ ≤ − It follows that x abj is a monic factor of xℓ c for every 0 j ℓ 1. Therefore the polynomial− xℓ c is given by− the product of≤ polynomials≤ − ℓ−1 − x a, x ab, , x ab over F l . − − ··· − q Further, let ℓ be an odd divisor of q + 1 with gcd(q,ℓ) = 1. Clearly j ∗ ℓ ∤ (q 1). Then b Fq2 for all 0 j ℓ 1. For any 1 j ℓ 1, we − j ∈ ≤ ≤ −j ≤ ≤ − note that b / Fq . On contrary, assume that b Fq for some 1 j ℓ 1, then qj j∈(mod ℓ). Since ℓ ∤ (q 1), so j∈ 0 (mod ℓ), which≤ ≤ is− not ≡ − ≡j possible because 1 j ℓ 1. This proves that b / Fq for all 1 j ℓ 1. Further, from Expression≤ ≤ − 2.3, it follows that ∈ ≤ ≤ −

(ℓ−1)/2 ℓ−1 xℓ c = (x a) (x abj) (x abj) − − − − Yj=1 j=(Yℓ+1)/2 (ℓ−1)/2 (ℓ−1)/2 = (x 1) (x abj) (x abℓ−j). − − − Yj=1 Yj=1

qj −j ℓ−j j qj Also, for all 1 j (ℓ 1)/2, since b = b = b and a(b + b ) Fq as 2 (a(bj + bqj))q =≤aq(≤bqj +−bq j)= a(bqj + bj), so that the minimal polynomial∈ of bj over F is (x abj)(x abqj)= x2 a(bj + bqj)x + a2 . Therefore, q − − − (ℓ−1)/2 xℓ c =(x a) (x2 a(bj + bqj)x + a2). − − − Yj=1

5 This completes the proof. In the following results we introduce two recursive sequences which are to be useful later on in describing the coefficients of irreducible factors of d Φ2k (x ) over Fq . Lemma 2.6. Let d be an odd divisor of q +1 and γ be a primitive dth root j qj j qj of unity in F 2 . Let δ = γ + γ and θ = γ γ for 0 j d 1. q j j − ≤ ≤ − Then the recursive sequences (δj)j≥0 of length (d + 1)/2 in Fq and (θj)j≥0 of length din Fq2 are given by: δ = δ δ δ with δ =2 and δ = γ + γ−1 and j 1 j−1 − j−2 0 1 θ = δ θ θ with θ =0 and θ = γ γ−1 . j 1 j−1 − j−2 0 1 − satisfying δ = d , θ = θ and δ2 2 = θ2 +2 = δ for 0 j d−j j d−j − j j − j 2j ≤ ≤ (d 1)/2. − ∗ Proof. Let γ be a primitive dth root of unity in F 2 and d (q +1). Let δ = q | j γj + γqj and θ = γj γqj for 0 j d 1. Since γqj = γ−j = γd−j for all j − ≤ ≤ − 0 j (d 1)/2, we get that δj = δd−j and θj = θd−j for 1 j (d 1)/2 with≤ δ≤ =− 2 and θ = 0. It is easy to verify that− δ = δ δ≤ ≤δ − and 0 0 j 1 j−1 − j−2 θj = δ1θj−1 θj−2 for j 2. Also, for any 0 j (d 1)/2, we establish the relations−δ2 2= γ2≥j + γ2qj = δ and θ2 ≤+2=≤ γ2j−+ γ2qj = δ . j − 2j j 2j Lemma 2.7. Let ∆ = δ : 1 j (d 1)/2 . Then ∆ has (d 1)/2 d { j ≤ ≤ − } d − distinct elements of F∗ and the mapping σ : ∆ ∆ defined by σ (δ )= q d d → d d j δ2j is a permutation. Proof. Let δ , δ in ∆ such that δ = δ , where 1 j , j (d 1)/2. j1 j2 d j1 j2 ≤ 1 2 ≤ − By Lemma 2.6, we have j2 = d j1 , a contradiction to the choice of j1 and j . Thus ∆ has (d 1)/2 distinct− elements. By Lemma 2.6, δ2 2 = δ 2 d − j − 2j for 1 j (d 1)/2. If 1 2j (d 1)/2, then δ2j ∆d , and if (d + 1)≤/2 ≤2j −d 1, then 1≤ t ≤(d −1)/2 for t = d ∈2j and hence δ = δ ≤= δ ≤∆ −. It follows that≤ ≤δ −∆ for every 1 −j (d 1)/2. 2j d−t t ∈ d 2j ∈ d ≤ ≤ − Now, define σd : ∆d ∆d such that σd(δj) = δ2j . Obviously, σd is well defined mapping. Further,→ let σ (δ ) = σ(δ ) for 1 j , j (d 1)/2, d j1 j2 ≤ 1 2 ≤ − then δ = δ . This gives either δ = δ or δ = δ . If δ = δ , 2j1 2j2 j1 j2 j1 − j2 j1 − j2 then γj1+j2 = 1 or γj1−j2 = 1, which is not possible because the order of γ is odd. Thus− for any 1 j −, j (d 1)/2, σ (δ )= σ (δ ) implies that ≤ 1 2 ≤ − d j1 d j2 δj1 = δj2 and hence σd is a permutation.

6 d 3 Factorization of Φ n(x ) F [x] 2 ∈ q In this section, we study a factorization of decomposable cyclotomic polyno- d mials of the type Φ2n (x ) over Fq , where d is an odd divisor of q + 1 and integer n 1. We begin with the more general factorization: ≥ Lemma 3.1. Let q and d be odd integers. Then, for any integer n 2, ≥ n d d d  (x 1)(x + 1) Φ2k (x ) for 2 n m 1 n − ≤ ≤ − 2 d  Yk=2 x 1=  n−m −  2m−1d 2rd (x 1) Φ m (x ) for n m. − 2 ≥  Yr=0  Proof. For any integer n 2, it is trivial to note that ≥ 2n x 1=(x 1)(x + 1) Φ k (x). − − 2 2≤Yk≤n For a moment if n m, then we express ≥ 2n m−1 x 1=(x 1) Φ k (x). − − 2 m≤Yk≤n r Let n = m+r, then by Lemma 2.2, Φ n (x) = Φ m (x2 ), where 0 r n m. 2 2 ≤ ≤ − By applying x xd , we can easily meet the desired conclusion. → In view of the above, our idea for obtaining the explicit factorization of 2nd x 1 over Fq is to factorize decomposable cyclotomic polynomials of the − d form Φ2k (x ) of smaller orders into the product of irreducible factors over Fq . d d d d d Note, Φ1(x )= x 1, Φ2(x ) = Φ1( x )= x +1 and for 2 k m 1, − − d 2k−1d≤ ≤ − from Lemma 2.4, the cyclotomic polynomial Φ2k (x ) = x + 1 over Fq2 can be written as: 2k−2 d d d(2i−1) Φ k (x )= (x β k ). (3.1) 2 ± 2 Yi=1 Since 2k (q + 1) if q 3 (mod 4), and 2k (q 1) if q 1 (mod 4), so we have two| distinguishable≡ cases as follows: | − ≡ Theorem 3.2. Let q 1 (mod 4) and q 1 (mod d). Then ≡ ≡ − d k−2 (i) If 2 k s, the factorization of Φ k (x ) into 2 (d + 1) irreducible ≤ ≤ 2 factors over Fq is given by:

2k−2 d 2i−1 2 2i−1 2i−1 Φ k (x )= (x α k )(x α k δ x + α k−1 ). (3.2) 2 ± 2 ± 2 j 2 Yi=1 1≤j≤(d−1)/2

7 d s−1 (ii) For any integer r 1, the factorization of Φ s+r (x ) into 2 d irre- ≥ 2 ducible factors over Fq is given by:

2s−2 d 2r 2i−1 2r l 2i−1 2r−1 l 2i−1 Φ s+r (x )= (x α k )(x α β s+1 θ x α α s ) 2 ± 2 ± 4 2 j − 2 2 Yi=1 1≤j≤(d−1)/2 0≤l≤1 (3.3).

Proof. If q 1 (mod 4), then m = s + 1. It follows that 2k (q 1) and ≡ s+1 | − β2k = α2k for 2 k s and 2 ∤ (q 1) as s = ν2(q 1). Now we have the following two≤ cases:≤ − − d Case (i) For 2 k s, from Expression 3.1, a factorization of Φ k (x ) over ≤ ≤ 2 Fq is given by:

2k−2 d d d(2i−1) Φ k (x )= (x α k ). 2 ± 2 Yi=1 For each 1 i 2k−2 , from Expression 2.4, we find the following factor- ≤d ≤d(2i−1) ization of x α k into the product of d + 1 irreducible factors over ± 2 Fq

(d−1)/2 d d(2i−1) 2i−1 2 2i−1 2i−1 x α k = (x α k ) (x α k δ x + α k−1 ) (3.4) ± 2 ± 2 ± 2 j 2 Yj=1

d k−2 and hence the number of irreducible factors of Φ2k (x ) over Fq is 2 (d+1). Case (ii) If q 1 (mod 4), then 2s+1 ∤ (q 1), but 2s+1 (q2 1). Applying ≡ − | −d 2d Lemma 2.2, Expression 3.2 yields a factorization of Φ2s+1 (x ) = Φ2s (x ) s−2 over Fq with 2 (d + 1) factors as follows:

2s−2 (d−1)/2 d 2 2i−1 4 2i−1 2 2i−1 Φ s+1 (x )= (x α s ) (x α s δ x + α s−1 ). 2  ± 2 ± 2 j 2  Yi=1 Yj=1 (3.5)

2i−1 ∗ 2 2i−1 Since α2s is a non-square element in Fq , so x α2s is irreducible over − 4 2i−1 2 2i−1 s Fq . Now, we check the irreducibility of trinomials x α2 δjx + α2s−1 , where 1 i 2s−2 and 1 j (d 1)/2 by applying conditions− of Lemma ≤ ≤ 2 2i−1 ≤ ≤2i−1− s−2 s 2.1. Let f(x)= x α2 δj + α2s−1 , where fixed i and j are as 1 i 2 − 2i−1 2i−1 ≤ ≤ 2 4 s 2 and 1 j (d 1)/2. Then f(x )= x α2 δjx + α2s−1 . Observe that ≤ ≤ − 2i−1 −j f(x) is the minimal polynomial of α2s γ over Fq so the order of f(x) is

8 s 2 2 d1 , where d1 = d/ gcd(j, d). Recall m = ν2(q 1) and s = ν2(q 1). Since q 1 (mod 4), so that m = s+1 and hence gcd(2− , (q2 1)/2sd )− = 2. This ≡ − 1 goes against the condition (ii) of Lemma 2.1. Therefore f(x2) is reducible 2 2i−1 over Fq . Using the similar argument given in [7, Theorem 3.14], x + α2s 2 4 2i−1 2 2i−1 is irreducible while f( x )= x + α s δ x + α s−1 is reducible over F . − 2 j 2 q It follows that Expression 3.5 contains 2s−1 irreducible binomials and 2s−2(d 1) reducible trinomials of degree 4 over F . Our aim is to split these − q 2s−2(d 1) trinomials into a product of 2s−1(d 1) trinomials of degree 2 − − over Fq . (d−1)/2 4 2i−1 2 2i−1 s−2 First consider x α s δ x + α s−1 , a product of 2 (d 1)/2 − 2 j 2 − Yj=1
trinomials from Expression 3.5. Applying Lemma 2.7, we have that

(d−1)/2 (d−1)/2 4 2i−1 2 2i−1 4 2i−1 2 2i−1 x α s δ x + α s−1 = x α s σ (δ )x + α s−1 − 2 j 2 − 2 d j 2 Yj=1
Yj=1
(d−1)/2 4 2i−1 2 2 2i−1 = x α s (θ + 2)x + α s−1 − 2 j 2 Yj=1
(d−1)/2 2 2i−1 2 2i−1 2 = (x α s ) (β s+1 θ x) − 2 − 2 j Yj=1
(d−1)/2 2 2i−1 2i−1 = (x β s+1 θ x α s ). ± 2 j − 2 Yj=1

Applying the substitution x α x, it is quite easy to see → 4

(d−1)/2 (d−1)/2 4 2i−1 2 2i−1 2 2i−1 2i−1 x + α s δ x + α s−1 = (x α β s+1 θ x + α s ). 2 j 2 ± 4 2 j 2 Yj=1
Yj=1 Combining the above two expressions, we get

(d−1)/2 (d−1)/2 4 2i−1 2 2i−1 2 l 2i−1 l 2i−1 x α s δ x + α s−1 = (x α β s+1 θ x α α s ). ± 2 j 2 ± 4 2 j − 2 2 Yj=1
Yj=1 0≤l≤1

q−1 q 2i−1 q 2i−1 q−1 2i−1 Since β2s+1 = 1 and θj = θj so that (β2s+1 θj) =(β2s+1 ) β2s+1 ( θj)= 2i−1 − 2i−1 q −2i−1 q 2i−1 2i−1 −2i−1 β2s+1 θj and (α4β2s+1 θj) =(β2s+1 θj) = β2s+1 θj , hence β2s+1 θj and α4β2s+1 θj

9 ∗ s d are elements of Fq . Using the above discussion, Φ2 +1 (x ) can be written as s−2 d−1 s−1 a product of 2 (2+( 2 )4) = 2 d irreducible factors over Fq as follows:

2s−2 d 2 2 l 2i−1 l 2i−1 Φ s+1 (x ) = Φ s (x ) (x α β s+1 θ x α α s ) (3.6) 2 2 ± 4 2 j − 2 2 Yi=1 1≤j≤(d−1)/2 0≤l≤1

r−1 Applying x x2 in Expression 3.6, we get → 2s−2 d 2r 2r l 2i−1 2r−1 l 2i−1 Φ s+r (x ) = Φ s (x ) (x α β s+1 θ x α α s ). (3.7) 2 2 ± 4 2 j − 2 2 Yi=1 1≤j≤(d−1)/2 0≤l≤1

2 2i−1 2i−1 s−2 s Let g(x)= x β2s+1 θjx + α2 , where 1 i 2 and 1 j (d 1)/2. − ≤ ≤ 2i−1 j ≤ ≤ − Observe that g(x) is the minimal polynomial of β2s+1 γ over Fq and hence s+1 the order of g(x) is 2 d2 , where d2 is a divisor of d. Since s +1 = ν (q2 1), 2 ∤ (q2 1)/2s+1 and hence gcd(2r, (q2 1)/2s+1d ) = 1 for 2 − − − 2 every r 1. From Lemma 2.1 and [7, Theorem 3.14] for s 2, it follows ≥2r−1 2r−1 ≥ s+r that g(x ), g( x ) are irreducible over Fq of same orders 2 d2 . Since − 2r−1 s+r s 2, so the order of g( α4x ) is 2 d2 , It follows that the irreducibility ≥ 2r−1 ± of g( α4x ) over Fq can be verified in similar fashion as discussed for r−1 g( x±2 ). Therefore, Expression 3.7 has 2s−1d irreducible factors over F ± q for every r 1. This completes the proof. ≥

Example 3.1. For q = 29, n =3 and d =5, we obtain α4 = 12, β8 =2√3, −1 γ = 3+15√3, γ = 3 15√3, θ1 = √3 and θ2 = 6√3. By Theorem − − − 5 − 3.2, the explicit factorization of Φ8(x ) over F29 is given by

Φ (x5) = Φ (x) (x2 (12)l(2√3)θ x ( 1)l12) 8 8 ± j − − 1≤Yj≤2 0≤l≤1 = Φ (x)(x2 (2√3)θ x 12)(x2 (2√3)θ x 12) 8 ± 1 − ± 2 − (x2 (12)(2√3)θ x + 12)(x2 (12)(2√3)θ x + 12) ± 1 ± 2 = (x2 + 12)(x2 + 17)(x2 6x + 17)(x2 7x + 17) ± ± (x2 14x + 12)(x2 3x + 12). ± ± 5 Φ8(x ) 2 2 As we know that Φ40(x)= and Φ8(x)=(x 12)(x + 12), so that Φ8(x) −

Φ (x)=(x2 6x + 17)(x2 7x + 17)(x2 14x + 12)(x2 3x + 12). 40 ± ± ± ± 10 Moreover, for any integer n 3, ≥ 2n−3 Φ2n5˙ (x) = Φ40(x ) n−2 n−3 n−2 n−3 = (x2 6x2 + 17)(x2 7x2 + 17) n−2 ± n−3 n−2± n−3 (x2 14x2 + 12)(x2 3x2 + 12). ± ± Theorem 3.3. Let q 3 (mod 4) and q 1 (mod d). Then the factor- d ≡ k−2 ≡ − ization of Φ2k (x ) into 2 d irreducible factors over Fq is given by:

2k−2 2  (x θi,j,kx + 1) for 2 k m 1 − ≤ ≤ −  Yi=1 d  0≤j≤d−1 Φ k (x )=  m−3 2  2  k−m+1 k−m (x2 θ x2 1) for k m 3. ± i,j − ≥ ≥  Yi=1  0≤j≤d−1   (3.8)

2i−1 j q(2i−1) qj k−2 2i−1 j m where θi,j,k = β2k γ + β2k γ for 1 i 2 and θi,j = β2 γ + q(2i−1) qj m−3 ≤ ≤ β m γ for 1 i 2 . The complete factorization of Φ k (x) F [x] 2 ≤ ≤ 2 ∈ q is given in [11, Lemma 2.6].

Proof. Let q 3 (mod 4) and q 1 (mod d). Then q2 1 (mod d) and 2 ≡ ≡ − ≡ d m = ν (q 1) and from Expression 3.1, the cyclotomic polynomial Φ k (x ) 2 − 2 over Fq2 in the following form:

2k−2 d d d(2i−1) d d(2i−1) Φ k (x ) = (x β k )(x + β k ). 2 − 2 2 Yi=1

Using the permutation i 2k−2 i + 1, we have that 7→ − 2k−2 2k−2 d d(2i−1) d d(−2i+1) (x + β k ) = (x β k ). 2 − 2 Yi=1 Yi=1

k q+1 For each fixed k, where 2 k m 1, since 2 (q + 1), we have β2k =1 qd(2i−1) d(−≤2i+1) ≤ − | k d 2 and hence β2k = β2k . Thus the factorization of Φ2 (x ) over Fq reduces to

2k−2 d d d(2i−1) d qd(2i−1) Φ k (x ) = (x β k )(x β k ), 2 − 2 − 2 Yi=1

11 In view of Expression 2.3 of Lemma 2.5, we have

2k−2 d 2i−1 j q(2i−1) j Φ k (x ) = (x β k γ )(x β k γ ). 2 − 2 − 2 Yi=1 0≤j≤d−1 Since gcd(q,d) = 1, so the least reduced residue system qj (mod d):0 j d 1 is a rearrangement of integers 0, 1, ,d 1.{ Therefore ≤ ≤ − } ··· − 2k−2 d 2i−1 j q(2i−1) qj Φ k (x )= (x β k γ )(x β k γ ). 2 − 2 − 2 Yi=1 0≤j≤d−1

k−2 2i−1 j For any 1 i 2 , the minimal polynomial of β k γ is ≤ ≤ 2 2 2i−1 j q(2i−1) qj (q+1)(2i−1) x (β k γ + β k γ )x + β k . − 2 2 2 d Therefore, the irreducible factorization of Φ2k (x ) over Fq is

2k−2 d 2 Φ k (x ) = (x θ x + 1) (3.9) 2 − i,j,k Yi=1 0≤j≤d−1

2i−1 j q(2i−1) qj k−2 with θi,j,k = β2k γ + β2k γ for 1 i 2 , 1 j d 1 and d 2d≤ ≤ ≤ ≤ − 2 k m 1. Since Φ m (x ) = Φ m−1 (x ). It follows that, from Expression ≤ ≤ − 2 2 3.9,

2m−3 d 4 2 Φ m (x )= (x θ x + 1). 2 − i,j,m−1 Yi=1 0≤j≤d−1 Since d is odd, so the least residue system 2j : 0 j d 1 is a rearrangement of integers 0, 1, ,d 1. Therefore{ ≤ ≤ − } ··· − 2m−3 d 4 2 Φ m (x ) = (x θ x + 1). 2 − i,2j,m−1 Yi=1 0≤j≤d−1

2i−1 j q(2i−1) qj m−3 Now for m 3, denote θ = β m γ + β m γ for 1 i 2 and ≥ i,j 2 2 ≤ ≤ 0 j d 1. Since m 1 = ν2(q + 1) 2 for q 3 (mod 4), so that q≤+1 ≤ − −2 ≥ ≡ m−3 β m = 1. Note that θ = θ 2 for every 1 i 2 and 2 − i,j i,2j,m−1 − ≤ ≤ 0 j d 1. It follows that ≤ ≤ − x4 θ x2 +1=(x2 1)2 θ2 x2 =(x2 θ x 1)(x2 + θ x 1). − i,2j,m−1 − − i,j − i,j − i,j − 12 Therefore

2m−3 d 2 2 Φ m (x )= (x θ x 1)(x + θ x 1). (3.10) 2 − i,j − i,j − Yi=1 0≤j≤d−1

This proves the result.

n 4 The factorization of x2 d 1 when d (q + 1)

− n | This section determines the explicit factorization of x2 d 1 over F , where d − q is an odd divisor of q + 1 and integer n 1. We begin with the factorization of xd 1 and xd + 1 over F . ≥ − q

Lemma 4.1. Let Fq be a finite field and, let d be an odd integer such that d (q + 1). Then | (d−1)/2 xd 1=(x 1) (x2 δ x + 1) − − − j Yj=1 and

(d−1)/2 d 2 x +1=(x + 1) (x + δjx + 1), Yj=1

j qj ∗ where δj = γ + γ Fq for every 1 j d 1 and γ be a primitive dth ∗ ∈ ≤ ≤ − root of unity in Fq2 . Proof. The result follows directly by substituting b = γ and c = 1 in Expression 2.4 of Lemma 2.5. ± In view of Lemma 4.1,

(d−1)/2 x2d 1=(x 1) (x2 δ x + 1). − ± ± j Yj=1 where q 1 (mod d). Further, for each 2 n m 1, since 2n (q 1) if q 1 (mod≡ − 4) and 2n (q + 1) if q 3 (mod≤ 4),≤ therefore− the factorization| − n of≡x2 d 1 over F will| be determined≡ for q 1 (mod 4) and q 3 (mod 4) − q ≡ ≡ separately.

13 4.1 Case: q 1 (mod 4) ≡ Theorem 4.2. Let Fq be a finite field with q 1 (mod 4) elements and d be an odd integer such that q 1 (mod d). ≡ ≡ − n (i) If 2 n s, then the factorization of x2 d 1 into the product of n−1 ≤ ≤ − 2 (d + 1) monic irreducible factors over Fq is given by:

2k−2 2nd 2i−1 x 1 = (x 1)(x + 1) (x α k ) − − ± 2 × Yi=1 2≤k≤n 2k−2 2 2 2i−1 2i−1 (x δ x + 1)(x α k δ x + α k−1 ) .  ± j ± 2 j 2  Yi=1 1≤j≤(d−1)/2 2≤k≤n

n (ii) If n>s 2, the factorization of x2 d 1 into the product of 2s−1((n ≥ − − s + 1)d + 1) monic irreducible factors over Fq is given by:

2s−2 2nd 2sd 2r 2i−1 2r l 2i−1 2r−1 l 2i−1 x 1 = (x 1) (x α s ) x α β s+1 θ x α α s − −  ± 2 ± 4 2 j − 2 2  Yi=1
1≤r≤n−s 1≤j≤(d−1)/2 0≤l≤1

s where the complete factorization of x2 d 1 is given in Case (i). − Proof. If q 1 (mod 4), then m = s + 1. By Lemma 3.1, the factorization n ≡ of x2 d 1 over F can be viewed as: − q n d d d  (x 1)(x + 1) Φ2k (x ) for 2 n s n − ≤ ≤ 2 d  Yk=2 x 1=  n−s −  2sd 2rd (x 1) Φ s (x ) for n>s 2. − 2 ≥  Yr=1   Case (i) For any 2 n s, by Lemma 4.1 and Theorem 3.2, the factor- n ≤ ≤ ization of x2 d 1 into the product of 2n−1(d + 1) monic irreducible factors can be obtained.− Case (ii) For n>s, by Theorem 3.2, we obtain the required form of the 2nd factorization of x 1 over Fq with the number of monic irreducible factors 2nd − 2sd of x 1 over Fq is the sum of the number of irreducible factors of x 1 − (d 1) − and 2s−2(n s)2+2s−2(n s) − 4.˙ In view of case (i), the number of − − 2 14 2sd s−1 monic irreducible factors of x 1 over Fq is 2 (d + 1) and hence the n number of monic irreducible factors− of x2 d 1 over F is − q 2s−1(d +1)+2s−1(n s)d =2s−1((n s + 1)d + 1). − − This completes the proof.

4.2 Case: q 3 (mod 4) ≡ Theorem 4.3. Let q 3 (mod 4) and d be an odd integer such that q 1 (mod d). Then ≡ ≡ −

n (i) If 2 n m 1, the factorization of x2 d 1 into the product of n−1 ≤ ≤ − − 2 d +1 irreducible factors over Fq is given by:

(d−1)/2 2k−2 n x2 d 1 = (x 1)(x + 1) (x2 δ x + 1) (x2 θ x + 1), − − ± j − i,j,k Yj=1 Yi=1 2≤k≤n 0≤j≤d−1

2i−1 j q(2i−1) qj k−2 where θ = β k γ + β k γ for 1 i 2 and 0 j d 1. i,j,k 2 2 ≤ ≤ ≤ ≤ − n (ii) If n m 3, the factorization of x2 d 1 into the product of ≥ ≥ − 2m−2d(n m +2)+1 irreducible factors over F is given by: − q 2m−3 n m−1 r+1 r x2 d 1=(x2 d 1) (x2 θ x2 1). − − ± i,j − Yi=1 0≤r≤n−m 0≤j≤d−1

2i−1 j q(2i−1) qj m−3 where θi,j = β2m γ + β2m γ for 1 i 2 , 0 j d 1. m−1 The complete factorization of x2 d 1≤over≤ F is a special≤ ≤ case− of − q the factorization given above in (i).

Proof. Let d be an odd divisor of q + 1. Then, by Lemma 3.1,

n d d d  (x 1)(x + 1) Φ2k (x ) for 2 n m 1 n − ≤ ≤ − 2 d  Yk=2 x 1=  n−m −  2m−1d 2rd (x 1) Φ m (x ) for 3 m n. − 2 ≤ ≤  Yr=0  

15 2nd Case (i) For any 2 n m 1, the factorization of x 1 over Fq follows from Lemma 4.1 and≤ Theorem≤ − 3.3 such as: − (d−1)/2 2k−2 n x2 d 1 = (x 1)(x + 1) (x2 δ x + 1) (x2 θ x + 1), − − ± j − i,j,k Yj=1 Yi=1 2≤k≤n 0≤j≤d−1

2i−1 j q(2i−1) qj k−2 where θi,j,k = β k γ + β k γ for 1 i 2 , 0 j d 1 and 2 2 n 2 k m 1. Further, the number of irreducible≤ ≤ factors≤ of x≤2 d −1 is ≤ ≤ − − 1+1+(d 1) + 2k−2d =(d +1)+ d(2n−1 1)=2n−1d +1. − − 2≤Xk≤n 2nd Case (ii) For any n m 3, the factorization of x 1 over Fq follows from Case (i) for n =≥m ≥1 and Theorem 3.3 such as: − − 2m−3 n m−1 r+1 r x2 d 1 = (x2 d 1) (x2 θ x2 1). − − ± i,j − Yi=1 0≤j≤d−1 0≤r≤n−m

2r+1 2r By Lemma 2.1, it is quite easy to verify that trinomials x θi,jx 1 n over F are irreducible over F . The number of irreducible factors± of x2 d− 1 q q − is 2m−2d +1+2m−2d(n m +1)=2m−2d(n m +2)+1. − − This finishes the proof.

Worked examples

Example 4.1. Let q = 29 and n 3. Then s =2, d = 15, α4 = 12, α8 = −1 ≥ 2√3, γ =2+ √3, γ =2 √3, δ1 =4 and θ1 =2√3. By Lemma 2.6, we obtain δ = δ δ 2=14, δ−= δ δ δ = 23, δ =4 23 14 = 20, δ = 28, 2 1 1 − 3 1 2 − 1 4 · − 5 δ6 = 5 and δ7 = 21. Further, again by Lemma 2.6, θ2 = δ1θ1 = 8√3, θ3 = δ1θ2 θ1 = √3, θ4 = 4√3, θ5 = 12√3, θ6 = 6√3, θ7 = 7√3. By n Theorem− 4.2 the explicit factorization− of x2 15 1 over− F is given− by: − 29 7 n r r r−1 x2 15 1 = (x60 1) (x2 α ))(x2 αl α θ x2 αl α − − ± 4 ± 4 8 j − 2 4 Yj=1
1≤r≤n−2 0≤l≤1 with 7 x60 1 = (x 1)(x + 1)(x α )(x + α ) (x2 δ x + 1) x2 α δ x + α . − − − 4 4 ± j ± 4 j 2 Yj=1

16 Example 4.2. Let q = 59 and n = 4. Then m = 3 and d = 15. As q 3 (mod 8), √2 / F∗ , we first notice that β = 18√2, β−1 = ≡ ∈ 59 4 4 1 −1 1 18√2, β8 = 18+ , and hence β = 18 + . Further, γ = − √2 8 − √2 5 22√2, γ−1 = 5 +22√2, γ2 = 10 16√2 and δ : 1 j − − − − − { j ≤ ≤ 7 = 10, 20, 26, 15, 1, 25, 13 . It is easy to see that β4, β8,γ are } {− − − − − ∗− } elements in F 2 F as √2 / F . By Theorem 4.3, the factorization of 59 \ 59 ∈ 59 x240 1 over F is given by − 59 r+1 r x240 1=(x60 1) (x2 θ x2 1), − − ± 1,j − 0≤Yr≤1 0≤j≤14 where θ = β γj β−1γ−j for 0 j 14 and 1,j 8 − 8 ≤ ≤ 7 x60 1=(x4 1) (x2 δ x + 1) (x2 θ x + 1) − − ± j − 1,j,2 Yj=1 1≤Yj≤14 with θ = β γj + β−1γ−j for every 1 j 14. 1,j,2 4 4 ≤ ≤ Acknowledgments

The author would like to thank the anonymous referee of an earlier version that allowed us to discover a gap in the proof of Theorem 3.2.

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