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Research article

Permutational behavior of reversed Dickson polynomials over finite fields II

  • Received: 06 June 2017 Accepted: 16 October 2017 Published: 06 November 2017
  • In this paper, we study the special reversed Dickson polynomial of the form Dpe1+...+pes+,k(1,x), where s,e1,...,es are positive integers, is an integer with 0<p. In fact, by using Hermite criterion we first give an answer to the question that the reversed Dickson polynomials of the forms Dps+1,k(1,x), Dps+2,k(1,x), Dps+3,k(1,x), Dps+4,k(1,x), Dps+pt,k(1,x) and Dps+pt+1,k(1,x) are permutation polynomials of Fq or not. Finally, utilizing the recursive formula of the reversed Dickson polynomials, we represent Dpe1+...+pes+,k(1,x) as the linear combination of the elementary symmetric polynomials with the power of 14x being the variables. From this, we present a necessary and sufficient condition for Dpe1+...+pes+,k(1,x) to be a permutation polynomial of Fq.

    Citation: Kaimin Cheng. Permutational behavior of reversed Dickson polynomials over finite fields II[J]. AIMS Mathematics, 2017, 2(4): 586-609. doi: 10.3934/Math.2017.4.586

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  • In this paper, we study the special reversed Dickson polynomial of the form Dpe1+...+pes+,k(1,x), where s,e1,...,es are positive integers, is an integer with 0<p. In fact, by using Hermite criterion we first give an answer to the question that the reversed Dickson polynomials of the forms Dps+1,k(1,x), Dps+2,k(1,x), Dps+3,k(1,x), Dps+4,k(1,x), Dps+pt,k(1,x) and Dps+pt+1,k(1,x) are permutation polynomials of Fq or not. Finally, utilizing the recursive formula of the reversed Dickson polynomials, we represent Dpe1+...+pes+,k(1,x) as the linear combination of the elementary symmetric polynomials with the power of 14x being the variables. From this, we present a necessary and sufficient condition for Dpe1+...+pes+,k(1,x) to be a permutation polynomial of Fq.


    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 (denoted as PP for convenience) of Fq. Properties, constructions and applications of permutation polynomials may be found in [5], [6] and [7]. The reversed Dickson polynomial of the first kind, denoted by Dn(a,x), was introduced in [4] 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 [8] 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 [4]. Hong, Qin and Zhao [3] studied the revered Dickson polynomial En(a,x) of the second kind. Very recently, the author [1] investigated the reversed Dickson polynomial Dn,k(a,x) of the (k+1)-th kind and obtained some properties and permutational behaviors of them.

    In this paper, we study the special reversed Dickson polynomial of the form Dpe1+...+pes+,k(1,x), where s,e1,...,es are positive integers, is an integer with 0<p. In fact, by using Hermite criterion we first give an answer to the question that the reversed Dickson polynomials of the forms Dps+1,k(1,x), Dps+2,k(1,x), Dps+3,k(1,x), Dps+4,k(1,x), Dps+pt,k(1,x) and Dps+pt+1,k(1,x) are permutation polynomials of Fq or not. Finally, utilizing the recursive formula of the reversed Dickson polynomials, we represent Dpe1+...+pes+,k(1,x) as the linear combination of the elementary symmetric polynomials with the power of 14x being the variables. From this, we present a necessary and sufficient condition for Dpe1+...+pes+,k(1,x) to be a permutation polynomial of Fq.

    Throughout this paper, as usual, for any given prime number p, we let vp(x) denote the p-adic valuation of any positive integer x, i.e., vp(x) is the largest nonnegative integer k such that pk divides x. We also assume p=char(Fq)3 and restrict 0k<p.


    2. Preliminary lemmas

    In this section, we list several properties of the reversed Dickson polynomials Dn,k(a,x) of the (k+1)-th kind and some useful lemmas.

    Lemma 2.1. [5] Let f(x)Fq[x]. Then f(x) is a PP of Fq if and only if cf(dx)+b is a PP of Fq for any given c,dFq and bFq.

    Lemma 2.2. Let s0 be an integer and a,b be in Fq. Then the binomial axps12+bxps+12 is a PP of Fq if and only if s=0.

    Proof. First we assume that the binomial axps12+bxps+12 is a PP of Fq. If s>0, then the equation axps12+bxps+12=xps12(a+bx)=0 has two distinct roots 0,ba which are in Fq. This is a contradiction. So the integer s must be zero. Conversely, if s=0, then it is easy to check that axps12+bxps+12 is a PP of Fq. Therefore Lemma 2.2 is proved.

    Lemma 2.3. [1] 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.

    Lemma 2.4. [1] Let n2 be an integer. Then the recursion

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

    holds for any xFq.

    Lemma 2.5. [1] Let p=char(Fq)3 and s be a positive integer. Then

    2Dps,k(1,x)+k2=k(14x)ps12.

    Lemma 2.6. [2] Let α and e be positive integers. Let d=gcd(α,e) and p be an odd prime. Then

    gcd(pα+1,pe1)={2,if ed  is  odd,pd+1,if  ed  is  even.

    Lemma 2.7. [5] Let f(x)Fq[x]. Then f(x) is permutation polynomial of Fq if and only if the following conditions hold:

    (i) f(x) has exactly one root in Fq;

    (ii) For each integer t with 0<t<q1 and t0(modp), the reduction of f(x)t(modxqx) has degree less than q1.

    Lemma 2.8. Let p be a prime with p>3 and a be a nonzero element in Fp. Then the binomial xps12+ax is a PP of Fpe if and only if s=0.

    Proof. Let p>3,aFp. Clearly, if s=0, then w(x):=xps12+ax=1+ax is a PP of Fpe. In what follows, we show that w(x)=xps12+ax is not a PP of Fpe when s>0. Let s>0 and ss0(mod2e) with 0s02e1. Then

    w(x)xps012+ax(modxpex)

    for any xFpe since ps12ps012(modpe1), i.e.,

    w(x)=xps012+ax (2.1)

    for any xFpe. We consider the following three cases.

    CASE 1. s>0 and s0=0. Then by (2.1) one has w(x)=1+ax for any xFpe. So f2(1a)=0. It then follows from f2(0)=0 that w(x)=xps12+ax is not a PP of Fpe.

    CASE 2. s>0 and s0 is a positive even number. Then xps012=1 for each xFp. By (2.1) one get w(x)=1+ax for any xFp. Therefore w(x)=0 has one nonzero root 1aFp. Hence w(x)=xps12+ax does not permute Fp since f2(0)=0. Note that f2(Fp)Fp. So one has that w(x)=xps12+ax does not permute Fpe.

    CASE 3. s>0 and s0 is an odd number. Then xps012=xp12 for each xFp. It follows from (2.1) that

    w(x)=xp12+ax

    for any xFp. Then we have

    (w(x))2=xp1+the  terms  of  x  with  the  degree  less  than  p1(modxpx).

    Then by Lemma 2.7, we know that w(x) is not a PP of Fp. Also note that f2(Fp)Fp. So w(x) is not a PP of Fpe.

    The above three cases tell us that w(x)=xps12+ax is not a PP of Fpe when s>0. This finishes the proof of Lemma 2.8.


    3. Reversed Dickson polynomials Dps+,k(1,x)

    In this section, we present an explicit formula for Dn,k(1,x) when n=ps+ with s0 and 0<p. Then we characterize Dn,k(1,x) to be a PP of Fq in this case.

    Theorem 3.1. Let p=char(Fq)3 and s be a positive integer. Then

    Dps+1,k(1,x)=2k4(14x)ps+12+k4(14x)ps12+12. (3.1)

    Furthermore, we have

    Dps+2,k(1,x)=i=0A2,ps+2i1(14x)ps+2i12+j=0A2,2j(14x)j,0

    and

    Dps+2+1,k(1,x)=+1i=0A2+1,ps+2i1(14x)ps+2i12+j=0A2+1,2j(14x)j,0,

    where all the coefficients Ai,j are given as follows:

    A0,ps1=k2,A0,0=2k2,A1,ps+1=2k4,A1,ps1=k4,A1,0=12,

    and

    {A2m+2,ps+2m+1=A2m+1,ps+2m+1+14A2m,ps+2m1,if m0A2m+2,ps+2i1=A2m+1,ps+2i114A2m,ps+2i1+14A2m,ps+2i3,if 1imA2m+2,ps1=A2m+1,ps114A2m,ps1,if m0A2m+2,0=A2m+1,014A2m,0,if m0A2m+2,2j=A2m+1,2j14A2m,2j+14A2m,2j2,if 1jmA2m+2,2m+2=14A2m,2m,if m0 (3.2)

    as well as

    {A2m+1,ps+2m+1=14A2m1,ps+2m1,if m0A2m+1,ps+2i1=A2m,ps+2i114A2m1,ps+2i1+14A2m1,ps+2i3,if 1imA2m+1,ps1=A2m,ps114A2m1,ps1,if m0A2m+1,0=A2m,014A2m1,0,if m0A2m+1,2j=A2m,2j14A2m1,2j+14A2m1,2j2,if 1jm1A2m+1,2m=A2m,2m+14A2m1,2m2,if m0. (3.3)

    Proof. First of all, we show (3.1) is true. We consider the following two cases.

    CASE 1. x14. For this case, putting x=y(1y) in second identity of Lemma 2.3 gives us that

    Dps+1,k(1,x)=Dps+1,k(1,y(1y))=k+(2k)u2(u+12)ps+1k+(k2)u2(1u2)ps+1u=2k8((u+1)ps(u+1)+(1u)ps(1u))+k8u((u+1)ps(u+1)(1u)ps(1u))=2k4(ups+1+1)+k4(ups1+1)=2k4((u2)ps+12)+k4((u2)ps12)+12,

    where u=2y1 and u2=14x. So (3.1) follows if x14.

    CASE 2. x=14. By the first identity of Lemma 2.3, one has

    Dps+1,k(1,14)=k(ps+1)k+22ps+1=2k4(14×14)ps+12+k4(14×14)ps12+12

    as required.

    Thus (3.1) is true for any xFq.

    Now we give the the remainder proof of Theorem 3.1. By Lemmas 2.4-2.5 and (3.1), we readily find that there exists coefficients Ai,jFq such that

    Dps+2,k(1,x)=i=0A2,ps+2i1(14x)ps+2i12+j=0A2,2j(14x)j (3.4)

    with 0p12 and

    Dps+2+1,k(1,x)=+1i=0A2+1,ps+2i1(14x)ps+2i12+j=0A2+1,2j(14x)j (3.5)

    with 0<p12. Therefore we now only need to determine all the coefficients Ai,j. Let u2=14x. On the one hand, by (3.4) and (3.5), one then has

    Dps+2,k(1,x)xDps+21,k(1,x)=Dps+2,k(1,x)1u24Dps+21,k(1,x)=+1i=0A2,ps+2i1ups+2i1+j=0A2,2ju2j14i=0A21,ps+2i1ups+2i1   141j=0A21,2ju2j+14i=0A21,ps+2i1ups+2i+1+141j=0A21,2ju2j+2=14A21,ps+21ups+2+1+i=1(A2,ps+2i114A21,ps+2i1+14A21,ps+2i3)ups+2i1   +(A2,ps114A21,ps1)ups1+(A2,2+14A21,22)u2   +1j=1(A2,2j14A21,2j+14A21,2j2)u2j+A2,014A21,0. (3.6)

    On the other hand, Lemma 2.4 tells us that

    Dps+2+1,k(1,x)=Dps+2,k(1,x)xDps+21,k(1,x).

    So by comparing the coefficient of the term ui in the right hand side of (3.6) and (3.5), one can get the desired results as (3.3). Following the similar way, one also obtain the recursions of Ai,j as (3.2). So the proof Theorem 3.1 is complete.

    For any nonzero integer x, let v2(x) be the 2-adic valuation of x. By Theorem 3.1, the following results are established.

    Theorem 3.2. Let q=pe with p being an odd prime and e being a positive integer. Let s be a nonnegative integer. Then each of following is true.

    (ⅰ). If k=0, then Dps+1,k(1,x) is a PP of Fq if and only if either p1(mod4) and v2(s)v2(e), or p3(mod4) and v2(s)max{v2(e),1}.

    (ⅱ). If k=2, then Dps+1,k(1,x) is a PP of Fq if and only if p=3, v2(s)=0 and \gcd(s, e)=1.

    (ⅲ). If k\ne 2 and k\ne 0, then D_{p^s+1, k}(1, x) is a PP of {\mathbb F_{q}} if and only if s=0.

    Proof. By (3.1) of Theorem 3.1, we have that D_{p^s+1, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

    (2-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}

    is a PP of {\mathbb F_{q}}.

    (ⅰ). Let k=0. Then D_{p^s+1, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the monomial x^{\frac{p^s+1}{2}} is a PP of {\mathbb F_{q}}, namely,

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

    So we consider the following two cases on the odd prime p.

    CASE 1. p\equiv 1\pmod{4}. Then \frac{p^s+1}{2} must be odd. It then follows that

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

    So in this case, by Lemma 2.6 we get that \gcd(\frac{p^s+1}{2}, p^e-1)=1 if and only if \frac{e}{\gcd(s, e)} is odd which is equivalent to v_2(e)\le v_2(s).

    CASE 2. p\equiv 3\pmod{4}. Then v_2(\frac{p^s+1}{2})\ge 1 when s is odd. In this case we have 2|\gcd(\frac{p^s+1}{2}, p^e-1) which is not allowed. So in the case of p\equiv 3\pmod{4}, s must be even. Then \frac{p^s+1}{2} is an odd number. It follows from Lemma 2.6 that \gcd(\frac{p^s+1}{2}, p^e-1)=1 if and only if \frac{e}{\gcd(s, e)} is odd which is equivalent to v_2(e)\le v_2(s) and v_2(s)\ge 1, i.e., v_2(s)\ge \max\{1, v_2(e)\} as desired. Part (ⅰ) is proved.

    (ⅱ). Let k=2. Assume that D_{p^s+1, k}(1, x) is a PP of {\mathbb F_{p^e}}. Then D_{p^s+1, k}(1, x) is a PP of {\mathbb F_{p^e}} if and only if x^{\frac{p^s-1}{2}} is a PP of {\mathbb F_{p^e}}. Clearly, s > 0 in this case. Suppose p > 3, then x^{\frac{p^s-1}{2}} 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 and s > 0 in what follows. 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 number and then so is \frac{3^s-1}{2}. Thus we have that x^{\frac{3^s-1}{2}} is a PP of {\mathbb F_{3^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 Part (ⅱ) is proved.

    (ⅲ). k\ne 0 and k\ne 2. Then the desired result follows from Lemma 2.2 that (2-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}} is a PP of {\mathbb F_{q}} if and only if s=0. Part (ⅲ) is proved.

    Theorem 3.3. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a nonnegative integer and s_0 be the least nonnegative residue of s modulo 2e. Then each of following is true.

    (ⅰ). If k=0, p=3, then D_{p^s+2, k}(1, x) is not a PP of {\mathbb F_{3^e}}.

    (ⅱ). If k=0, p > 3, s_0=0, then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{p^e}}.

    (ⅲ). If k=0, p > 3, s=e, then D_{p^e+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if q=p^e\equiv 1\pmod{3}.

    (ⅳ). If k=2, then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if s=0.

    (ⅴ). Let k=4, p=3. If s=0 or s_0=1, then the binomial D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{3^e}}. If s > 0 and s_0 is even, then D_{p^s+2, k}(1, x) is not a PP of {\mathbb F_{3^e}}.

    (ⅵ). Let k=4, p > 3. Then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{p^e}} if and only if s=0.

    (ⅶ). If k\ne 0, 2, 4 and p\nmid (4-k), then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if s=0 and k\ne 3.

    Proof. By Theorem 3.1, we have that D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

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

    is a PP of {\mathbb F_{q}}.

    (ⅰ). Let k=0, p=3. Then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the monomial x^{\frac{p^s+1}{2}}+\frac{1}{2}x is a PP of {\mathbb F_{q}}. Let

    f_1(x):=x^{\frac{p^s+1}{2}}+\frac{1}{2}x.

    It is easy to see that f_1(x) is not a PP of {\mathbb F_{3^e}} since f_1(0)=f_1(1)=0. So in this case D_{p^s+2, k}(1, x) is not a PP of {\mathbb F_{3^e}}.

    (ⅱ). Let k=0, p > 3, s_0=0. Then \frac{p^s+1}{2}\equiv 1\pmod{p^e-1} which implies that

    f_1(x)\equiv \frac{3}{2}x \pmod{x^{p^e}-x}

    for any x\in {\mathbb F_{q}^*}. Note that f_1(0)=\frac{3}{2}\times 0=0 and the monomial \frac{3}{2}x is a PP of {\mathbb F_{q}}. So f_1(x) is a PP of {\mathbb F_{q}}. That is to say D_{p^s+2, k}(1, x) is not a PP of {\mathbb F_{p^e}}.

    (ⅲ). Let k=0, p > 3, s=e. Then by Theorem 7.11 in [5] we have f_1(x) is a PP of {\mathbb F_{q}} if and only if \eta\big(\big(\frac{1}{2}\big)^2-1\big)=1, i.e., \eta(-3)=1, where \eta(\cdot) denotes the quadratic character of {\mathbb F_{q}}. One can also find that \eta(-3)=1 if and only if q=p^e\equiv 1\pmod{3}, as desired.

    (ⅳ). If k=2, then the desired result follows from Lemma 2.2 that the binomial 2x^{\frac{p^s+1}{2}}+2x^{\frac{p^s-1}{2}} is a PP of {\mathbb F_{q}} if and only if s=0. So D_{p^s+2, k}(1, x) is not a PP of {\mathbb F_{p^e}} if and only if s=0.

    (ⅴ). Let k=4, p=3. Then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if 2x^{\frac{p^s-1}{2}}-x is a PP of {\mathbb F_{q}}. Let

    f_2(x):=x^{\frac{p^s-1}{2}}+\frac{1}{2}x.

    Obviously f_2(x)=x^{\frac{3^s-1}{2}}+x=1+x is a PP of {\mathbb F_{3^e}} when s=0. Now let 0 < s\equiv s_0\pmod{2e} with 0\le {s_0}\le 2e-1. Then \frac{p^s-1}{2}\equiv \frac{p^{s_0}-1}{2}\pmod{p^e-1}. Therefore

    f_2(x)\equiv x^{\frac{3^{s_0}-1}{2}}+x \pmod{x^{3^e}-x}

    for any x\in {\mathbb F_{q}^*}. If s_0=1, f_2(x)\equiv 2x \pmod{x^{3^e}-x} and f_2(0)=2\times 0=0. This means that f_2(x)=2x for any x\in {\mathbb F_{q}}. So f_2(x) is a PP of {\mathbb F_{q}} when s_0=1. If s > 0 and s_0 is even, then x^{\frac{3^{s_0}-1}{2}}+x=1+x for any x\in {\mathbb F_{3}^*}. Note that f_2({\mathbb F_{3}})\subseteq {\mathbb F_{3}} and f_2(0)=f_2(-1)=0, which tells us that f_2(x) is not a PP of {\mathbb F_{3}}. Thus the desired results follows. Unfortunately, following the similar way, we cannot say anything for the case of s > 0 and s_0 being odd with s_0\ge 3.

    (ⅵ). Let k=4, p > 3. Then D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if 2x^{\frac{p^s-1}{2}}-x is a PP of {\mathbb F_{q}}. It then follows from Lemma 2.8 that D_{p^s+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if s=0, as required.

    (ⅶ). Let p\ge 3, k\ne 0, 2, 4 and p\nmid (k-4). Denote

    f_3(x):=(4-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}+(2-k)x.

    First, if s=0, then f_3(x)=(6-2k)x+k which is a PP of {\mathbb F_{p^e}} if and only if k\ne 3. In what follows we will show that f_3(x) is not a PP of {\mathbb F_{p^e}} when s > 0. Let 0 < s\equiv s_0\pmod{2e} with 0\le s_0\le 2e-1. Then

    f_3(x)\equiv (4-k)x^{\frac{p^{s_0}+1}{2}}+kx^{\frac{p^{s_0}-1}{2}}+(2-k)x\pmod{x^{p^e}-x}. (3.7)

    We consider the following cases.

    CASE 1. s > 0, s_0=0. By (3.7) we have f_3(x)\equiv (6-2k)x+k\pmod{x^{p^e}-x}, which means that f_3(x)=(6-2k)x+k for any x\in {\mathbb F_{p^e}^*}. If k=3, then \forall x\in {\mathbb F_{p^e}^*} f_3(x)=k. Obviously, f_3(x) is not a PP of {\mathbb F_{p^e}}. If k\ne 3, then f_3(x)=0 has one nonzero root \frac{k}{2k-6}\in {\mathbb F_{p^e}^*} since k\ne 0. But f_3(0)=0. So f_3(x) is not a PP of {\mathbb F_{p^e}} in this case.

    CASE 2. s > 0 and s_0 is a positive even number. Then x^{\frac{p^{s_0}+1}{2}}=x and x^{\frac{p^{s_0}-1}{2}}=1 for each x\in {\mathbb F_{p}^*}, which together with (3.7) imply that f_3(x)=(6-2k)x+k for any x\in {\mathbb F_{p}^*}. If k=3, then \forall x\in {\mathbb F_{p}^*}, f_3(x)=k. Obviously, f_3(x) is not a PP of {\mathbb F_{p}}. If k\ne 3, then f_3(x)=0 has one nonzero root \frac{k}{2k-6}\in {\mathbb F_{p}^*} since k\ne 0. But f_3(0)=0. Therefore f_3(x) is not a PP of {\mathbb F_{p}} in this case. So is f_3(x) of {\mathbb F_{p^e}} since f_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}.

    CASE 3. s > 0 and s_0 is odd. Then x^{\frac{p^{s_0}+1}{2}}=x^{\frac{p+1}{2}} and x^{\frac{p^{s_0}-1}{2}}=x^{\frac{p-1}{2}} for each x\in {\mathbb F_{p}^*}, which together with (3.7) imply that f_3(x)=(4-k)x^{\frac{p+1}{2}}+kx^{\frac{p-1}{2}}+(2-k)x for any x\in {\mathbb F_{p}^*}. If p=3, then k must equal 1 since 0\le k < p and k\ne 0, 2, 4, which contradicts to the condition p\nmid (4-k). So one has p > 3. Then

    [f_3(x)]^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_3(x) is not a PP of {\mathbb F_{p}}. Also note that f_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_3(x) is not a PP of {\mathbb F_{p^e}}.

    Combining the above cases, we verify that f_3(x) is not a PP of {\mathbb F_{p^e}} when s > 0. Thus Part (vii) is proved. So the proof of Theorem 3.3 is complete.

    Theorem 3.4. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a nonnegative integer and s_0 be the least nonnegative residue of s modulo 2e. Then each of following is true.

    (ⅰ). If k=0, p=3, then D_{3^s+3, 0}(1, x) is a PP of {\mathbb F_{3^e}} if and only if v_2(s-1)\ge \max\{1, v_2(e)\}.

    (ⅱ). Let k=0, p > 3. If s_0 is an even number, then D_{p^s+3, 0}(1, x) is not a PP of {\mathbb F_{p^e}}.

    (ⅲ). If k=2, s=0, then D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}}

    (ⅳ). Let k=2, s > 0, p=3. If s_0=1, then D_{3^s+3, 2}(1, x) is a PP of {\mathbb F_{3^e}}. If s_0 is even, then D_{3^s+3, 2}(1, x) is not a PP of {\mathbb F_{3^e}}.

    (ⅴ). Let k=2, p > 3. Then D_{p^s+3, 2}(1, x) is a PP of {\mathbb F_{p^e}} if and if s=0.

    (ⅵ). If k=3, then D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if p=3 and v_2(s-1)\ge \max\{1, v_2(e)\}.

    (ⅶ). If k\ne 0, 2, 3, then D_{p^s+3, k}(1, x) is not a PP of {\mathbb F_{p^e}}.

    Proof. By Theorem 3.1, we have that D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

    (2-k)x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}}+ kx^{\frac{p^s-1}{2}}+(6-2k)x (3.8)

    is a PP of {\mathbb F_{q}}.

    (ⅰ). Letting k=0, we have D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if (3.8) is a PP of {\mathbb F_{q}}, i.e., the trinomial x^{\frac{p^s+3}{2}}+3x^{\frac{p^s+1}{2}}+3x is a PP of {\mathbb F_{q}}. Let

    f_4(x):=x^{\frac{p^s+3}{2}}+3x^{\frac{p^s+1}{2}}+3x.

    Now let p=3. Then f_4(x) is a PP of {\mathbb F_{q}} if and only if x^{\frac{3^s+3}{2}} is a PP of {\mathbb F_{q}}. The latter is equivalent to \gcd(\frac{3^s+3}{2}, 3^e-1)=1. Now we let \gcd(\frac{3^s+3}{2}, 3^e-1)=1. If s is even, then one has v_2(\frac{3^s+3}{2})\ge 1. It follows that 2|\gcd(\frac{3^s+3}{2}, 3^e-1), which is a contradiction. So s must be odd. Then \frac{3^s+3}{2} is an odd integer. It follows from Lemma 2.6 that \gcd(\frac{3^s+3}{2}, 3^e-1)=\gcd(\frac{3^{s-1}+1}{2}, \frac{3^e-1}{2})= \frac{1}{2}\gcd(3^{s-1}+1, 3^e-1)=1 if and only if \frac{e}{\gcd(s-1, e)} is odd. This means that \gcd(\frac{3^s+3}{2}, 3^e-1)=1 if and only if v_2(e)\le v_2(s-1) and v_2(s-1)\ge 1, namely, v_2(s-1)\ge \max\{1, v_2(e)\}, as desired.

    (ⅱ). Let k=0, p > 3. Then \frac{p^s+3}{2}\equiv \frac{p^{s_0}+3}{2}\pmod{p^e-1} and \frac{p^s+1}{2}\equiv \frac{p^{s_0}+1}{2}\pmod{p^e-1}. So

    f_4(x)\equiv x^{\frac{p^{s_0}+3}{2}}+3x^{\frac{p^{s_0}+1}{2}}+3x\pmod{x^{p^e}-x}. (3.9)

    Clearly, if s_0 is even, then x^{\frac{p^{s_0}+3}{2}}=x^2 and x^{\frac{p^{s_0}+1}{2}}=x for any x\in {\mathbb F_{p}^*}. Then by (3.9) we have f_4(x)=x^2+6x=x(x+6) for any x\in {\mathbb F_{p}^*}. Hence f_4(x)=0 has one nonzero root -6 in {\mathbb F_{p}^*}. But f_4(0)=0. It then follows that f_4(x) is not a PP of {\mathbb F_{p}}. One notes that f_4({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_4(x) is not a PP of {\mathbb F_{p^e}}. It follows that D_{p^s+3, k}(1, x) is not a PP of {\mathbb F_{q}} when s_0 is even.

    (ⅲ). Letting k=2, we have D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if (3.8) is a PP of {\mathbb F_{q}}, i.e., 3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x is a PP of {\mathbb F_{q}}. Let

    f_5(x):=3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x.

    Let s=0. Then f_5=4x+1, which clearly is a PP of {\mathbb F_{q}}. So D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}}.

    (ⅳ). Let k=2, p=3, s > 0. Then the desired result follows from the proof of Part (v) of Theorem 3.3.

    (ⅴ). Let k=2, p > 3. By Part (iii) we only need to show that D_{p^s+3, k}(1, x) is not a PP of {\mathbb F_{q}} when s > 0. Let 0 < s\equiv s_0\pmod{2e} with 0\le s_0\le 2e-1. Then one has

    f_5(x)\equiv 3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x\pmod{x^{p^e}-x}.

    If s_0 is even, then f_5(x)=4x+1 for any x\in {\mathbb F_{p}^*}. In this situation, f_5(x)=0 has one nonzero root -\frac{1}{4}\in {\mathbb F_{p}^*}. So f_5(x) is not a PP of {\mathbb F_{p}} since f_5(0)=0. Also note that f_5({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Thus f_5(x) is not a PP of {\mathbb F_{p^e}} in this case. If s_0 is odd, then f_5(x)=3x^{\frac{p+1}{2}}+x^{\frac{p-1}{2}}+x for any x\in {\mathbb F_{p}^*}. So in {\mathbb F_{p}^*}, we have

    (f_5(x))^2\equiv x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_5(x) is not a PP of {\mathbb F_{p}}. We note that f_5({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_5(x) is not a PP of {\mathbb F_{p^e}}. It infers that D_{p^s+3, k}(1, x) is not a PP of {\mathbb F_{q}} when s > 0. Part (v) is proved.

    (ⅵ). Let k=3. Then D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if (3.8) is a PP of {\mathbb F_{q}}, i.e., the trinomial

    f_6(x):=-x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}}+3x^{\frac{p^s-1}{2}}

    is a PP of {\mathbb F_{q}}. By the result of Part (ⅰ), we then have from the fact D_{3^s+3, 3}(1, x)=D_{3^s+3, 0}(1, x) that f_6(x) is a PP of of {\mathbb F_{3^e}} if and only if v_2(s-1)\ge \max\{1, v_2(e)\}. Then we only need to show that f_6(x) is not a PP of {\mathbb F_{p^e}} when p > 3. Let s\equiv s_0\pmod{2e} with 0\le s_0\le 2e-1. Then one has

    f_6(x)\equiv -x^{\frac{p^{s_0}+3}{2}}+6x^{\frac{p^{s_0}+1}{2}}+3x^{\frac{p^{s_0}-1}{2}} \pmod{x^{p^e}-x}.

    If s_0 is even, then f_6(x)=-x^2+6x+3 for any x\in {\mathbb F_{p}^*}. Then f_6(x) is not a PP of {\mathbb F_{p}} since f_6(2)=f_6(4)=11. Also note that f_6({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Thus f_6(x) is not a PP of {\mathbb F_{p^e}} for p > 3 and s_0 being even. If s_0 is odd, then f_6(x)=-x^{\frac{p+3}{2}}+6x^{\frac{p+1}{2}}+3x^{\frac{p-1}{2}} for any x\in {\mathbb F_{p}^*}. So in {\mathbb F_{p}^*}, we have

    (f_6(x))^2\equiv 9x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_6(x) is not a PP of {\mathbb F_{p}}. We note that f_6({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_6(x) is not a PP of {\mathbb F_{p^e}} when p > 3 and s_0 is odd. So f_6(x) is a PP of {\mathbb F_{q}} if and only if p=3 and v_2(s-1)\ge \max\{1, v_2(e)\}, that is, D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if p=3 and v_2(s-1)\ge \max\{1, v_2(e)\}. Part (ⅵ) is proved.

    (ⅳ). Let k\ne 0, 2, 3 and 0\le k < p. Then D_{p^s+3, k}(1, x) is a PP of {\mathbb F_{q}} if and only if (3.8) is a PP of {\mathbb F_{q}}, i.e., if and only if

    f_7(x):=(2-k)x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}} +kx^{\frac{p^s-1}{2}}+(6-2k)x

    is a PP of {\mathbb F_{q}}. Let s\equiv s_0\pmod{2e} with 0\le s_0\le 2e-1. Then one has

    f_7(x)\equiv (2-k)x^{\frac{p^{s_0}+3}{2}}+6x^{\frac{p^{s_0}+1}{2}} +kx^{\frac{p^{s_0}-1}{2}}+(6-2k)x \pmod{x^{p^e}-x}.

    If s_0 is even, then f_7(x)=(2-k)x^2+(12-2k)x+k for any x\in {\mathbb F_{p}^*}. One then finds that f_7(\frac{4}{k-2})=f_7(\frac{8-2k}{k-2}) and \frac{4}{k-2}\ne \frac{8-2k}{k-2} when k\ne 4. If k=4, then p\ge 5. In this case, f_7(x)=-2x^2+4x+4 for any x\in {\mathbb F_{p}^*}, which implies f_7(-1)=f_7(3) when k=4. Therefore f_7(x) is not a PP of {\mathbb F_{p}}. Also note that f_7({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Thus f_7(x) is not a PP of {\mathbb F_{p^e}} when s_0 is even. If s_0 is odd, then

    f_7(x)=(2-k)x^{\frac{p+3}{2}}+6x^{\frac{p+1}{2}}+kx^{\frac{p-1}{2}}+(6-2k)x (3.10)

    for any x\in {\mathbb F_{p}^*}. We consider the following two cases.

    CASE 1. Let p=3. Then k=1 since k < p and k\ne 0, 2. Hence \forall x\in{\mathbb F_{3}^*}, f_7(x)=x^3+2x. It then follows from f_7(0)=f_7(1)=0 that f_7(x) is not a PP of {\mathbb F_{3}}. We note that f_7({\mathbb F_{3}})\subseteq {\mathbb F_{3}}. Therefore f_7(x) is not a PP of {\mathbb F_{3^e}}.

    CASE 2. Let p > 3. By (3.10), in {\mathbb F_{p}} we have

    (f_7(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_7(x) is not a PP of {\mathbb F_{p}}. We note that f_7({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_7(x) is not a PP of {\mathbb F_{p^e}} when p > 3 and s_0 is odd.

    Hence f_7(x) is not a PP of {\mathbb F_{p^e}} when k\ne 0, 2, 3, from which we deduce immediately that D_{p^s+3, k}(1, x) is not a PP of {\mathbb F_{p^e}} when k\ne 0, 2, 3. Part (vii) is proved. So we completes the proof of Theorem 3.4.

    Theorem 3.5. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a nonnegative integer and s_0 be the least nonnegative residue of s modulo 2e. If D_{p^s+4, k}(1, x) is a PP of {\mathbb F_{q}}, then either k=0 and s_0 is odd, or k > 0, k\ne 2 and s=0.

    Proof. It is sufficient to show that D_{p^s+4, k}(1, x) is not a PP of {\mathbb F_{q}} when k=0, s_0 is even, or k > 0, s > 0. By Theorem 3.1, we get

    \begin{align*} 32D_{p^s+4, k}(1, x)&=k(1-4x)^{\frac{p^s-1}{2}}+(8+2k)(1-4x)^{\frac{p^s+1}{2}}\\ &+(8-3k)(1-4x)^{\frac{p^s+3}{2}}+2+3k+(12-2k)(1-4x)+(2-k)(1-4x)^2. \end{align*}

    Then D_{p^s+4, k}(1, x) is a PP of {\mathbb F_{q}} if and only if kx^{\frac{p^s-1}{2}}+(8+2k)x^{\frac{p^s+1}{2}} +(8-3k)x^{\frac{p^s+3}{2}}+(12-2k)x+(2-k)x^2 is a PP of {\mathbb F_{q}}. Let

    f_8(x):=kx^{\frac{p^s-1}{2}}+(8+2k)x^{\frac{p^s+1}{2}} +(8-3k)x^{\frac{p^s+3}{2}}+(12-2k)x+(2-k)x^2.

    Now we show that f_8(x) is not a PP of {\mathbb F_{q}} when k=0, s_0 is even, or k > 0, s > 0. Then the following cases are considered.

    CASE 1. k=0 and s_0 is an even. Then f_8(x)=4x^{\frac{p^{s_0}+1}{2}} +4x^{\frac{p^{s_0}+3}{2}}+6x+x^2. It infers that

    f_8(x)\equiv 4x^{\frac{p^{s_0}+1}{2}} +4x^{\frac{p^{s_0}+3}{2}}+6x+x^2\pmod{x^{q}-x}.

    Additionally, \forall x\in{\mathbb F_{p}^*}, x^{\frac{p^{s_0}+1}{2}}=x and x^{\frac{p^{s_0}+3}{2}}=x^2 since s_0 is an even. Therefore

    f_8(x)=5x(x+2)

    for any x\in{\mathbb F_{p}^*}. Then f_8(0)=f_8(-2)=0. So f_8(x) is not a PP of {\mathbb F_{p}}. Also f_8(x) is not a PP of {\mathbb F_{p^e}} since f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}.

    CASE 2. k=2. Then

    f_8(x)=2x^{\frac{p^{s}-1}{2}} +12x^{\frac{p^{s}+1}{2}}+2x^{\frac{p^{s}+3}{2}}+8x.

    SUBCASE 2-1. p=3. Then f_8(x)=2x^{\frac{p^{s}-1}{2}} +2x^{\frac{p^{s}+3}{2}}+2x. So f_8(x)=2x^{2} +2x+2 when s=0, which then follows that f_8(0)=f_8(2)=2. If s > 0, we have easily that f_8(0)=f_8(1)=0. Thus f_8(x) is not a PP of {\mathbb F_{3^e}} whenever.

    SUBCASE 2-2. p > 3. Then

    f_8(x)\equiv 2x^{\frac{p^{s_0}-1}{2}} +12x^{\frac{p^{s_0}+1}{2}}+2x^{\frac{p^{s_0}+3}{2}} +8x\pmod{x^{p^e}-x}.

    If s_0 is even, then f_8(x)=2x^2+20x+2 for any x\in {\mathbb F_{p}^*}. This implies that f_8(-4)=f_8(-6), which then follows that f_8(x) is not a PP of {\mathbb F_{p}}. Note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_8(x) is not a PP of {\mathbb F_{q}} when s_0 is even. If s_0 is odd, then f_8(x)=2x^{\frac{p-1}{2}}+12x^{\frac{p+1}{2}} +2x^{\frac{p+3}{2}}+8x for any x\in {\mathbb F_{p}^*}. We then deduces that

    (f_8(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_8(x) is not a PP of {\mathbb F_{p}}. We note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_8(x) is not a PP of {\mathbb F_{p^e}} when s_0 is odd.

    Thus D_{p^s+4, 2}(1, x) is not a PP of {\mathbb F_{p^e}} for any nonnegative integer s and odd prime p.

    CASE 3. k=6, s > 0. Then p\ge 7 and

    f_8(x)=6x^{\frac{p^{s}-1}{2}} +20x^{\frac{p^{s}+1}{2}}-10x^{\frac{p^{s}+3}{2}}-4x^2.

    If s > 0 and s_0 is even, then f_8(x)=-14x^2+20x+6 for any x\in {\mathbb F_{p}^*}. This implies that f_8(0)=f_8(-1)=0 if p=7, or f_8(\frac{4}{7})=f_8(\frac{6}{7}) if p > 7. This means that f_8(x) is not a PP of {\mathbb F_{p}}. Note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_8(x) is not a PP of {\mathbb F_{q}} when s_0 is even. If s > 0 and s_0 is odd, then f_8(x)=6x^{\frac{p-1}{2}}+20x^{\frac{p+1}{2}} -10x^{\frac{p+3}{2}}-4x^2 for any x\in {\mathbb F_{p}^*}, which implies that

    (f_8(x))^2\equiv 36x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_8(x) is not a PP of {\mathbb F_{p}}. We note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_8(x) is not a PP of {\mathbb F_{p^e}} when s_0 is odd.

    Thus D_{p^s+4, 6}(1, x) is not a PP of {\mathbb F_{p^e}} when s > 0.

    CASE 4. k=p-4, s > 0. Then p\ge 5 and

    f_8(x)=(p-4)x^{\frac{p^{s}-1}{2}} +(20-3p)x^{\frac{p^{s}+3}{2}}+(20-2p)x+(6-p)x^2.

    If s > 0, s_0 is even, then f_8(x)=26x^2+20x-4 for any x\in {\mathbb F_{p}^*}. This implies that f_8(0)=f_8(\frac{1}{5})=0 if p=13, or f_8(\frac{-4}{13})=f_8(\frac{-6}{13}) if p\ne 13. This means that f_8(x) is not a PP of {\mathbb F_{p}}. Note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_8(x) is not a PP of {\mathbb F_{q}} when s_0 is even. If s > 0, s_0 is odd, then f_8(x)=-4x^{\frac{p-1}{2}}+20x^{\frac{p+3}{2}} +20x+6x^2 for any x\in {\mathbb F_{p}^*}, which implies that

    (f_8(x))^2\equiv 16x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_8(x) is not a PP of {\mathbb F_{p}}. We note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_8(x) is not a PP of {\mathbb F_{p^e}} when s_0 is odd.

    Thus D_{p^s+4, p-4}(1, x) is not a PP of {\mathbb F_{p^e}} when s > 0.

    CASE 5. p\mid (3k-8), s > 0. Then p\ge 5, p\nmid (2-k) and

    f_8(x)=kx^{\frac{p^{s}-1}{2}} +(8+2k)x^{\frac{p^{s}+1}{2}}+(12-2k)x+(2-k)x^2.

    If s > 0, s_0 is even, then f_8(x)=(2-k)x^2+20x+k for any x\in {\mathbb F_{p}^*}. This implies that f_8(\frac{-11}{2-k})=f_8(\frac{-9}{2-k})=0. This means that f_8(x) is not a PP of {\mathbb F_{p}}. Note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_8(x) is not a PP of {\mathbb F_{q}} when s_0 is even. If s > 0, s_0 is odd, then f_8(x)=kx^{\frac{p-1}{2}}+(8+2k)x^{\frac{p+1}{2}} +(12-2k)x+(2-k)x^2 for any x\in {\mathbb F_{p}^*}, which implies that

    (f_8(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_8(x) is not a PP of {\mathbb F_{p}}. We note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_8(x) is not a PP of {\mathbb F_{p^e}} when s_0 is odd.

    Thus D_{p^s+4, k}(1, x) is not a PP of {\mathbb F_{p^e}} when p\mid (3k-8) and s > 0.

    CASE 6. k\ne 0, 2, 6, p-4, s > 0 and p\nmid (3k-8). Then

    f_8(x)=kx^{\frac{p^{s}-1}{2}} +(8+2k)x^{\frac{p^{s}+1}{2}}+(8-3k)x^{\frac{p^{s}+3}{2}}+(12-2k)x+(2-k)x^2.

    If s > 0, s_0 is even, then f_8(x)=(10-4k)x^2+20x+k for any x\in {\mathbb F_{p}^*}. If p\mid (2k-5), then p\ne 5 and f_8(x)=20x+k, \forall x\in\mathbb{F}_p^*. It implies that f_8(0)=f_8(\frac{-k}{20})=0. So f_8(x) is not a PP of {\mathbb F_{p}} when p\mid (2k-5). If p\nmid (2k-5), then f_8(\frac{4}{2k-5})=f_8(\frac{6}{2k-5}), which means that f_8(x) is not a PP of {\mathbb F_{p}} when p\nmid (2k-5). Thus f_8(x) is not a PP of {\mathbb F_{p}} when s > 0, s_0 is even. Note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So f_8(x) is not a PP of {\mathbb F_{q}} when s_0 is even. If s > 0, s_0 is odd, then f_8(x)=kx^{\frac{p-1}{2}}+(8+2k)x^{\frac{p+1}{2}} +(8-3k)x^{\frac{p+3}{2}}+(12-2k)x+(2-k)x^2 for any x\in {\mathbb F_{p}^*}. If p=3, then k=1. In this case f_8(x)=2x+2x^2+2x^3, which implies that f_8(0)=f_8(1)=0. It then follows that f_8(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (f_8(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Then by Lemma 2.7, we know that f_8(x) is not a PP of {\mathbb F_{p}}. Thus f_8(x) is not a PP of {\mathbb F_{p}} when s_0 is odd. We note that f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Therefore f_8(x) is not a PP of {\mathbb F_{p^e}} when s_0 is odd.

    Thus D_{p^s+4, k}(1, x) is not a PP of {\mathbb F_{p^e}} when k\ne 0, 2, 6, p-4, s > 0 and p\nmid (3k-8). Combining all of the above cases, we have the desired result. Therefore Theorem 3.5 is proved.

    Corollary 3.6. Let q=p^e with p being an odd prime and e being a positive integer. Let s and k be nonnegative integers with 0 < k < p. Then D_{p^s+4, k}(1, x) is a PP of {\mathbb F_{q}} if and only if s=0 and p\mid(2k-5).

    Proof. The desired result follows immediately from the proof of Theorem 3.5.


    4. Reversed Dickson polynomials D_{p^s+p^t+\ell, k}(1, x)

    In this section, we present an explicit formula for D_{n, k}(1, x) when n=p^s+p^t+\ell with \le s < t and 0\le \ell < p. Then we characterize D_{n, k}(1, x) to be a PP of {\mathbb F}_q in this case.

    Theorem 4.1. Let p={\rm char}({\mathbb F_{q}}) be an odd prime. Let s and t be integers such that 0\le s < t. Then

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

    Proof. We consider the following two cases.

    CASE 1. x\ne \frac{1}{4}. For this case, putting x=y(1-y) in the second identity of Lemma 2.3 gives us that

    \begin{align*} D_{p^s+p^{t}, k}(1, x)=&D_{p^s+p^{t}, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^s+p^{t}}-\big(1+(k-2)y\big)(1-y)^{p^s+p^{t}}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s+p^{t}} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s+p^{t}}}{u}\\ =&\frac{k}{4}(u^{p^s-1}+u^{p^{t}-1})-\frac{k-2}{4}(1+u^{p^s+p^{t}})\\ =&\frac{k}{4}\big((u^2)^{\frac{p^s-1}{2}}+(u^2)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}(1+(u^2)^{\frac{p^s+p^{t}}{2}}), \end{align*}

    where u=2y-1 and u^2=1-4x. So we obtain that

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

    as desired.

    CASE 2. x=\frac{1}{4}. By the first identity of Lemma 2.3, one has

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

    Besides,

    \frac{k}{4}\big((1-4\times \frac{1}{4})^{\frac{p^s-1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}\big(1 +(1-4\times \frac{1}{4})^{\frac{p^{s}+p^{t}}{2}}\big)=\frac{-k+2}{4}.

    Thus the required result follows. So Theorem 4.1 is proved.

    Theorem 4.2. Let q=p^e with p being an odd prime and e being a positive integer. Let s and t be positive integers with s < t. Then each of following is true.

    (ⅰ). If k=0, then D_{p^s+p^t, k}(1, x) is a PP of {\mathbb F_{q}} if and only if either p\equiv 1\pmod{4} and v_2(t-s)\ge v_2(e) , or p\equiv 3\pmod{4} and v_2(t-s)\ge \max\{v_2(e), 1\}.

    (ⅱ). Let k=2. If p > 3, then D_{p^s+p^t, k}(1, x) is not a PP of {\mathbb F_{p^e}}. If p=3 and st is even, then D_{p^s+p^t, k}(1, x) is not a PP of {\mathbb F_{p^e}}.

    (ⅲ). If k\ne 0, 2, then D_{p^s+p^t, k}(1, x) is not a PP of {\mathbb F_{q}}.

    Proof. By Theorem 4.1, we have that D_{p^s+p^{t}, k}(1, x) is a PP of {\mathbb F_{q}} if and only if

    k\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)x^{\frac{p^{s}+p^{t}}{2}}

    is a PP of {\mathbb F_{q}}.

    (ⅰ). Let k=0. Then D_{p^s+p^{t}, k}(1, x) is a PP of {\mathbb F_{q}} if and only if x^{\frac{p^{s}+p^{t}}{2}} is a PP of {\mathbb F_{q}} if and only if

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

    Additionally, \gcd\big(\frac{p^s+p^t}{2}, p^e-1\big)= \gcd\big(\frac{p^{t-s}+1}{2}, p^e-1\big). Then the desired result follows from the same way as proving Part (ⅰ) of Theorem 3.2.

    (ⅱ). Let k=2. Then D_{p^s+p^{t}, k}(1, x) is a PP of {\mathbb F_{q}} if and only if x^{\frac{p^{s}-1}{2}}+x^{\frac{p^{t}-1}{2}} is a PP of {\mathbb F_{q}}. Let

    g_1(x):=x^{\frac{p^{s}-1}{2}}+x^{\frac{p^{t}-1}{2}}.

    So

    g_1(x)\equiv x^{\frac{p^{s_0}-1}{2}}+x^{\frac{p^{t_0}-1}{2}}\pmod{x^{p^e}-x}.

    Then the following cases are considered.

    CASE 1. s > 0 and both s_0 and t_0 are even. Then g_1(x)=2 for any x\in {\mathbb F_{p}^*}. So g_1(x) is not a PP of {\mathbb F_{p}}. One also notices that g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Thus g_1(x) is not a PP of {\mathbb F_{q}}.

    CASE 2. s > 0 and one of s_0 and t_0 is even, the other is odd. Then g_1(x)=x^{\frac{p-1}{2}}+1 for any x\in {\mathbb F_{p}^*}. If p=3, then \forall x\in{\mathbb F_{p}^*}, g_1(x)=x+1, which implies g_1(0)=g_1(-1)=0. So g_1(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_1(x))^2\equiv x^{p-1}+2x^{\frac{p-1}{2}}+1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_1(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_1(x) is not a PP of {\mathbb F_{p}}. Obviously, g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Hence g_1(x) is not a PP of {\mathbb F_{q}} in this case.

    CASE 3. s > 0, p > 3 and both s_0 and t_0 are odd. Then g_1(x)=2x^{\frac{p-1}{2}} for any x\in {\mathbb F_{p}}. But \gcd(\frac{p-1}{2}, p-1)=\frac{p-1}{2} > 1. Therefore g_1(x) is not a PP of {\mathbb F_{p}}. Note that g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Hence g_1(x) is not a PP of {\mathbb F_{q}} in this case.

    Combining the above cases, we know that part (ⅱ) is true.

    (ⅲ). Let k\ne 0 and k\ne 2. Let

    g_2(x):=k\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)x^{\frac{p^{s}+p^{t}}{2}}.

    Then

    g_2(x)\equiv k\big(x^{\frac{p^{s_0}-1}{2}}+ x^{\frac{p^{{t_0}}-1}{2}}\big)-(k-2)x^{\frac{p^{{s_0}}+p^{{t_0}}}{2}}\pmod{x^q-x}.

    Then we divide the proof into the following three cases.

    CASE 1. Both s_0 and t_0 are even. Then g_2(x)=2k-(k-2)x for any x\in {\mathbb F_{p}^*}. So g_2(x) is not a PP of {\mathbb F_{p}} since g_2(0)=g_2(\frac{2k}{k-2}). One also notices that g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Thus g_2(x) is not a PP of {\mathbb F_{q}}.

    CASE 2. One of s_0 and t_0 is even, the other is odd. Then g_2(x)=k+kx^{\frac{p-1}{2}}-(k-2)x^{\frac{p+1}{2}} for any x\in {\mathbb F_{p}^*}. If p=3, then k=1 and so g_2(0)=g_2(1)=0, which implies g_2(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_2(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_2(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_2(x) is not a PP of {\mathbb F_{p}}. Obviously, g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Hence g_2(x) is not a PP of {\mathbb F_{q}} in this case.

    CASE 3. Both s_0 and t_0 are odd. Then g_2(x)=2kx^{\frac{p-1}{2}}-(k-2)x for any x\in {\mathbb F_{p}^*}. If p=3, then k=1 and so g_2(x)=0, \forall x\in\mathbb{F}_p, which implies g_2(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_2(x))^2\equiv 4k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    Since 4k^2\in \mathbb{F}_p^*, it then follows from Lemma 2.7 that g_2(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_2(x) is not a PP of {\mathbb F_{p}}. Obviously, g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. Hence g_2(x) is not a PP of {\mathbb F_{q}} in this case.

    Combining the above cases, we deduce that g_2(x) is not a PP of {\mathbb F_{q}} in the condition of k\ne 0, 2. Thus D_{p^s+p^t, k}(1, x) is not a PP of {\mathbb F_{q}}. The proof of Theorem 4.2 is completed.

    Theorem 4.3. Let q=p^e with p being an odd prime and e being a positive integer. Let s and t be positive integers with s < t. Then

    \begin{align} D_{p^s+p^{t}+1, k}(1, x)&=\frac{1}{4}(1-4x)^{\frac{p^s+p^{t}}{2}}+ \frac{1}{4}+\frac{1}{8}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)\nonumber\\ &-\frac{k-2}{8}\big((1-4x)^{\frac{p^s+1}{2}}+ (1-4x)^{\frac{p^{t}+1}{2}}\big).\label{cheng5} \end{align} (4.1)

    Furthermore, D_{p^s+p^{t}+1, k}(1, x) is not a PP of {\mathbb F_{q}}.

    Proof We consider the following two cases.

    CASE 1. x\ne \frac{1}{4}. For this case, putting x=y(1-y) in the second identity of Lemma 2.3 gives us that

    \begin{align*} &D_{p^s+p^{t}+1, k}(1, x)=D_{p^s+p^{t}+1, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^s+p^{t}+1}-\big(1+(k-2)y\big)(1-y)^{p^s+p^{t}+1}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s+p^{t}+1} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s+p^{t}+1}}{u}\\ =&\frac{k}{8}(1+u^{p^s-1}+u^{p^{t}-1}+u^{p^s+p^{t}})-\frac{k-2}{8}(1+u^{p^s+1}+u^{p^{t}+1}+u^{p^s+p^{t}})\\ =&\frac{1}{4}u^{p^s+p^{t}}+\frac{1}{4}+\frac{k}{8}(u^{p^s-1}+u^{p^{t}-1})-\frac{k-2}{8}(u^{p^s+1}+u^{p^{t}+1}), \end{align*}

    where u=2y-1 and u^2=1-4x. Then we have that

    \begin{align*} &D_{p^s+p^{t}+1, k}(1, x)=\frac{1}{4}(1-4x)^{\frac{p^s+p^{t}}{2}}+\frac{1}{4}\\ &+\frac{1}{8}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{8}\big((1-4x)^{\frac{p^s+1}{2}}+ (1-4x)^{\frac{p^{t}+1}{2}}\big) \end{align*}

    as desired.

    CASE 2. x=\frac{1}{4}. On the one hand, by the first identity of Lemma 2.3, one has

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

    On the other hand,

    \frac{1}{4}(1-4\times \frac{1}{4})^{\frac{p^s+p^{t}}{2}}+\frac{1}{4}+ \frac{1}{8}\big((1-4\times \frac{1}{4})^{\frac{p^s-1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}-1}{2}}\big)- \frac{k-2}{8}\big((1-4\times \frac{1}{4})^{\frac{p^s+1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}+1}{2}}\big)=\frac{1}{4}.

    Combing Case 1 and Case 2, we know that (4.1) always holds. So D_{p^s+p^{t}+1, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

    g_3(x):=2x^{\frac{p^s+p^{t}}{2}}+2\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)\big(x^{\frac{p^s+1}{2}}+ x^{\frac{p^{t}+1}{2}}\big)

    is a PP of {\mathbb F_{q}}.

    In what follows, we show that g_3(x) is not a PP of {\mathbb F_{q}}. Now let s\equiv s_0\pmod{2e} and t\equiv t_0\pmod{2e} with 0\le s_0\le 2e-1, 0\le t_0\le 2e-1. Then

    g_3(x)\equiv 2x^{\frac{p^{s_0}+p^{{t_0}}}{2}}+2\big(x^{\frac{p^{s_0}-1}{2}}+ x^{\frac{p^{{t_0}}-1}{2}}\big)-(k-2)\big(x^{\frac{p^{s_0}+1}{2}}+ x^{\frac{p^{{t_0}}+1}{2}}\big)\pmod{x^q-x}.

    First we let k=2. In this case we have

    g_3(x)\equiv 2x^{\frac{p^{s_0}+p^{{t_0}}}{2}}+2x^{\frac{p^{s_0}-1}{2}}+ 2x^{\frac{p^{{t_0}}-1}{2}}\pmod{x^q-x}.

    If both s_0 and t_0 are even, then \forall x\in{\mathbb F_{p}^*}, g_3(x)=2x+4. It follows that g_3(x) is not a PP of {\mathbb F_{p}} since g_3(0)=g_3(-2)=0.

    If exactly one of s_0 and t_0 is even, then g_3(x)=2x^{\frac{p-1}{2}}+2x^{\frac{p+1}{2}}+2 for any x\in {\mathbb F_{p}^*}. In this case if p=3, then g_3(0)=g_3(1)=0, which implies g_3(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_3(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_3(x) is not a PP of {\mathbb F_{p}}.

    If both s_0 and t_0 are odd, then g_3(x)=2x+4x^{\frac{p-1}{2}} for any x\in {\mathbb F_{p}^*}. If p=3, then g_3(x)=0, \forall x\in {\mathbb F_{p}^*}, which implies g_3(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_3(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_3(x) is not a PP of {\mathbb F_{p}}.

    Combining the above discussions, we derive that g_3(x) is not a PP of {\mathbb F_{p}}. Note that g_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So g_3(x) is not a PP of {\mathbb F_{q}} when k=2.

    Now let k\ne 2. The following cases are considered.

    If both s_0 and t_0 are even, then \forall x\in{\mathbb F_{p}^*}, g_3(x)=(6-2k)x+4. Clearly, if k=3, then g_3(x) is not a PP of {\mathbb F_{p}}. If k\ne 3, then g_3(0)=g_3(\frac{2}{k-3})=0. This implies that g_3(x) is not a PP of {\mathbb F_{p}}.

    If exactly one of s_0 and t_0 is even then g_3(x)=(4-k)x^{\frac{p+1}{2}}+2x^{\frac{p-1}{2}}-(k-2)x+2 for any x\in {\mathbb F_{p}^*}. If p=3, then k=0 or k=1. And g_3(0)=0, g_3(1)=k+2, g_3(-1)=2. So in this case, either g_3(1)=g_3(-1)=2 if k=0, or g_3(1)=g_3(0)=0 if k=1, which implies g_3(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_3(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_3(x) is not a PP of {\mathbb F_{p}}.

    If both s_0 and t_0 are odd then g_3(x)=2x^p+4x^{\frac{p-1}{2}}-2(k-2)x^{\frac{p+1}{2}} for any x\in {\mathbb F_{p}^*}. If p=3, then g_3(1)=g_3(-1)=k-2, which implies g_3(x) is not a PP of {\mathbb F_{3}}. If p > 3, then

    (g_3(x))^2\equiv 16x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.

    It follows from Lemma 2.7 that g_3(x) is not a PP of {\mathbb F_{p}} when p > 3. Therefore g_3(x) is not a PP of {\mathbb F_{p}}.

    From them, we derive that g_3(x) is not a PP of {\mathbb F_{p}} when k\ne 2. Note that g_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}. So g_3(x) is not a PP of {\mathbb F_{q}} when k\ne 2. Hence g_3(x) is not a PP of {\mathbb F_{q}}. Thus D_{p^s+p^{t}+1, k}(1, x) is not a PP of {\mathbb F_{q}}.

    By Lemma 2.4, Theorem 4.1 and Theorem 4.3, we have the following general result.

    Theorem 4.4. Let q=p^e with p being an odd prime and e being a positive integer. Let s and t be positive integers with s < t. Then

    \begin{align} D_{p^s+p^t+2, k}(1, x)&=\frac{2-k}{16}(1-4x)^{\frac{p^s+p^t+2}{2}}+\frac{2+k}{16}(1-4x)^{\frac{p^s+p^t}{2}} +\frac{4-k}{16}\Big((1-4x)^{\frac{p^s+1}{2}}+(1-4x)^{\frac{p^t+1}{2}}\Big)\nonumber\\ &\ \ \ +\frac{2-k}{16}\Big((1-4x)^{\frac{p^s-1}{2}}+(1-4x)^{\frac{p^t11}{2}}\Big) +\frac{2-k}{16}(1-4x)+\frac{2-k}{16}. \end{align} (4.2)

    Consequently, D_{p^s+p^{t}+2, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

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

    is a PP of {\mathbb F_{q}}. Furthermore, let \ell\ge 0 be an integer. Then

    \begin{align*} &D_{p^s+p^t+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}(1-4x)^{\frac{p^s+p^t+2i}{2}} +\sum\limits_{j=0}^{\ell}B_{2\ell, 2j}(1-4x)^{j}\\ &\ \ \ +\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+2i-1}\big((1-4x)^{\frac{p^s+2i-1}{2}}+ (1-4x)^{\frac{p^t+2i-1}{2}}\big), \ \ 0\le\ell\le\frac{p-1}{2}, \end{align*}

    and

    \begin{align*} &D_{p^s+p^t+2\ell+1, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell+1, p^s+p^t+2i}(1-4x)^{\frac{p^s+p^t+2i}{2}} +\sum\limits_{j=0}^{\ell}B_{2\ell+1, 2j}(1-4x)^{j}\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell+1, p^s+2i-1} \big((1-4x)^{\frac{p^s+2i-1}{2}}+(1-4x)^{\frac{p^t+2i-1}{2}}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align*}

    where all the coefficients B_{i, j} are given as follows:

    B_{0, p^s+p^t}=\frac{2-k}{4}, B_{0, p^s-1}=\frac{k}{4}, B_{0, 0}=\frac{2-k}{4},
    B_{1, p^s+p^t}=\frac{1}{4}, B_{1, p^s+1}=\frac{2-k}{8}, B_{1, p^s-1}=\frac{1}{8}, B_{1, 0}=\frac{1}{4},

    and

    \begin{align}\label{ckm7} {\left\{ \begin{array}{ll} B_{2m+2, p^s+p^t+2m+2}=\frac{1}{4}B_{2m, p^s+p^t+2m}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+p^t+2i}=B_{2m+1, p^s+2i}-\frac{1}{4}B_{2m, p^s+p^t+2i}+\frac{1}{4}B_{2m, p^s+2i-2}, & {\rm if} \ 1\le i\le m\\ B_{2m+2, p^s+p^t}=B_{2m+1, p^s+p^t}-\frac{1}{4}B_{2m, p^s+p^t}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+2m+1}=B_{2m+1, p^s+2m+1}+\frac{1}{4}B_{2m, p^s+2m-1}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+2i-1}=B_{2m+1, p^s+2i-1}-\frac{1}{4}B_{2m, p^s+2i-1}+\frac{1}{4}B_{2m, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ B_{2m+2, p^s-1}=B_{2m+1, p^s-1}-\frac{1}{4}B_{2m, p^s-1}, & {\rm if} \ m\ge 0\\ B_{2m+2, 0}=B_{2m+1, 0}-\frac{1}{4}B_{2m, 0}, & {\rm if} \ m\ge 0\\ B_{2m+2, 2j}=B_{2m+1, 2j}-\frac{1}{4}B_{2m, 2j}+\frac{1}{4}B_{2m, 2j-2}, & {\rm if} \ 1\le j\le m\\ B_{2m+2, 2m+2}=\frac{1}{4}B_{2m, 2m}, & {\rm if} \ m\ge 0\\ \end{array} \right.} \end{align} (4.3)

    as well as

    \begin{align}\label{ckm8} {\left\{ \begin{array}{ll} B_{2m+1, p^s+p^t+2m}=B_{2m, p^s+p^t+2m}+\frac{1}{4}B_{2m-1, p^s+p^t+2m-2}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+p^t+2i}=B_{2m, p^s+2i}-\frac{1}{4}B_{2m-1, p^s+p^t+2i}+\frac{1}{4}B_{2m-1, p^s+2i-2}, & {\rm if} \ 1\le i\le m-1\\ B_{2m+1, p^s+p^t}=B_{2m, p^s+p^t}-\frac{1}{4}B_{2m-1, p^s+p^t}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+2m+1}=\frac{1}{4}B_{2m-1, p^s+2m-1}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+2i-1}=B_{2m, p^s+2i-1}-\frac{1}{4}B_{2m-1, p^s+2i-1}+\frac{1}{4}B_{2m-1, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ B_{2m+1, p^s-1}=B_{2m, p^s-1}-\frac{1}{4}B_{2m-1, p^s-1}, & {\rm if} \ m\ge 0\\ B_{2m+1, 0}=B_{2m, 0}-\frac{1}{4}B_{2m-1, 0}, & {\rm if} \ m\ge 0\\ B_{2m+1, 2j}=B_{2m, 2j}-\frac{1}{4}B_{2m-1, 2j}+\frac{1}{4}B_{2m-1, 2j-2}, & {\rm if} \ 1\le j\le m-1\\ B_{2m+1, 2m}=B_{2m, 2m}+\frac{1}{4}B_{2m-1, 2m-2}, & {\rm if}\ m\ge 0\\ \end{array} \right.} \end{align} (4.4)

    Proof. The identity immediately follows from Lemma 2.4, Theorem 4.1 and Theorem 4.3. Moreover we readily find that there exists coefficients B_{i, j}\in {\mathbb F_{q}} such that

    \begin{align}\label{ckm9} &D_{p^s+p^t+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{j=0}^{\ell}B_{2\ell, 2j}u^{2j}\nonumber\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell, p^s+2i-1} \big(u^{p^s+2i-1}+u^{p^t+2i-1}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align} (4.5)

    and

    \begin{align}\label{ckm10} &D_{p^s+p^t+2\ell-1, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{j=0}^{\ell}B_{2\ell-1, 2j}u^{2j}\nonumber\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell-1, p^s+2i-1} \big(u^{p^s+2i-1}+u^{p^t+2i-1}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align} (4.6)

    where u^2=1-4x. Now let's determine all the coefficients B_{i, j}. On the one hand, by (4.5) and (4.6), one then has

    \begin{align}\label{ckm11} &D_{p^s+p^t+2\ell, k}(1, x)-xD_{p^s+p^t+2\ell-1, k}(1, x)=D_{p^s+p^t+2\ell, k}(1, x)- \frac{1-u^2}{4}D_{p^s+p^t+2\ell-1, k}(1, x)\nonumber\\ &=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+2i-1}\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)+ \sum\limits_{j=0}^{\ell}B_{2\ell, 2j}u^{2j}\nonumber\\ &\ \ \ -\frac{1}{4}\sum\limits_{i=0}^{\ell-1}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i} -\frac{1}{4}\sum\limits_{j=0}^{\ell}B_{2\ell-1, p^s+2i-1}\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)\nonumber\\ &\ \ \ -\frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, 2j}u^{2j}+ \frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i+2}\nonumber\\ &\ \ \ +\frac{1}{4}\sum\limits_{j=0}^{\ell}B_{2\ell-1, p^s+2i-1}\big(u^{p^s+2i+1}+u^{p^t+2i+1}\big) +\frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, 2j}u^{2j+2}\nonumber\\ &=\big(B_{2\ell, p^s+p^t+2l}+\frac{1}{4}\big)u^{p^s+p^t+2\ell}+\sum\limits_{i=1}^{\ell-1}\big( B_{2\ell, p^s+p^t+2i}-\frac{1}{4}B_{2\ell-1, p^s+p^t+2i}+\frac{1}{4}B_{2\ell-1, p^s +p^t+2i-2}\big)u^{p^s+p^t+2i}\nonumber\\ &\ \ \ +\big(B_{2\ell, p^s+p^t}-\frac{1}{4}B_{2\ell-1, p^s+p^t}\big)u^{p^s+p^t}+\frac{1}{4} B_{2\ell-1, p^s+2\ell-1}\big(u^{p^s+2\ell+1}+u^{p^t+2\ell+1}\big)\nonumber\\ &+\sum\limits_{i=1}^{\ell}\big(B_{2\ell, p^s+2i-1}-\frac{1}{4}B_{2\ell-1, p^s+2i-1}+ \frac{1}{4}B_{2\ell-1, p^s+2i-3}\big)\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)\nonumber\\ &+\big(B_{2\ell, p^s-1}-\frac{1}{4}B_{2\ell-1, p^s-1}\big)\big(u^{p^s-1}+u^{p^t-1}\big) +\big(B_{2\ell, 2\ell}+\frac{1}{4}B_{2\ell-1, 2\ell-2}\big)u^{2\ell}\nonumber\\ &+\sum\limits_{j=1}^{\ell-1}\big(B_{2\ell2j}-\frac{1}{4}B_{2\ell-1, 2j}+ \frac{1}{4}B_{2\ell-1, 2j-2}\big)u^{2j}+B_{2\ell, 0}-\frac{1}{4}B_{2\ell-1, 0}. \end{align} (4.7)

    On the other hand, Lemma 2.4 tells us that

    D_{p^s+p^t+2\ell+1, k}(1, x)=D_{p^s+p^t+2\ell, k}(1, x)-xD_{p^s+p^t+2\ell-1, k}(1, x).

    So by comparing the coefficient of the term u^i in the right hand side of (4.6) and (4.7), one can get the desired results as (4.4). Following the similar way, one also obtain the recursions of B_{i, j} as (4.3). So the proof Theorem 4.4 is complete.


    5. Reversed Dickson polynomials D_{p^{e_1}+p^{e_2}+\cdots+p^{e_s}+\ell, k}(1, x)

    Let s\ge 1 be an integer. Let e_1, e_2, \cdots, e_s, \ell be integers with 0\le e_1 < e_2 < \cdots < e_s and 0\le \ell < p. In this section, we present an explicit formula for D_{n, k}(1, x) presented by elementary symmetric polynomials in terms of the power of (1-4x) when n=p^{e_1}+p^{e_2}+\cdots+p^{e_s}+\ell. Then we characterize D_{n, k}(1, x) to be a PP of {\mathbb F}_q in this case.

    Let \sigma_i(x_1, x_2, \cdots, x_s) be the elementary polynomials in s variables x_1, x_2, \cdots, x_s which are defined by

    \begin{align*} &\sigma_0(x_1, x_2, \cdots, x_s)=1, \\ &\sigma_1(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j\le n}x_j, \\ &\sigma_2(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j<k\le n}x_jx_k, \\ &\sigma_3(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j<k<\ell \le n}x_jx_kx_{\ell}, \\ \end{align*}

    and so forth, ending with

    \sigma_s(x_1, x_2, \cdots, x_s)=x_1x_2\cdots x_s.

    Now we give the first result of this section.

    Theorem 5.1. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a positive integer. Let e_1, \cdots, e_s be nonnegative integers with e_1 < \cdots < e_s. Then

    \begin{align*} D_{p^{e_1}+...+p^{e_s}, k}(1, x)&=\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}

    Consequently, D_{p^{e_1}+...+p^{e_s}, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

    (2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}}{2}}, \cdots, x^{\frac{p^{e_s}}{2}}\big) +k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}-1/i}{2}}, \cdots, x^{\frac{p^{e_s}-1/i}{2}}\big)

    is a PP of {\mathbb F_{q}}.

    Proof. We divide the proof into the following two cases.

    CASE 1. x\ne \frac{1}{4}. For this case, putting x=y(1-y) in the second identity of Lemma 2.3 gives us that

    \begin{align*} &D_{p^{e_1}+...+p^{e_s}, k}(1, x)=D_{p^{e_1}+...+p^{e_s}, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^{e_1}+...+p^{e_s}}-\big(1+(k-2)y\big)(1-y)^{p^{e_1}+...+p^{e_s}}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^{e_1}+\cdots+p^{e_s}} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^{e_1}+\cdots+p^{e_s}}}{u}\\ =&\frac{k+(2-k)u}{2^{p^{e_1}+\cdots+p^{e_s}+1}u}\prod\limits_{i=1}^{s}(u^{p^{e_i}}+1) -\frac{k+(k-2)u}{2^{p^{e_1}+\cdots+p^{e_s}+1}u}\prod\limits_{i=1}^{s}(1-u^{p^{e_i}})\\ =&\frac{1}{2^{s+1}u}\Big((k+(2-k)u)\sum\limits_{0\le i\le s}\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}}) -(k+(k-2)u)\sum\limits_{0\le i\le s}(-1)^i\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}})\Big)\\ =&\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big)+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(u^{p^{e_1}-1/i}, \cdots, u^{p^{e_s}-1/i}\big)\Big), \end{align*}

    where u=2y-1 and u^2=1-4x. Then we have that

    \begin{align*} D_{p^{e_1}+...+p^{e_s}, k}(1, x)&=\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}

    as desired.

    CASE 2. x=\frac{1}{4}. On the one hand, by the first identity of Lemma 2.3, one has

    D_{p^{e_1}+...+p^{e_s}, k}\big(1, \frac{1}{4}\big) =\frac{k(p^{e_1}+\cdots+p^{e_s})-k+2}{2^{p^{e_1}+\cdots+p^{e_s}}}=\frac{2-k}{2^s}.

    On the other hand,

    \begin{align*} &\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^ {\frac{p^{e_s}-1/i}{2}}\big)\Big)\Big |_{x=1/4}=\frac{2-k}{2^s}. \end{align*}

    Thus the required result follows. So Theorem 5.1 is proved.

    heorem 5.2. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a positive integer. Let e_1, \cdots, e_s be nonnegative integers with e_1 < \cdots < e_s. Then

    \begin{align*} D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)&=\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}

    Consequently, D_{p^{e_1}+...+p^{e_s}+1, k}(1, x) is a PP of {\mathbb F_{q}} if and only if the polynomial

    2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}}{2}}, \cdots, x^{\frac{p^{e_s}}{2}}\big) +\big((2-k)x+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}-1/i}{2}}, \cdots, x^{\frac{p^{e_s}-1/i}{2}}\big)

    is a PP of {\mathbb F_{q}}.

    Proof. We consider the following two cases.

    CASE 1. x\ne \frac{1}{4}. For this case, putting x=y(1-y) in the second identity of Lemma 2.3 gives us that

    \begin{align*} &D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)=D_{p^{e_1}+...+p^{e_s}+1, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^{e_1}+...+p^{e_s}+1}-\big(1+(k-2)y\big)(1-y)^{p^{e_1}+...+p^{e_s}+1}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^{e_1}+\cdots+p^{e_s}+1} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^{e_1}+\cdots+p^{e_s}+1}}{u}\\ =&\frac{k+(2-k)u}{2^{p^{e_1}+\cdots+p^{e_s}+2}u}\prod\limits_{i=1}^{s}(u^{p^{e_i}}+1)(u+1) -\frac{k+(k-2)u}{2^{p^{e_1}+\cdots+p^{e_s}+2}u}\prod\limits_{i=1}^{s}(1-u^{p^{e_i}})(1-u)\\ =&\frac{1}{2^{s+2}u}\Big((k+2+(2-k)u^2)\sum\limits_{0\le i\le s}\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}}) -(2k+(4-2k)u^2)\sum\limits_{0\le i\le s}(-1)^i\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}})\Big)\\ =&\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big)+\big((2-k)u^2+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(u^{p^{e_1}-1/i}, \cdots, u^{p^{e_s}-1/i}\big)\Big), \end{align*}

    where u=2y-1 and u^2=1-4x. Then we have

    \begin{align*} D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)&=\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}

    as desired.

    CASE 2. x=\frac{1}{4}. On the one hand, by the first identity of Lemma 2.3, one has

    D_{p^{e_1}+...+p^{e_s}+1, k}\big(1, \frac{1}{4}\big) =\frac{k(p^{e_1}+\cdots+p^{e_s}+1)-k+2}{2^{p^{e_1}+\cdots+p^{e_s}+1}}=\frac{2}{2^{s+1}}.

    On the other hand,

    \begin{align*} &\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big)\Big|_{x=1/4} =\frac{2}{2^{s+1}}. \end{align*}

    Thus the required result follows. So Theorem 5.3 is proved.

    Then Theorems 5.1-5.2 together with Lemma 2.4 show that the general result is true.

    Theorem 5.3. Let q=p^e with p being an odd prime and e being a positive integer. Let s be a positive integer. Let e_1, \cdots, e_s be nonnegative integers with e_1 < \cdots < e_s. Then for any \ell \ge 0 each of the identities is true.

    \begin{align*} D_{p^{e_1}+...+p^{e_s}+2\ell, k}(1, x)=\frac{1}{2^{s+2\ell}}\Big(\sum\limits_{j=0}^{\ell} C_{2\ell, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm even}} \sigma_i\big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big) +\sum\limits_{j=0}^{\ell}Q_{2\ell, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm odd}} \sigma_i \big(u^{p^{e_1}-\frac{1}{i}}, \cdots, u^{p^{e_s}-\frac{1}{i}}\big)\Big), \end{align*}
    \begin{align*} D_{p^{e_1}+...+p^{e_s}+2\ell+1, k}(1, x)&=\frac{1}{2^{s+2\ell+1}}\Big(\sum\limits_{j=0}^{\ell} C_{2\ell+1, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm even}} \sigma_i\big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big) +\sum\limits_{j=0}^{\ell+1} Q_{2\ell+1, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm odd}} \sigma_i \big(u^{p^{e_1}-\frac{1}{i}}, \cdots, u^{p^{e_s}-\frac{1}{i}}\big)\Big), \end{align*}

    where u^2=1-4x, and the coefficients C_{a, 2b} and Q_{a, 2b} can be determined as follows:

    C_{0, 0}=2-k, Q_{0, 0}=k, C_{1, 0}=k, Q_{1, 2}=2-k,
    \begin{align}\label{ch4} {\left\{ \begin{array}{ll} C_{2m+2, 0}=2C_{2m+1, 0}-C_{2m, 0}, & {\rm if} \ m\ge 0\\ C_{2m+2, 2j}=2C_{2m+1, 2j}+C_{2m, 2j-2}-C_{2m, 2j}, & {\rm if} \ 1\le j\le m\\ C_{2m+2, 2m+2}=C_{2m, 2m}, & {\rm if} \ m\ge 0\\ Q_{2m+2, 0}=2Q_{2m+1, 0}-Q_{2m, 0}, & {\rm if} \ m\ge 0\\ Q_{2m+2, 2j}=2Q_{2m+1, 2j}+Q_{2m, 2j-2}-Q_{2m, 2j}, & {\rm if} \ 1\le j\le m\\ Q_{2m+2, 2m+2}=2Q_{2m+1, 2m+2}+Q_{2m, 2m}, & {\rm if} \ m\ge 0 \end{array} \right.} \end{align} (5.1)

    as well as

    \begin{align}\label{ch5} {\left\{ \begin{array}{ll} C_{2m+1, 0}=2C_{2m, 0}-C_{2m-1, 0}, & {\rm if} \ m\ge 1\\ C_{2m+1, 2j}=2C_{2m, 2j}+C_{2m-1, 2j-2}-C_{2m-1, 2j}, & {\rm if} \ 1\le j\le m-1\\ C_{2m+1, 2m}=2C_{2m, 2m}+C_{2m-1, 2m-2}, & {\rm if} \ m\ge 1\\ Q_{2m+1, 0}=2Q_{2m, 2j}-Q_{2m-1, 0}, & {\rm if} \ m\ge 0\\ Q_{2m+1, 2j}=2Q_{2m, 2j}+Q_{2m-1, 2j-2}-Q_{2m-1, 2j}, & {\rm if} \ 1\le j\le m\\ Q_{2m+1, 2m+2}=2Q_{2m-1, 2m}, & {\rm if} \ m\ge 1 \end{array} \right.} \end{align} (5.2)

    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.


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  • This article has been cited by:

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