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2 Divisibility of generalized reciprocal polynomi- als

In this section we summarize some properties for the divisibility of generalized reciprocal and generalized self-reciprocal polynomials over Fq. Theorem 1. a-reciprocal of a product of two polynomials is the product of a- reciprocals of given polynomials.

n i m j Proof. Let f(x) = i=0 bix , g(x) = j=0 cj x , (b0 = 0,c0 = 0). From the definition, 6 6 P P n n 1 i n−i 1 n−i i fˆa(x)= bia x = bn−ia x , b0 b0 i=0 i=0 mX mX 1 j m−j 1 m−j j gˆ (x)= c a x = c − a x , a c j c m j 0 j=0 0 j=0 X X

n+m 1 n−i m−j k fˆa(x) gˆa(x)= bn−ia cm−j a x · b0c0   k=0 i+j=k X X   n+m   1 n+m−k k = bn−i cm−j a x b0c0  · ·  Xk=0 i+Xj=k   and

n+m n+m f(x) g(x)= b c xk = d xk d := b c . ·  i j  k  k i j  Xk=0 i+Xj=k Xk=0 i+Xj=k     Therefore n+m n+m 1 k n+m−k 1 n+m−k k fga(x)= dka x = dn+m−ka x b0c0 b0c0 kX=0 Xk=0 c 2 We set i′ := n i, j′ := m j, then − −

dn+m−k = bn−i′ cm−j′ ′ ′ i +Xj =k and thus n+m 1 − ′ ′ n+m k k fga(x)= bn−i cm−j a x . b0c0  ′ ′ · ·  kX=0 i +Xj =k c  

It might be fˆa(x)= fˆb(x) for nonzero distinct elements a,b in Fq. For example, if f(x)= x2 + c F [x], c = 0, then fˆ (x)= fˆ (x). ∈ 5 6 2 3 A monic polynomial f(x) Fq[x] is said to be a-self reciprocal when fˆa(x) = f(x). In [4], the notion of a-self∈ reciprocal polynomial has been defined for only n i even degree. Note that f(x)= i=0 bix is a-self reciprocal if and only if for each i 2 n i, bn−ib0 = bia . When i = n, we see that b0 = a . P n First consider a case when n is odd. If a and thus a is not a square in Fq, then there is not any a-self reciprocal monic (srm) polynomial of degree n. When n 0 = a Fq is a square in Fq, if f(x) is a-srm polynomial then either b0 = √a or b 6 = ∈√an. 0 −

Theorem 2. Let n be an odd, 0 = a Fq be a square in Fq and f(x) Fq[x] be a-srm polynomial of degree n. 6 ∈ ∈ n (1) If b0 = √a then f(x) is divided by x + √a. (2) If b = √an then f(x) is divided by x √a. 0 − − n Proof. Suppose b0 = √a . Then we have

n n−1 n f √a = √a + b − √a + + b √a + √a − − n 1 − ··· 1 − n−1  = bn−1  √a + + b1 √a ,   − ··· − 1 n n n−1 n fˆa √a = √a √a + +bn−1a √a + a − −( √a)n − ··· − −  n   n−1  n  = √a + b √a + + b − √a + √a − 1 − ··· n 1 − − h n−1 i = b √a + + b − √a = f √a  − 1 − ··· n 1 − − − h   i  and from f = fˆa, f ( √a) = f ( √a) which implies that f ( √a) = 0. The second case follows similarly.− − − −

Corollary 1. If 0 = a F is a square in F , then x + √a and x √a are the 6 ∈ q q − only a-srim polynomials of odd degree over Fq.

2m i Next let n be an even, that is, n = 2m. f(x) = i=0 bix is said to be trivial m m or nontrivial respectively, according to b0 = a or b0 = a [4]. If f(x) is a trivial a-srm polynomial, then f (√a) = f ( √−a) =P 0, hence f(x) is a multiple of x2 a. Therefore x2 a is the only trivial a−-srim polynomial. From the definition, − m − m bmb0 = bma and b0 = a , so that bm = 0, hence trivial a-srm polynomials have always even terms. −

3 2 Lemma 1. x2 a is a trivial a-srm polynomial and x2 a is a nontrivial one. − − Lemma 2. The product of two a-srm polynomials satisfies the following multipli- cation table. trivial nontrivial trivial· nontrivial trivial nontrivial trivial nontrivial

The proof of the above lemmas are very simple so we omit.

Theorem 3. If f(x) Fq[x] is a-srm polynomial of even degree, then it can be ∈ k written as f(x)= x2 a g(x), where g(x) is a nontrivial a-srm polynomial not divided by x2 a and − · −  odd, f(x) is trivial, k = . even, f(x) is nontrivial  2 2 Proof. Suppose that x a divides f(x). Then f(x) = x a f1(x) for some f (x) F [x]. If f(x) is− trivial, then f (x) is nontrivial by− Theorem· 1 and the 1 ∈ q 1  above lemmas. And if f(x) is nontrivial, then f1(x) is trivial, so it is divided by 2 2 2 x a and it can be written as f(x) = x a f2(x). Again from the above lemmas− f (x) is nontrivial. Continuing this procedure− · implies the claim. 2  If f(x) F [x] is divided by x2 a, then √a and √a are roots of f(x). ∈ q − − If a is a square in Fq[x], then there might be an a-srm polynomial which is not divided by x2 a but has a root √a or √a. For example, if q = 5,a = 4, then √a =2, √a =− 3, − − f(x)= x6 + x5 +3x4 +4x3 +2x2 + x +4 F [x] ∈ 5 is a nontrivial 4-srm polynomial which is not divided by x2 4 but f(2) = 0 and − g(x)= x6 +3x4 +3x3 +2x2 +4 F [x] ∈ 5 is a nontrivial 4-srm polynomial which is not divided by x2 4 but g(3) = 0. − Theorem 4. Let f(x) Fq[x] be a nontrivial a-srm polynomial of degree n =2m and a be a square in F∈. If f(x) is not divided by x2 a and f (√a)=0, then it q − can be written as f(x) = (x √a)k g(x), where k is even and g(x) is a nontrivial a-srm polynomial with g (√a−) =0. · 6 n i m n Proof. Let f(x)= i=0 bix then by assumption b0 = f(0) = a = √a . Set

Pf(x) n−1 n−2 f (x) := = x + c − x + + c x + c F [x]. 1 x √a n 2 ··· 1 0 ∈ q − n n−1 Clearly ( √a) c0 = b0 = √a which implies that c0 = √a . Then f1(x) is a-srm− polynomial· since x √a is a-srm polynomial. By− Theorem 2, f (x) is − 1 divided by x √a and thus we can write as f(x) = (x √a)2 f (x). It is clear − − · 2 that (x √a)2 is a nontrivial a-srm polynomial, hence f (x) is a nontrivial a-srm − 2 polynomial of degree n 2. Continuing this procedure for f2(x) completes the proof. −

4 When a is a square in Fq, from this theorem the consideration of nontrivial a-srm polynomials not divided by x2 a is reduced to the case of ones without roots √a and √a. − − 3 The parity of the number of irreducible factors of generalized self-reciprocal polynomials

In this section we characterize the parity of the number of irreducible factors of the nontrivial a-srm polynomials over Fq.

Theorem 5. Let f(x) F [x] be a nontrivial a-srm polynomial of degree 2n with ∈ q r pairwise distinct irreducible factors over Fq. Then r 0 (mod 2) if and only if ( 1)nan(n−2)f (√a) f ( √a) is a square in F , where √≡a is a square root of a in − − q an extension of Fq.

2n i n a n−1 Proof. Let f(x) = i=0 bix , then f(x) = x g x + x , where g(x) = bn + i=0 n b2n−iDn−i,a(x) and Dn−i,a(x) is the Dickson polynomial [4]. Clearly f(0) = a =0 and by properties ofP the , R(f,f ′)= R(f,nf  xf ′). Since P6 − ′ − ′ a a f (x)= nxn 1g(x + a/x)+ xng x + 1 x − x2 − a − ′ a  = nxn 1g x + + xn 2g x + (x2 a) x x −     and ′ − ′ a nf(x) xf (x)= xn 1g x + (x2 a), − − x − we have  

− ′ − ′ D(f) = ( 1)n(2n 1)R(f,f ) = ( 1)n f(0) 1 R(f,nf xf ) − − · · − − − ′ a = ( 1)n a n R(f, x2 a) R f, xn 1g x + − · · − · − x −  − ′ a = ( 1)n a n f √a f √a R f, xn 1g x + − · · − · − x    a a   Let x0, , xn−1, , , be the roots of f(x) in some extension of Fq. Then ··· x0 ··· xn−1

n−1 n−1 n−1 n−1 ′ a n−1 ′ a a ′ a R f, x g x + = xi g xi + g + xi − x xi xi xi    iY=0   iY=0     n−1 n−1 n−1 ′ a n−1 1−n ′ a = xi g xi + a xi g xi + xi xi iY=0   iY=0   n−1 2 n(n−1) ′ a = a g xi + . " xi # iY=0   a On the other hand, since xi is a root of f(x) if and only if xi + is a root of g(x) xi and n−1 n(n−1)/2 ′ n(n−1)/2 ′ a D(g) = ( 1) R(g,g ) = ( 1) g xi + , − − xi iY=0  

5 we have − D(f) = ( 1)nan(n 2)f √a f √a D(g)2. − − · Required result follows from Stickelberger theorem [9] (see also [2, Theorem 4]).

2n i Let f(x)= i=0 bix be as in Theorem 5, then

P n n−1 2i 2i+1 f(x)= b2ix + b2i+1x Xi=0 Xi=0 and

n n−1 i i f √a = b2ia + √a b2i+1a , Xi=0 Xi=0  n n−1 f √a = b ai √a b ai. − 2i − 2i+1 i=0 i=0  X X Denote n n−1 i i A := b2ia , B := b2i+1a Xi=0 Xi=0 2 2 then f (√a) f ( √a) = A aB , which means f (√a) f ( √a) Fq. Since f(x) − n−i− − ∈ is nontrivial, bi = b2n−ia . Therefore if n is odd, then

(n−1)/2 (n−3)/2 n−i n−i−1 (n−1)/2 A =2 b2n−2ia , B =2 b2n−2i−1a + bna , Xi=0 Xi=0 and if n is even then

n/2−1 n/2−1 n−i n/2 n−i−1 A =2 b2n−2ia + bna , B =2 b2n−2i−1a . Xi=0 Xi=0 Q n a For any monic polynomial f(x) Fq[x] of degree n, denote fa (x) := x f x + x . Then f Q(x) is a nontrivial a-srm polynomial∈ of degree 2n. This is a natural gener- a  alization of f Q(x) in [7] and it has been mentioned in [4], too.

Theorem 6. Suppose that f(x) Fq[x] is a monic of degree ∈ Q n( 1) and f (√a) f ( √a) = 0. Then fa (x) is either a-srim polynomial or a product≥ of a-reciprocal− pair of6 irreducible polynomials of degree n which are not a-self reciprocal.

Q a qn Proof. Let α be any root of fa (x). Then β = α + α is a root of f(x), so β = β. k n Here n is the smallest positive k with βq = β. Multiply αq to both sides of the identity n a n a αq + = βq = β = α + , αqn α then n n n− α2q + a = αq +1 + aαq 1

6 and so n n− αq +1 a αq 1 1 =0. − − n n−  n  Hence αq +1 = a or αq 1 = 1. When αq +1 = a, the irreducible factor g(x) of f Q(x) with a root α has degree 2 because if g(x) has degree 1, then a is a square a ≥ in Fq and g(x) = x √a which contradicts to f (√a) f ( √a) = 0. Therefore g(x) is a nontrivial a±-srm polynomial of degree 2d for a positive− 6 divisor of n. If d d we assume d

Finally we prove that Theorem 5 is still valid when f(x) has repeated irreducible factors.

Theorem 7. Let f(x) Fq[x] be a nontrivial a-srm polynomial of degree 2n with f (√a) f ( √a) = 0, and∈ let r be the number of irreducible factors counted with multiplicity− of f(6 x). Then r is even if and only if ( 1)nan(n−2)f (√a) f ( √a) is − − a square in Fq.

Proof. As in the proof of Theorem 5, there exists g(x) Fq[x] such that f(x) = xng x + a . Suppose that g(x) = g (x) g (x) where∈ g (x) F [x] is monic x 1 ··· k i ∈ q irreducible polynomial of degree ni and n1 + + nk = n. Denote fi(x) := ni a ··· x gi x + x , then f(x) = f1(x) fk(x) where every fi(x) is a nontrivial a- srm polynomial of degree 2n over···F . By Theorem 6, every f (x) is either ir-  i q i reducible or a product of two distinct monic irreducible polynomials of degree ni. Now suppose that the claim holds for two nontrivial a-srm polynomials g(x), h(x) of degree 2s, 2t with rs, rh irreducible factors counted with multiplicity respec- s s(s−2) tively over Fq. If both rs and rh are odd, then neither ( 1) a g(√a)g( √a) t t(t−2) − s+t (s+t)(s+t−2) − nor ( 1) a h(√a)h( √a) is a square in Fq, hence ( 1) a g(√a) g( √−a)h(√a)h( √a) is− a square in F . Similar observation− about other cases − − q for rs and rh tells us the claim is still valid for g(x)h(x). It is sufficient to apply Theorem 5 to complete the proof.

Acknowledgement. We would like to thank anonymous referees for their valuable comments and suggestions.

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