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Research article

Permutational behavior of reversed Dickson polynomials over finite fields

  • Received: 15 March 2017 Accepted: 18 April 2017 Published: 20 April 2017
  • In this paper, we develop the method presented previously by Hong, Qin and Zhao to obtain several results on the permutational behavior of the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind over the finite field Fq. Particularly, we present the explicit evaluation of the first moment aFqDn,k(1,a). Our results extend the results of Hong, Qin and Zhao to the general k0 case.

    Citation: Kaimin Cheng. Permutational behavior of reversed Dickson polynomials over finite fields[J]. AIMS Mathematics, 2017, 2(2): 244-259. doi: 10.3934/Math.2017.2.244

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  • In this paper, we develop the method presented previously by Hong, Qin and Zhao to obtain several results on the permutational behavior of the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind over the finite field Fq. Particularly, we present the explicit evaluation of the first moment aFqDn,k(1,a). Our results extend the results of Hong, Qin and Zhao to the general k0 case.


    1. Introduction

    Permutation polynomials and Dickson polynomials are two of the most important topics in the area of finite fields. Let Fq be the finite field of characteristic p with q elements. Let Fq[x] be the ring of polynomials over Fq in the indeterminate x. If the polynomial f(x)Fq[x] induces a bijective map from Fq to itself, then f(x)Fq[x] is called a permutation polynomial of Fq. Properties, constructions and applications of permutation polynomials may be found in [4], [5] and [6]. Associated to any integer n0 and a parameter aFq, the n-th Dickson polynomials of the first kind and of the second kind, denoted by Dn(x,a) and En(x,a), are defined for n1 by

    Dn(x,a):=[n2]i=0nni(nii)(a)ixn2i

    and

    En(x,a):=[n2]i=0(nii)(a)ixn2i,

    respectively, and D0(x,a):=2,E0(x,a):=1. It is well known that Dn(x,0) is a permutation polynomial of Fq if and only if gcd(n,q1)=1, and if a0, then Dn(x,a) induces a permutation of Fq if and only if gcd(n,q21)=1. There are lots of published results on permutational properties of Dickson polynomial En(x,a) of the second kind (see, for example, [1]).

    The reversed Dickson polynomial of the first kind, denoted by Dn(a,x), was introduced in [3] and defined as follows

    Dn(a,x):=[n2]i=0nni(nii)(x)ian2i

    if n1 and D0(a,x)=2, where [n2] means the largest integer no more than n2. Wang and Yucas [7] extended this concept to that of the n-th reversed Dickson polynomial of (k+1)-th kind Dn,k(a,x)Fq[x], which is defined for n1 by

    Dn,k(a,x):=[n2]i=0nkini(nii)(x)ian2i (1.1)

    and D0,k(a,x)=2k. Some families of permutation polynomials from the revered Dickson polynomials of the first kind were obtained in [3]. Hong, Qin and Zhao [2] studied the revered Dickson polynomial En(a,x) of the second kind that is defined for n1 by

    En(a,x):=[n2]i=0(nii)(x)ian2i

    and E0(a,x)=1. In fact, they gave some necessary conditions for the revered Dickson polynomial En(1,x) of the second kind to be a permutation polynomial of Fq. Regarding the revered Dickson polynomial Dn,2(a,x)Fq[x] of the third kind, from its definition one can derive that

    Dn,2(a,x)=aEn1(a,x) (1.2)

    for each xFq. Using (1.2), one can deduce immediately from [2] the similar results on the permutational behavior of the reversed Dickson polynomial Dn,2(a,x) of the third kind.

    In this paper, our main goal is to develop the method presented by Hong, Qin and Zhao in [2] to investigate the reversed Dickson polynomial Dn,k(a,x) of the (k+1)-th kind which is defined by (1.1) if n1 and D0,k(a,x):=2k. For a0, we write x=y(ay) with an indeterminate ya2. Then one can rewrite Dn,k(a,x) as

    Dn,k(a,x)=((k1)a(k2)y)yn(a+(k2)y)(ay)n2ya. (1.3)

    We have

    Dn,k(a,a24)=(knk+2)an2n. (1.4)

    In fact, (1.3) and (1.4) follow from Theorem 2.2 (ⅰ) and Theorem 2.4 (ⅰ) below. It is easy to see that if char(Fq)=2, then Dn,k(a,x)=En(a,x) if k is odd and Dn,k(a,x)=Dn(a,x) if k is even. We also find that Dn,k(a,x)=Dn,k+p(a,x), so we can restrict p>k. Thus we always assume p=char(Fq)3 in what follows.

    The paper is organized as follows. First in section 2, we study the properties of the reversed Dickson polynomial Dn,k(a,x) of the (k+1)-th kind. Subsequently, in Section 3, we prove a necessary condition for the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind to be a permutation polynomial of Fq and then introduce an auxiliary polynomial to present a characterization for Dn,k(1,x) to be a permutation of Fq. From the Hermite criterion [4] one knows that a function f:FqFq is a permutation polynomial of Fq if and only if the i-th moment

    aFqf(a)i={0,if 0iq2,1,if i=q1.

    Thus to understand well the permutational behavior of the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind, we would like to know if the i-th moment aFqDn,k(1,a)i is computable. We are able to treat with this sum when i=1. The final section is devoted to the computation of the first moment aFqDn,k(1,a).


    2. Reversed Dickson polynomials of the (k+1)-th kind

    In this section, we study the properties of the reversed Dickson polynomials Dn,k(a,x) of the (k+1)-th kind. Clearly, if a=0, then

    Dn,k(0,x)={0,if n is odd,(1)n2+1(k2)xn2,if n is even.

    Therefore, Dn,k(0,x) is a PP (permutation polynomial) of Fq if and only if n is an even integer with gcd(n2,q1)=1. In what follows, we always let aFq. First, we give a basic fact as follows.

    Lemma 2.1. [4] Let f(x)Fq[x]. Then f(x) is a PP of Fq if and only if cf(dx) is a PP of Fq for any given c,dFq.

    Then we can deduce the following result.

    Theorem 2.2. Let a,bFq. Then the following are true.

    (ⅰ). One has Dn,k(a,x)=anbnDn,k(b,b2a2x).

    (ⅱ). We have that Dn,k(a,x) is a PP of Fq if and only if Dn,k(1,x) is a PP of Fq.

    Proof. (ⅰ). By the definition of Dn,k(a,x), we have

    anbnDn,k(b,b2a2x)=anbn[n2]i=0nkini(nii)(1)ibn2ib2ia2ixi=[n2]i=0nkini(nii)(1)ian2ixi=Dn,k(a,x)

    as required. Part (ⅰ) is proved.

    (ⅱ). Taking b=1 in part (ⅰ), we have

    Dn,k(a,x)=anDn,k(1,xa2).

    It then follows from Lemma 2.1 that Dn,k(a,x) is a PP of Fq if and only if Dn,k(1,x) is a PP of Fq. This completes the proof of part (ⅱ). So Theorem 2.2 is proved.

    Theorem 2.2 tells us that to study the permutational behavior of Dn,k(a,x) over Fq, one only needs to consider that of Dn,k(1,x). In the following, we supply several basic properties on the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind. The following result is given in [2].

    Lemma 2.3. [2] Let n0 be an integer. Then

    Dn(1,x(1x))=xn+(1x)n

    and

    En(1,x(1x))=xn+1(1x)n+12x1

    if x12.

    Theorem 2.4. Each of the following is true.

    (ⅰ). For any integer n0, we have

    Dn,k(1,14)=knk+22n

    and

    Dn,k(1,x)=(k1(k2)y)yn(1+(k2)y)(1y)n2y1

    if x=y(1y)14.

    (ⅱ). If n1 and n2 are positive integers such that n1n2(modq21), then one has Dn1,k(1,x0)=Dn2,k(1,x0) for any x0Fq{14}.

    Proof. (ⅰ). First of all, it is easy to see that D0,k(1,14)=2k=k×0k+220 and D1,k(1,14)=1=k×1k+221. the first identity is true for the cases that n=0 and 1. Now let n2. Then one has

    Dn,k(1,14)=[n2]i=0nkini(nii)(14)i=[n2]i=0n(k1)ini(nii)(14)i+[n2]i=0ini(nii)(14)i=Dn,k1(1,14)+14[n2]1i=0(n2ii)(14)i=Dn,k1(1,14)+14En2(1,14),

    which follows from Theorem 2.2 (1) in [2] that

    Dn,k(1,14)=Dn,1(1,14)+(k1)14En2(1,14)=n+12n+(k1)n(k1)2n=knk+22n

    as desired. So the first identity is proved.

    Now we turn our attention to the second identity. Let x14, then there exists yFq2{12} such that x=y(1y). So by the definition of the n-th reversed Dickson polynomial of the (k+1)-th kind, one has

    Dn,k(1,y(1y))=[n2]i=0nkini(nii)(y(1y))i=[n2]i=0k(ni)knni(nii)(y(1y))i=k[n2]i=0(nii)(y(1y))i(k1)[n2]i=0nni(nii)(y(1y))i=kEn(1,y(1y))(k1)Dn(1,y(1y)). (2.1)

    But Lemma 2.3 gives us that

    Dn(1,y(1y))=yn+(1y)n (2.2)

    and

    En(1,y(1y))=ni=0yni(1y)i=yn+1(1y)n+12y1. (2.3)

    Thus it follows from (2.1) to (2.3) that

    Dn,k(1,x)=Dn,k(1,y(1y))=kEn(1,y(1y))(k1)Dn(1,y(1y))=kyn+1k(1y)n+12y1(k1)(yn+(1y)n)=(k1(k2)y)yn(1+(k2)y)(1y)n2y1

    as required. So the second identity holds. Part (ⅰ) is proved.

    (ⅱ). For each x0Fq{14}, one can choose an element y0Fq2{12} such that x0=y0(1y0). Since n1n2(modq21), one has yn10=yn20 and (1y0)n1=(1y0)n2. It then follows from part (ⅰ) that

    Dn1,k(1,x0)=Dn1,k(1,y0(1y0))=(k1(k2)y0)yn10(1+(k2)y0)(1y0)n12y01=(k1(k2)y0)yn20(1+(k2)y0)(1y0)n22y01=Dn2,k(1,x0)

    as desired. This ends the proof of Theorem 2.4.

    Evidently, by Theorem 2.2 (ⅰ) and Theorem 2.4 (ⅰ) one can derive that (1.3) and (1.4) are true.

    Proposition 2.5. Let n2 be an integer. Then the recursion

    Dn,k(1,x)=Dn1,k(1,x)xDn2,k(1,x)

    holds for any xFq.

    Proof. We consider the following two cases.

    CASE 1. x14. For this case, one may let x=y(1y) with yFq2{12}. Then by Theorem 2.4 (ⅰ), we have

    Dn1,k(1,x)xDn2,k(1,x)=Dn1,k(1,y(1y))y(1y)Dn2,k(1,y(1y))=(k1(k2)y)yn1(1+(k2)y)(1y)n12y1    y(1y)(k1(k2)y)yn2(1+(k2)y)(1y)n22y1=(k1(k2)y)yn(1+(k2)y)(1y)n2y1=Dn,k(1,x)

    as required.

    CASE 2. x=14. Then by Theorem 2.4 (ⅰ), we have

    Dn1,k(1,14)14Dn2,k(1,14)=k(n1)k+22n114k(n2)k+22n2=knk+22n=Dn,k(1,14).

    This concludes the proof of Proposition 2.5.

    By Proposition 2.5, we can obtain the generating function of the reversed Dickson polynomial Dn,k(1,x) of the (k+1)-th kind as follows.

    Proposition 2.6. The generating function of Dn,k(1,x) is given by

    n=0Dn,k(1,x)tn=(k1)tk+21t+xt2.

    Proof. By the recursion presented in Proposition 2.5, we have

    (1t+xt2)n=0Dn,k(1,x)tn=n=0Dn,k(1,x)tnn=0Dn,k(1,x)tn+1+xn=0Dn,k(1,x)tn+2=(k1)tk+2+n=0(Dn+2,k(1,x)Dn+1,k(1,x)+xDn,k(1,x))tn+2=(k1)tk+2.

    Thus the desired result follows immediately.

    Lemma 2.7. [3] Let xFq2. Then x(1x)Fq if and only if xq=x or xq=1x.

    Let V be defined by

    V:={xFq2:xq=1x}.

    Clearly, FqV={12}. Then we obtain a characterization for Dn,k(1,x) to be a PP of Fq as follows.

    Theorem 2.8. Let q=pe with p>3 being a prime and e being a positive integer. Let

    f:y(k1(k2)y)yn(1+(k2)y)(1y)n2y1

    be a mapping on (FqV){12}. Then Dn,k(1,x) is a PP of Fq if and only if f is 2-to-1 and f(y)knk+22n for any y(FqV){12}.

    Proof. First, we show the sufficiency part. Let f be 2-to-1 and f(y)knk+22n for any y(FqV){12}. Let Dn,k(1,x1)=Dn,k(1,x2) for x1,x2Fq. To show that Dn,k(1,x) is a PP of Fq, it suffices to show that x1=x2, which will be done in what follows.

    First of all, one can find y1,y2Fq2 satisfying x1=y1(1y1) and x2=y2(1y2). By Lemma 2.7, we know that y1,y2FqV. We divide the proof into the following two cases.

    CASE 1. At least one of x1 and x2 is equal to 14. Without loss of any generality, we may let x1=14. So by Theorem 2.4 (ⅰ), one derives that

    Dn,k(1,x2)=Dn,k(1,x1)=Dn,k(1,14)=knk+22n. (2.4)

    We claim that x2=14. Assume that x214. Then y212. Since f(y)knk+22n for any y(FqV){12}, by Theorem 2.4 (ⅰ), we get that

    Dn,k(1,x2)=(k1(k2)y2)yn2(1+(k2)y2)(1y2)n2y21=f(y2)knk+22n,

    which contradicts to (2.4}). Hence the claim is true, and so we have x1=x2 as required.

    CASE 2. Both of x1 and x2 are not equal to 14. Then y112 and y212. Since Dn,k(1,x1)=Dn,k(1,x2), by Theorem 2.4 (ⅰ), one has

    (k1(k2)y1)yn1(1+(k2)y1)(1y1)n2y11=(k1(k2)y2)yn2(1+(k2)y2)(1y2)n2y21,

    which is equivalent to f(y1)=f(y2). However, f is a 2-to-1 mapping on (FqV){12}, and f(y2)=f(1y2) by the definition of f. It then follows that y_1=y_2 or y_1=1-y_2. Thus x_1=x_2 as desired. Hence the sufficiency part is proved.

    Now we prove the necessity part. Let D_{n, k}(1, x) be a PP of {\mathbb F_{q}}. Choose two elements y_1, y_2\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\} such that f (y_1)=f (y_2), that is,

    \frac{(k-1-(k-2)y_1)y_1^n-(1+(k-2)y_1)(1-y_1)^n}{2y_1-1} =\frac{(k-1-(k-2)y_2)y_2^n-(1+(k-2)y_2)(1-y_2)^n}{2y_2-1}. (2.5)

    Since y_1, y_2\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}, it follows from Lemma 2.7 that y_1(1-y_1)\in{\mathbb F_{q}} and y_2(1-y_2)\in{\mathbb F_{q}}. So by Theorem 2.4 (ⅰ), (2.5) implies that

    D_{n, k}(1, y_1(1-y_1))=D_{n, k}(1, y_2(1-y_2)).

    Thus y_1(1-y_1)=y_2(1-y_2) since D_{n, k}(1, x) is a PP of {\mathbb F_{q}}, which infers that y_1=y_2 or y_1=1-y_2. Since y_2\ne \frac{1}{2}, one has y_2\ne 1-y_2. Therefore f is a 2-to-1 mapping on ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}.

    Now take y'\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}. Then from Lemma 2.7 it follows that y'(1-y')\in{\mathbb F_{q}} and

    y'(1-y')\ne \frac{1}{2}\Big(1-\frac{1}{2}\Big).

    Notice that D_{n, k}(1, x) is a PP of {\mathbb F_{q}}. Hence one has

    D_{n, k}(1, y'(1-y'))\ne D_{n, k}\Big(1, \frac{1}{2}\Big(1-\frac{1}{2}\Big)\Big).

    But Theorem 2.4 (ⅰ) tells us that

    D_{n, k}\Big(1, \frac{1}{2}\Big(1-\frac{1}{2}\Big)\Big)=\frac{kn-k-2}{2^n}.

    Then by Theorem 2.4 (ⅰ) and noting that y'\ne \frac{1}{2}, we have

    \frac{\big(k-1-(k-2)y'\big)y'^{n}-\big(1+(k-2)y'\big)(1-y')^{n}}{2y'-1},

    which infers that f (y')\ne \frac{kn-k-2}{2^n} for any y'\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}. So the necessity part is proved.

    The proof of Theorem 2.8 is complete.

    Now we can use Theorem 2.4 to present an explicit formula for D_{n, k}(1, x) when n is a power of the characteristic p. Then we derive the detailed characterization for D_{n, k}(1, x) being a PP of {\mathbb F}_q in this case.

    Proposition 2.9. Let p={\rm char}({\mathbb F_{q}})\ge 3 and s\ge 0 be an integer. Then

    2D_{p^s, k}(1, x)+k-2=k(1-4x)^{\frac{p^s-1}{2}}.

    Proof. We consider the following two cases.

    CASE 1. x\ne \frac{1}{4}. For this case, putting x=y (1-y) in Theorem 2.4 (ⅰ) gives us that

    D_{p^s, k}(1, x)=D_{p^s, k}(1, y(1-y))\\ =\frac{\big(k-1-(k-2)y\big)y^{p^s}-\big(1+(k-2)y\big)(1-y)^{p^s}}{2y-1}\\ =\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s}}{u}\\ =\frac{1}{2^{p^s+1}u}\Big((k+(2-k)u)(u+1)^{p^s}-(k+(k-2)u)(1-u)^{p^s}\Big)\\ =\frac{1}{2}(ku^{p^s-1}-k+2),

    where u=2y-1. So we obtain that

    2D_{p^s, k}(1, x) =k(u^2)^{\frac{p^s-1}{2}}-k+2 =k\big((2y-1)^2\big)^{\frac{p^s-1}{2}}-k+2,

    which infers that

    2D_{p^s, k}(1, x)+k-2=k(1-4x)^{\frac{p^s-1}{2}}

    as desired.

    CASE 2. x=\frac{1}{4}. By Theorem 2.4 (ⅰ), one has

    2D_{p^s, k}\big(1, \frac{1}{4}\big)+k-2=2\times\frac{kp^s-k+2}{2^{p^s}}+k-2=0 =k(1-4\times\frac{1}{4})^{\frac{p^s-1}{2}}

    as required. So Proposition 2.9 is proved.

    It is well known that every linear polynomial over {\mathbb F_{q}} is a PP of {\mathbb F_{q}} and that the monomial x^n is a PP of {\mathbb F_{q}} if and only if \gcd (n, q-1)=1. Then by Proposition 2.9, we have the following result.

    Corollary 2.10. Let p\ge 3 be a prime, q=p^e with e\ge 1 and s\ge 0 be an integer. Then D_{p^s, k}(1, x) is a PP of {\mathbb F_{q}} if and only if k\ge 1, p=3, s is odd and \gcd (s, e)=1.

    Proof. First assume that D_{p^s, k}(1, x) is a PP of {\mathbb F_{p^e}}. It then follows from Proposition 2.9 that D_{p^s, k}(1, x) is a PP of {\mathbb F_{p^e}} if and only if

    k(1-4x)^{\frac{p^s-1}{2}} (2.6)

    is a PP of {\mathbb F_{p^e}}. Clearly, k\ge 1 and s>0 in this case. Suppose p>3, then (2.6) is a PP of {\mathbb F_{p^e}} if and only if

    \gcd\Big(\frac{p^s-1}{2}, p^e-1\Big)=1.

    This is impossible since \frac{p-1}{2}|\gcd\big (\frac{p^s-1}{2}, q-1\big) implies that

    \gcd\Big(\frac{p^s-1}{2}, q-1\Big)\ge\frac{p-1}{2}>1.

    So p=3, k\ge 1 and s>0 in what following. Now Suppose s>0 is even, then it is easy to see that 2|\gcd (\frac{3^s-1}{2}, 3^e-1) which is a contradiction. This means that s must be an odd integer and then so is \frac{3^s-1}{2}. Thus we have that (2.6) is a PP of {\mathbb F_{p^e}} if and only if

    \gcd\Big(\frac{3^s-1}{2}, 3^e-1\Big)=\frac{1}{2}\gcd\big(3^s-1, 3^e-1\big)=\frac{1}{2}(3^{\gcd(s, e)}-1)=1,

    which is equivalent to that s is odd and \gcd (s, e)=1. So Corollary 2.10 is proved.


    3. A necessary condition for D_{n, k}(1, x) to be permutational and an auxiliary polynomial

    In this section, we study a necessary condition on n for D_{n, k}(1, x) to be a PP of {\mathbb F_{q}}. On one hand, it is easy to check that

    D_{0, k}(1, 0)=2-k, D_{n, k}(1, 0)=1

    for any n\ge 1 and D_{0, k}(1, 1)=2-k, D_{1, k}(1, 1)=1. On the other hand, Proposition 2.5 tells us that

    D_{n+2, k}(1, 1)=D_{n+1, k}(1, 1)-D_{n, k}(1, 1)

    for n\ge 0. Then one can easily show that the sequence \{D_{n, k}(1, 1)|n\in \mathbb{N}\} is periodic with the smallest positive periods 6. In fact, one has

    D_{n, k}(1, 1)={\left\{ \begin{array}{rl} 2-k, & {\rm if} \ n\equiv 0\pmod{6}, \\ 1, & {\rm if} \ n\equiv 1\pmod{6}, \\ k-1, & {\rm if} \ n\equiv 2\pmod{6}, \\ k-2, & {\rm if} \ n\equiv 3\pmod{6}, \\ -1, & {\rm if} \ n\equiv 4\pmod{6}, \\ 1-k, & {\rm if} \ n\equiv 5\pmod{6} \end{array} \right.}

    So we have the following result.

    Theorem 3.1. Assume that D_{n, k}(1, x) is a PP of {\mathbb F_{q}} with q=p^e and p>3. Then n\not\equiv 1\pmod{6}.

    Proof. Let D_{n, k}(1, x) be a PP of {\mathbb F_{q}}. Then D_{n, k}(1, 0) and D_{n, k}(1, 1) are distinct. Then by the above results, the desired result n\not\equiv 1 \pmod{6} follows immediately.

    Let n, k be nonnegative integers. We define the following auxiliary polynomial p_{n, k}(x)\in \mathbb{Z}[x] by

    p_{n, k}(x):=k\sum_{j\ge 0}\binom{n}{2j+1}x^j-(k-2)\sum_{j\ge 0}\binom{n}{2j}x^j

    for n\ge 1, and

    p_{0, k}(x):=2^n(2-k).

    Then we have the following relation between D_{n, k}(1, x) and p_{n, k}(x).

    Theorem 3.2. Let p>3 be a prime and n\ge 0 be an integer. Then each of the following is true.

    (ⅰ). One has

    D_{n, k}(1, x)=\frac{1}{2^n}p_{n, k}(1-4x). (3.1)

    (ⅱ). We have that D_{n, k}(1, x) is a PP of {\mathbb F_{q}} if and only if p_{n, k}(x) is a PP of {\mathbb F_{q}}.

    Proof. (ⅰ). Clearly, (3.1) follows from the definitions of p_{0, k}(x) and D_{0, k}(1, x) if n=0. Then we assume that n\ge 1 in what follows.

    First, let x\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}. Then there exists y\in{\mathbb F_{q^2}}\setminus\{\frac{1}{2}\} such that x=y (1-y). Let u=2y-1. It then follows from Theorem 2.4 (ⅰ) that

    D_{n, k}(1, x)=D_{n, k}(1, y(1-y))\\ =\frac{\big(k-1-(k-2)y\big)y^{n}-\big(1+(k-2)y\big)(1-y)^{n}}{2y-1}\\ =\frac{1}{u}\Big(\frac{-(k-2)u+k}{2}\big(\frac{u+1}{2}\big)^{n}-\frac{(k-2)u+k}{2}\big(\frac{1-u}{2}\big)^{n}\Big)\\ =\frac{1}{2^{n+1}u}\Big(k\big((u+1)^n-(1-u)^n\big)-(k-2)u\big((u+1)^n+(1-u)^n\big)\Big)\\ =\frac{1}{2^{n}}\Bigg ( k\sum_{j\ge 0}\binom{n}{2j+1}x^j-(k-2)\sum_{j\ge 0}\binom{n}{2j}u^{2j}\Bigg)\\ =\frac{1}{2^n}p_{n, k}(u^2)\\ =\frac{1}{2^n}p_{n, k}(1-4y(1-y))\\ =\frac{1}{2^n}p_{n, k}(1-4x)

    as desired. So (3.1) holds in this case.

    Consequently, we let x=\frac{1}{4}. Then by Theorem 2.4 (ⅰ), we have

    D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{kn-k+2}{2^n}.

    On the other hand, we can easily check that

    p_{n, k}(0)=kn-k+2.

    Therefore

    D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{1}{2^n}p_{n, k}(0) =\frac{1}{2^n}p_{n, k}\Big(1-4\times \frac{1}{4}\Big)

    as one desires. So (3.1) is proved.

    (ⅱ). Notice that \frac{1}{2^n}\in{\mathbb F_{q}^*} and 1-4x is linear. So D_{n, k}(1, x) is a PP of {\mathbb F_{q}} if and only if p_{n, k}(x) is a PP of {\mathbb F_{q}}. This ends the proof of Theorem 3.2.


    4. The first moment \sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)

    In this section, we compute the first moment \sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a). By Proposition 2.6, one has

    \sum_{n=0}^{\infty}D_{n, k}(1, x)t^n=\frac{(k-1)t-k+2}{1-t+xt^2} =\frac{(k-1)t-k+2}{1-t}\frac{1}{1-\frac{t^2}{t-1}x} \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1}\sum_{\ell=0}^{\infty } \bigg(\frac{t^2}{t-1}\bigg)^{m+\ell (q-1)}x^{m+\ell (q-1)}\Big) \notag\\ \equiv\frac{2t-1}{1-t}\Big(1+\sum_{m=1}^{q-1}\sum_{\ell=0}^{\infty } \bigg(\frac{t^2}{t-1}\bigg)^{m+\ell (q-1)}x^{m}\Big)\pmod{x^q-x} \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1} \frac{(\frac{t^2}{t-1})^m}{1-(\frac{t^2}{t-1})^{q-1}}x^m\Big) \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1} \frac{(t-1)^{q-1-m}t^{2m}}{(t-1)^{q-1}-t^{2(q-1)}}x^m\Big). (4.1)

    Moreover, by Theorem 2.4 (ⅱ), it follows that for any x\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}, one has

    D_{n_1, k}(1, x)=D_{n_2, k}(1, x)

    when n_1\equiv n_2\pmod{q^2-1}. Thus if x\ne \frac{1}{4}, one has

    \sum_{n=0}^\infty D_{n, k}(1, x)t^n=1+\sum_{n=1}^{q^2-1} \sum_{\ell=0}^{\infty }D_{n+\ell(q^2-1), k}(1, x)t^{n+\ell(q^2-1)}\notag\\ =1+\sum_{n=1}^{q^2-1}D_{n, k}(1, x)\sum_{\ell=0}^{\infty }t^{n+\ell(q^2-1)}\notag\\ =1+\frac{1}{1-t^{q^2-1}}\sum_{n=1}^{q^2-1}D_{n, k}(1, x)t^{n}. (4.2)

    Then (4.1) together with (4.2) gives that for any x\ne \frac{1}{4}, we have

    \sum_{n=1}^{q^2-1}D_{n, k}(1, x)t^n= \Big(\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n-1\Big)(1-t^{q^2-1})\notag\\ \equiv \Big(\frac{(k-1)t-k+2}{1-t}-1\Big)(1-t^{q^2-1}) +\frac{(1-t^{q^2-1})((k-1)t-k+2)}{1-t}\sum_{m=1}^{q-1} \frac{(t-1)^{q-1-m}t^{2m}}{(t-1)^{q-1}-t^{2(q-1)}}x^m\pmod{x^q-x}\notag\\ =\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}+h(t)\sum_{m=1}^{q-1} (t-1)^{q-1-m}t^{2m}x^m, (4.3)

    where

    h(t):=\frac{(t^{q^2-1}-1)((k-1)t-k+2)}{(t-1)^q-(t-1)t^{2(q-1)}}.

    Lemma 4.1. [4] Let u_0, u_1, \cdots, u_{q-1} be the list of the all elements of {\mathbb F_{q}}. Then

    \sum_{i=0}^{q-1}u_i^k={\left\{ \begin{array}{rl} 0, & {\rm if} \ 0\le k\le q-2, \\ -1, & {\rm if} \ k=q-1. \end{array} \right.}

    Now by Theorem 2.4 (ⅰ), Lemma 4.1 and (4.3), we derive that

    \sum_{n=1}^{q^2-1}\sum_{a\in{\mathbb F_{q}}}D_{n, k}(1, a)t^n =\sum_{n=1}^{q^2-1}D_{n, k}\Big(1, \frac{1}{4}\Big)t^n+\sum_{n=1}^{q^2-1} \sum_{a\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}}D_{n, k}(1, a)t^n\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n+\sum_{a\in{\mathbb F_{q}} \setminus\{\frac{1}{4}\}}\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}+h(t)\sum_{m=1} ^{q-1}(t-1)^{q-1-m}t^{2m}\sum_{a\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}}a^m\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n+(q-1)\frac{(kt+1-k)(1-t^{q^2-1})}{1-t} +h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\sum_{a\in{\mathbb F_{q}}}a^m\notag\\ \ \ \ -h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n-\frac{(kt+1-k)(1-t^{q^2-1})}{1-t} -h(t)t^{2(q-1)}-h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m. (4.4)

    Since (t-1)^q=t^q-1 and q is odd, one has

    h(t)=\frac{(t^{q^2-1}-1)(2t-1)}{(t-1)^q-(t-1)t^{2(q-1)}}\notag\\ =\frac{(t^{q^2-1}-1)(2t-1)}{(1-t^{q-1})(t^q-t^{q-1}-1)} \notag\\ =\frac{(t^{q^2}-t)(2t-1)}{(t-t^{q})(t^q-t^{q-1}-1)}\notag\\ =\frac{(t^{q}-t)^q+t^q-t}{t-t^{q}}\cdot \frac{2t-1}{t^q-t^{q-1}-1}\notag\\ =\frac{(-1-(t-t^q)^{q-1})(2t-1)}{t^q-t^{q-1}-1}\notag\notag\\ =\frac{(2t-1)\sum_{i=0}^{q^2-q}b_it^i}{t^q-t^{q-1}-1}, (4.5)

    where

    \sum_{i=0}^{q^2-q}b_it^i:=-1-(t-t^q)^{q-1}.

    Then by the binomial theorem applied to (t-t^q)^{q-1}, we can derive the following expression for the coefficient b_i.

    Proposition 4.2. For each integer i with 0\le i\le q^2-q, write i=\alpha+\beta q with \alpha and \beta being integers such that 0\le \alpha, \beta \le q-1. Then

    b_i={\left\{ \begin{array}{ll} (-1)^{\beta +1}\binom{q-1}{\beta}, & {\it if} \ \alpha +\beta =q-1, \\ -1, & {\it if} \ \alpha =\beta =0, \\ 0, & {\it otherwise}. \end{array} \right.}

    For convenience, let

    a_n:=\sum_{a\in{\mathbb F_{q}}}D_{n, k}(1, a).

    Then by (4.4) and (4.5), we arrive at

    \sum_{n=1}^{q^2-1}\Big(a_n-\frac{kn-k+2}{2^n}\Big)t^n =-\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}-\frac{(2t-1)\sum_{i=0}^{q^2-q}b_it^i}{t^q-t^{q-1}-1} \Big(t^{2(q-1)}+\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m\Big),

    which implies that

    (t^q-t^{q-1}-1)\sum_{n=1}^{q^2-1}\Big(a_n-\frac{kn-k+2}{2^n}\Big)t^n\notag\\ =-(t^q-t^{q-1}-1)(kt+1-k)\sum_{i=0}^{q^2-2}t^i-(2t-1)\Big(t^{2(q-1)}+\sum_{k=1}^{q-1} (t-1)^{q-1-k}t^{2k}\Big(\frac{1}{4}\Big)^k\Big)\sum_{i=0}^{q^2-q}b_it^i. (4.6)

    Let

    \sum_{i=1}^{q^2+q-1}c_it^i

    denote the right-hand side of (4.6) and let

    d_n:=a_n-\frac{kn-k+2}{2^n}

    for each integer n with 1\le n\le q^2-1. Then (4.6) can be reduced to

    (t^q-t^{q-1}-1)\sum_{n=1}^{q^2-1}d_nt^n=\sum_{i=1}^{q^2+q-1}c_it^i. (4.7)

    Then by comparing the coefficient of t^i with 1\le i\le q^2+q-1 of the both sides in (4.7), we derive the following relations:

    {\left\{ \begin{array}{ll} c_j=-d_j, & {\rm if} \ 1\le j\le q-1, \\ c_q=-d_1-d_q, & \\ c_{q+j}=d_j-d_{j+1}-d_{q+j}, & {\rm if} \ 1\le j\le q^2-q-1, \\ c_{q^2+j}=d_{q^2-q+j}-d_{q^2-q+j+1}, & {\rm if} \ 0\le j\le q-2, \\ c_{q^2+q-1}=d_{q^2-1}, & \end{array} \right.}

    from which we can deduce that

    {\left\{ \begin{array}{ll} d_j=-c_j, & {\rm if} \ 1\le j\le q-1, \\ d_q=c_1-c_q, & \\ d_{\ell q+j}=d_{(\ell -1)q+j}-d_{(\ell-1)q+j+1}-c_{\ell q+j}, & {\rm if} \ 1\le \ell\le q-2\ \ {\rm and}\ \ 1\le j\le q-1, \\ d_{\ell q}=d_{(\ell -1)q}-d_{(\ell-1)q+1}-c_{\ell q}, & {\rm if} \ 2\le \ell\le q-2, \\ d_{q^2-q+j}=\sum_{i=j}^{q-1}c_{q^2+i}, & {\rm if} \ 0\le j\le q-1. \end{array} \right.} (4.8)

    Finally, (4.8) together with the following identity

    \sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)=d_n+\frac{kn-k+2}{2^n}

    shows that the last main result of this paper is true:

    Theorem 4.3. Let c_i be the coefficient of t^i in the right-hand side of (4.6) with i being an integer such that 1\le i\le q^2+q-1. Then we have

    \sum_{a\in {\mathbb F_{q}}}D_{j, k}(1, a)=-c_j+\frac{kj-k+2}{2^j}\ \ if\ \ 1\le j\le q-1, \\ \sum_{a\in {\mathbb F_{q}}}D_{q, k}(1, a)=c_1-c_q-\frac{k-2}{2}, \\ \sum_{a\in {\mathbb F_{q}}}D_{\ell q+j, k}(1, a)=\sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+j, k}(1, a)- \sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+j+1, k}(1, a)-c_{\ell q+j}+\frac{k}{2^{\ell +j}}\\ \ \ \ \ if\ \ 1\le \ell\le q-2 \ {\it and} \ 1\le j\le q-1, \\ \sum_{a\in {\mathbb F_{q}}}D_{\ell q, k}(1, a)=\sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q, k}(1, a)- \sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+1, k}(1, a)-c_{\ell q}+\frac{k}{2^{\ell}}\ \ if\ \ 2\le \ell\le q-2

    and

    \sum_{a\in {\mathbb F_{q}}}D_{q^2-q+j, k}(1, a) =\sum_{i=j}^{q-1}c_{q^2+i}+\frac{kj-k+2}{2^j}\ \ if\ \ 0\le j\le q-1.

    Acknowledgement

    Cheng was supported partially by the General Project of Department of Education of Sichuan Province 15ZB0434. [2000] Primary 11T06, 11T55, 11C08.


    Conflict of Interest

    The author declares no conflicts of interest in this paper.


    [1] S. D. Cohen, Dickson polynomials of the second kind that are permutations, Canad. J. Math., 46 (1994), 225-238.
    [2] S. Hong, X. Qin andW. Zhao, Necessary conditions for reversed Dickson polynomials of the second kind to be permutational, Finite Fields Appl., 37 (2016), 54-71.
    [3] X. Hou, G. L. Mullen, J.A. Sellers and J.L. Yucas, Reversed Dickson polynomials over finite fields, Finite Fields Appl., 15 (2009), 748-773.
    [4] R. Lidl and H. Niederreiter, Finite Fields, second ed. , Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 20 1997.
    [5] X. Qin and S. Hong, Constructing permutation polynomials over finite fields, Bull. Aust. Math. Soc., 89 (2014), 420-430.
    [6] X. Qin, G. Qian and S. Hong, New results on permutation polynomials over finite fields, Int. J. Number Theory, 11 (2015), 437-449.
    [7] Q. Wang and J. L. Yucas, Dickson polynomials over finite fields, Finite Fields Appl., 18 (2012), 814-831.
  • This article has been cited by:

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