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
[1] | Kaimin Cheng . Permutational behavior of reversed Dickson polynomials over finite fields. AIMS Mathematics, 2017, 2(2): 244-259. doi: 10.3934/Math.2017.2.244 |
[2] | Xiaoer Qin, Li Yan . Some specific classes of permutation polynomials over Fq3. AIMS Mathematics, 2022, 7(10): 17815-17828. doi: 10.3934/math.2022981 |
[3] | Qian Liu, Jianrui Xie, Ximeng Liu, Jian Zou . Further results on permutation polynomials and complete permutation polynomials over finite fields. AIMS Mathematics, 2021, 6(12): 13503-13514. doi: 10.3934/math.2021783 |
[4] | Varsha Jarali, Prasanna Poojary, G. R. Vadiraja Bhatta . A recent survey of permutation trinomials over finite fields. AIMS Mathematics, 2023, 8(12): 29182-29220. doi: 10.3934/math.20231495 |
[5] | Turki Alsuraiheed, Elif Segah Oztas, Shakir Ali, Merve Bulut Yilgor . Reversible codes and applications to DNA codes over F42t[u]/(u2−1). AIMS Mathematics, 2023, 8(11): 27762-27774. doi: 10.3934/math.20231421 |
[6] | Sami Alabiad, Yousef Alkhamees . On classification of finite commutative chain rings. AIMS Mathematics, 2022, 7(2): 1742-1757. doi: 10.3934/math.2022100 |
[7] | Xiaofan Xu, Yongchao Xu, Shaofang Hong . Some results on ordinary words of standard Reed-Solomon codes. AIMS Mathematics, 2019, 4(5): 1336-1347. doi: 10.3934/math.2019.5.1336 |
[8] | Weihua Li, Chengcheng Fang, Wei Cao . On the number of irreducible polynomials of special kinds in finite fields. AIMS Mathematics, 2020, 5(4): 2877-2887. doi: 10.3934/math.2020185 |
[9] | Ugur Duran, Can Kızılateş, William Ramírez, Clemente Cesarano . A new class of degenerate unified Bernoulli-Euler Hermite polynomials of Apostol type. AIMS Mathematics, 2025, 10(7): 16117-16138. doi: 10.3934/math.2025722 |
[10] | Jiankang Wang, Zhefeng Xu, Minmin Jia . Distribution of values of Hardy sums over Chebyshev polynomials. AIMS Mathematics, 2024, 9(2): 3788-3797. doi: 10.3934/math.2024186 |
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=0nn−i(n−ii)(−x)ian−2i |
if n≥1 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 n≥1 by
Dn,k(a,x):=[n2]∑i=0n−kin−i(n−ii)(−x)ian−2i | (1.1) |
and D0,k(a,x)=2−k. 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+ℓ,k(1,x), where s,e1,...,es are positive integers, ℓ is an integer with 0≤ℓ<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+ℓ,k(1,x) as the linear combination of the elementary symmetric polynomials with the power of 1−4x being the variables. From this, we present a necessary and sufficient condition for Dpe1+...+pes+ℓ,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 0≤k<p.
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,d∈F∗q and b∈Fq.
Lemma 2.2. Let s≥0 be an integer and a,b be in Fq∗. Then the binomial axps−12+bxps+12 is a PP of Fq if and only if s=0.
Proof. First we assume that the binomial axps−12+bxps+12 is a PP of Fq. If s>0, then the equation axps−12+bxps+12=xps−12(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 axps−12+bxps+12 is a PP of Fq. Therefore Lemma 2.2 is proved.
Lemma 2.3. [1] For any integer n≥0, we have
Dn,k(1,14)=kn−k+22n |
and
Dn,k(1,x)=(k−1−(k−2)y)yn−(1+(k−2)y)(1−y)n2y−1 |
if x=y(1−y)≠14.
Lemma 2.4. [1] Let n≥2 be an integer. Then the recursion
Dn,k(1,x)=Dn−1,k(1,x)−xDn−2,k(1,x) |
holds for any x∈Fq.
Lemma 2.5. [1] Let p=char(Fq)≥3 and s be a positive integer. Then
2Dps,k(1,x)+k−2=k(1−4x)ps−12. |
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,pe−1)={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<q−1 and t≢0(modp), the reduction of f(x)t(modxq−x) has degree less than q−1.
Lemma 2.8. Let p be a prime with p>3 and a be a nonzero element in Fp. Then the binomial xps−12+ax is a PP of Fpe if and only if s=0.
Proof. Let p>3,a∈F∗p. Clearly, if s=0, then w(x):=xps−12+ax=1+ax is a PP of Fpe. In what follows, we show that w(x)=xps−12+ax is not a PP of Fpe when s>0. Let s>0 and s≡s0(mod2e) with 0≤s0≤2e−1. Then
w(x)≡xps0−12+ax(modxpe−x) |
for any x∈F∗pe since ps−12≡ps0−12(modpe−1), i.e.,
w(x)=xps0−12+ax | (2.1) |
for any x∈F∗pe. 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 x∈F∗pe. So f2(1a)=0. It then follows from f2(0)=0 that w(x)=xps−12+ax is not a PP of Fpe.
CASE 2. s>0 and s0 is a positive even number. Then xps0−12=1 for each x∈F∗p. By (2.1) one get w(x)=1+ax for any x∈F∗p. Therefore w(x)=0 has one nonzero root −1a∈F∗p. Hence w(x)=xps−12+ax does not permute Fp since f2(0)=0. Note that f2(Fp)⊆Fp. So one has that w(x)=xps−12+ax does not permute Fpe.
CASE 3. s>0 and s0 is an odd number. Then xps0−12=xp−12 for each x∈F∗p. It follows from (2.1) that
w(x)=xp−12+ax |
for any x∈F∗p. Then we have
(w(x))2=xp−1+the terms of x with the degree less than p−1(modxp−x). |
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)=xps−12+ax is not a PP of Fpe when s>0. This finishes the proof of Lemma 2.8.
In this section, we present an explicit formula for Dn,k(1,x) when n=ps+ℓ with s≥0 and 0≤ℓ<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)=2−k4(1−4x)ps+12+k4(1−4x)ps−12+12. | (3.1) |
Furthermore, we have
Dps+2ℓ,k(1,x)=ℓ∑i=0A2ℓ,ps+2i−1(1−4x)ps+2i−12+ℓ∑j=0A2ℓ,2j(1−4x)j,ℓ≥0 |
and
Dps+2ℓ+1,k(1,x)=ℓ+1∑i=0A2ℓ+1,ps+2i−1(1−4x)ps+2i−12+ℓ∑j=0A2ℓ+1,2j(1−4x)j,ℓ≥0, |
where all the coefficients Ai,j are given as follows:
A0,ps−1=k2,A0,0=2−k2,A1,ps+1=2−k4,A1,ps−1=k4,A1,0=12, |
and
{A2m+2,ps+2m+1=A2m+1,ps+2m+1+14A2m,ps+2m−1,if m≥0A2m+2,ps+2i−1=A2m+1,ps+2i−1−14A2m,ps+2i−1+14A2m,ps+2i−3,if 1≤i≤mA2m+2,ps−1=A2m+1,ps−1−14A2m,ps−1,if m≥0A2m+2,0=A2m+1,0−14A2m,0,if m≥0A2m+2,2j=A2m+1,2j−14A2m,2j+14A2m,2j−2,if 1≤j≤mA2m+2,2m+2=14A2m,2m,if m≥0 | (3.2) |
as well as
{A2m+1,ps+2m+1=14A2m−1,ps+2m−1,if m≥0A2m+1,ps+2i−1=A2m,ps+2i−1−14A2m−1,ps+2i−1+14A2m−1,ps+2i−3,if 1≤i≤mA2m+1,ps−1=A2m,ps−1−14A2m−1,ps−1,if m≥0A2m+1,0=A2m,0−14A2m−1,0,if m≥0A2m+1,2j=A2m,2j−14A2m−1,2j+14A2m−1,2j−2,if 1≤j≤m−1A2m+1,2m=A2m,2m+14A2m−1,2m−2,if m≥0. | (3.3) |
Proof. First of all, we show (3.1) is true. We consider the following two cases.
CASE 1. x≠14. For this case, putting x=y(1−y) in second identity of Lemma 2.3 gives us that
Dps+1,k(1,x)=Dps+1,k(1,y(1−y))=k+(2−k)u2(u+12)ps+1−k+(k−2)u2(1−u2)ps+1u=2−k8((u+1)ps(u+1)+(1−u)ps(1−u))+k8u((u+1)ps(u+1)−(1−u)ps(1−u))=2−k4(ups+1+1)+k4(ups−1+1)=2−k4((u2)ps+12)+k4((u2)ps−12)+12, |
where u=2y−1 and u2=1−4x. So (3.1) follows if x≠14.
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=2−k4(1−4×14)ps+12+k4(1−4×14)ps−12+12 |
as required.
Thus (3.1) is true for any x∈Fq.
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,j∈Fq such that
Dps+2ℓ,k(1,x)=ℓ∑i=0A2ℓ,ps+2i−1(1−4x)ps+2i−12+ℓ∑j=0A2ℓ,2j(1−4x)j | (3.4) |
with 0≤ℓ≤p−12 and
Dps+2ℓ+1,k(1,x)=ℓ+1∑i=0A2ℓ+1,ps+2i−1(1−4x)ps+2i−12+ℓ∑j=0A2ℓ+1,2j(1−4x)j | (3.5) |
with 0≤ℓ<p−12. Therefore we now only need to determine all the coefficients Ai,j. Let u2=1−4x. On the one hand, by (3.4) and (3.5), one then has
Dps+2ℓ,k(1,x)−xDps+2ℓ−1,k(1,x)=Dps+2ℓ,k(1,x)−1−u24Dps+2ℓ−1,k(1,x)=ℓ+1∑i=0A2ℓ,ps+2i−1ups+2i−1+ℓ∑j=0A2ℓ,2ju2j−14ℓ∑i=0A2ℓ−1,ps+2i−1ups+2i−1 −14ℓ−1∑j=0A2ℓ−1,2ju2j+14ℓ∑i=0A2ℓ−1,ps+2i−1ups+2i+1+14ℓ−1∑j=0A2ℓ−1,2ju2j+2=14A2ℓ−1,ps+2ℓ−1ups+2ℓ+1+ℓ∑i=1(A2ℓ,ps+2i−1−14A2ℓ−1,ps+2i−1+14A2ℓ−1,ps+2i−3)ups+2i−1 +(A2ℓ,ps−1−14A2ℓ−1,ps−1)ups−1+(A2ℓ,2ℓ+14A2ℓ−1,2ℓ−2)u2ℓ +ℓ−1∑j=1(A2ℓ,2j−14A2ℓ−1,2j+14A2ℓ−1,2j−2)u2j+A2ℓ,0−14A2ℓ−1,0. | (3.6) |
On the other hand, Lemma 2.4 tells us that
Dps+2ℓ+1,k(1,x)=Dps+2ℓ,k(1,x)−xDps+2ℓ−1,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 p≡1(mod4) and v2(s)≥v2(e), or p≡3(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.
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.
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) |
Cheng was supported partially by the General Project of Department of Education of Sichuan Province 15ZB0434. [2000]Primary 11T06, 11T55, 11C08.
The author declares no conflicts of interest in this paper.
[1] | K. Cheng, Permutational Behavior of Reversed Dickson Polynomials over Finite Fields, AIMS Math., 2 (2017), 244-259. |
[2] | R. Coulter, Explicit evaluation of some Weil sums, Acta Arith., 83 (1998), 241-251. |
[3] | 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. |
[4] | 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. |
[5] | R. Lidl and H. Niederreiter, Finite Fields, second ed., Encyclopedia of Mathematics and its Applications, Cambridge University Press, Cambridge, 20,1997. |
[6] | X. Qin and S. Hong, Constructing permutation polynomials over finite fields, Bull. Aust. Math. Soc., 89 (2014), 420-430. |
[7] | X. Qin, G. Qian and S. Hong, New results on permutation polynomials over finite fields, Int. J. Number Theory, 11 (2015), 437-449. |
[8] | Q. Wang and J. Yucas, Dickson polynomials over finite fields, Finite Fields Appl., 18 (2012), 814-831. |
1. | Weihua Li, Chengcheng Fang, Wei Cao, On the number of irreducible polynomials of special kinds in finite fields, 2020, 5, 2473-6988, 2877, 10.3934/math.2020185 |