Characterization and Enumeration of Good Punctured Polynomials

Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2016, Article ID 6093219, 7 pages http://dx.doi.org/10.1155/2016/6093219

Research Article Characterization and Enumeration of Good Punctured Polynomials over Finite Fields Somphong Jitman,1 Aunyarut Bunyawat,2 Supanut Meesawat,2 Arithat Thanakulitthirat,2 and Napat Thumwanit2 1

Department of Mathematics, Faculty of Science, Silpakorn University, Nakhon Pathom 73000, Thailand Department of Mathematics, Mahidol Wittayanusorn School, Nakhon Pathom 73170, Thailand

2

Correspondence should be addressed to Somphong Jitman; [email protected] Received 25 November 2015; Revised 3 March 2016; Accepted 6 March 2016 Academic Editor: Shyam L. Kalla Copyright © 2016 Somphong Jitman et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A family of good punctured polynomials is introduced. The complete characterization and enumeration of such polynomials are given over the binary field F2 . Over a nonbinary finite field F𝑞 , the set of good punctured polynomials of degree less than or equal to 2 are completely determined. For 𝑛 ≥ 3, constructive lower bounds of the number of good punctured polynomials of degree 𝑛 over F𝑞 are given.

1. Introduction From the fundamental theorem of algebra, every polynomial over the rational numbers Q (or over the real numbers R) has a root in C. However, it is not guaranteed that a polynomial has a root in Q or in R. Therefore, for a given polynomial over Q (resp., R), it is of natural interest to determine whether it has a root in Q (resp., R). In general, determining whether a given polynomial has a root in a nonalgebraically closed field is an interesting problem and has been extensively studied (see, e.g., [1–4]). In this paper, we introduce punctured forms of a polynomial 𝑓(𝑥) over a field F (see the definition below) and focus on determining whether the punctured parts of 𝑓(𝑥) have a root in F. Due to the rich algebraic structures and various applications of polynomials over finite fields (see [5–9] and references therein), their properties such as factorization, root finding, and irreducibility have extensively been studied (see [10–13]). In this paper, we mainly focus on punctured polynomials over a finite field which is not an algebraically closed field. The readers may refer to [5] for more details on finite fields and polynomials over finite fields.

Let 𝑞 be a prime power and let F𝑞 denote the finite field of 𝑞 elements. Denote by F𝑞∗ the set of nonzero elements in F𝑞 . Let F𝑞 [𝑥] 𝑚

= {∑𝑎𝑖 𝑥𝑖 | 𝑎𝑖 ∈ F𝑞 ∀𝑖 = 0, 1, 2, . . . , 𝑚, 𝑚 ∈ N ∪ {0}}

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𝑖=0

be the set of all polynomials with indeterminate 𝑥 over F𝑞 . Let F𝑞 [𝑥]𝑛 = {𝑓 (𝑥) ∈ F𝑞 [𝑥] | deg (𝑓 (𝑥)) ≤ 𝑛} , ̂F 𝑞 [𝑥]𝑛 = {𝑓 (𝑥) ∈ F𝑞 [𝑥] | deg (𝑓 (𝑥)) = 𝑛}

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be the set of all polynomials of degree less than or equal to 𝑛 over F𝑞 and the set of all polynomials of degree 𝑛 over F𝑞 , respectively.

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International Journal of Mathematics and Mathematical Sciences

𝑖 Given a polynomial 𝑓(𝑥) = ∑𝑚 𝑖=0 𝑎𝑖 𝑥 ∈ F𝑞 [𝑥] of degree 𝑚, for each 𝑗 ∈ {0, 1, 2, . . . , 𝑚}, the 𝑗th punctured polynomial of 𝑓(𝑥) is defined to be 𝑗−1

𝑚

𝑖=0

𝑖=𝑗+1

𝑓(𝑗) (𝑥) = ∑𝑎𝑖 𝑥𝑖 + ∑ 𝑎𝑖 𝑥𝑖−1 .

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For convenience, by abuse of notation, the degree of zero polynomial is defined to be 0. Hence, we can write 𝑓(0) (𝑥) = 0 for all constant polynomials 𝑓(𝑥) = 𝑎 ∈ F𝑞 . A polynomial 𝑓(𝑥) ∈ F𝑞 [𝑥] of degree 𝑚 is said to be good punctured if 𝑓(𝑗) (𝑥) has a root in F𝑞 for all 𝑗 ∈ {0, 1, 2, . . . , 𝑚}. Otherwise, 𝑓(𝑥) is said to be bad punctured. The constant polynomials 𝑓(𝑥) = 𝑎 ∈ F𝑞 are always good punctured and referred to as trivial good punctured polynomials. A good punctured polynomial is called nontrivial if it is not a trivial good punctured polynomial. Example 1. Let 𝑓(𝑥) = 𝑥2 + 2𝑥 + 2 be a polynomial in F3 [𝑥]. Then 𝑓(2) (𝑥) = 2𝑥 + 2, 𝑓(1) (𝑥) = 𝑥 + 2, and 𝑓(0) (𝑥) = 𝑥 + 2. It is not difficult to see that 𝑓(𝑥) is good punctured. Given a positive integer 𝑛 and a prime power 𝑞, let 𝑇(𝑞,𝑛) ̂ (𝑞,𝑛) denote the set of good punctured polynomials of and 𝑇 degree less than or equal to 𝑛 over F𝑞 and the set of all good punctured polynomials of degree 𝑛 over F𝑞 , respectively. Precisely, 𝑇(𝑞,𝑛) = {𝑓 (𝑥) ∈ F𝑞 [𝑥]𝑛 | 𝑓(𝑗) (𝑥) has a root in F𝑞 ∀0 ≤ 𝑗 ≤ deg (𝑓 (𝑥))} , (4)

̂ (𝑞,𝑛) = {𝑓 (𝑥) 𝑇 ∈ ̂F 𝑞 [𝑥]𝑛 | 𝑓(𝑗) (𝑥) has a root in F𝑞 ∀0 ≤ 𝑗 ≤ 𝑛} .

̂ (𝑞,0) = F𝑞 By convention, since deg(0) = 0, we have 𝑇(𝑞,0) = 𝑇 ̂ (𝑞,0) | = 𝑞. and |𝑇(𝑞,0) | = |𝑇 ̂ (𝑞,𝑛) , we have Remark 2. From the definitions of 𝑇(𝑞,𝑛) and 𝑇 the following facts: ̂ (𝑞,𝑛) is a subset of 𝑇(𝑞,𝑛) . (1) 𝑇 ̂ (𝑞,𝑛) is a disjoint union for all 𝑛 ≥ 2. (2) 𝑇(𝑞,𝑛) = 𝑇(𝑞,𝑛−1) ∪ 𝑇 ̂ (𝑞,𝑛) | for all 𝑛 ≥ 2. (3) |𝑇(𝑞,𝑛) | = |𝑇(𝑞,𝑛−1) | + |𝑇 Example 3. Over the finite fields F2 , we have 2

2

̂ (2,2) = {𝑥2 , 𝑥2 + 𝑥, 𝑥2 + 𝑥 + 1} . 𝑇

2. Good Punctured Polynomials over the Binary Field In this section, we focus on good punctured polynomials over the finite field F2 . The characterization and enumeration of such polynomials are completely determined. First, we determine the set 𝑇(2,𝑛) of good punctured polynomials of degree less than or equal to 𝑛 over the binary field F2 . It is not difficult to see that 𝑇(2,1) = F2 . For 𝑛 ≥ 2, the set 𝑇(2,𝑛) is given as follows. Theorem 4. Let 𝑛 ≥ 2 be a positive integer. Then 𝑇(2,𝑛) = 𝐴 ∪ 𝐵 ∪ 𝐶,

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𝑖 where 𝐴 = {∑2𝑘 𝑖=0 𝑥 | 0 ≤ 𝑘 ≤ 𝑛/2}, 𝐵 = {𝑥𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−1 𝑎𝑛𝑑 𝑓(1) = 0}, and 𝐶 = {𝑥2 𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−2 }.

Proof. First, we prove that 𝐴 ∪ 𝐵 ∪ 𝐶 ⊆ 𝑇(2,𝑛) . Let 𝑔(𝑥) ∈ 𝐴 ∪ 𝐵 ∪ 𝐶. We distinguish the proof into three cases. 𝑖 Case 1 (𝑔(𝑥) ∈ 𝐴). Then 𝑔(𝑥) = ∑2𝑘 𝑖=0 𝑥 for some 0 ≤ 𝑘 ≤ 𝑛/2. 𝑖 It follows that 1 is a root of 𝑔(𝑗) (𝑥) = ∑2𝑘−1 𝑖=0 𝑥 for all 0 ≤ 𝑗 ≤ 2𝑘. Hence, 𝑔(𝑥) ∈ 𝑇(2,𝑛) .

Case 2 (𝑔(𝑥) ∈ 𝐵). Then 𝑔(𝑥) = 𝑥𝑓(𝑥) for some 𝑓(𝑥) ∈ F2 [𝑥]𝑛−1 . We have 𝑔(0) (1) = 𝑓(1) = 0 and 𝑔(𝑖) (0) = 𝑔(0) = 0 for all 0 < 𝑖 ≤ deg(𝑔(𝑥)). Hence, 𝑔(𝑖) (𝑥) has a root in F2 for all 0 ≤ 𝑖 ≤ deg(𝑔(𝑥)). Therefore, 𝑔(𝑥) ∈ 𝑇(2,𝑛) . Case 3 (𝑔(𝑥) ∈ 𝐶). Then 𝑔(𝑥) = 𝑥2 ℎ(𝑥) for some ℎ(𝑥) ∈ F2 [𝑥]𝑛−2 . It follows that 0 is a root of 𝑔(0) (𝑥) = 𝑔(1) (𝑥) = 𝑥ℎ(𝑥) and 𝑔(𝑖) (𝑥) = 𝑥2 ℎ(𝑖−2) (𝑥) for all 2 ≤ 𝑖 ≤ deg(𝑔(𝑥)). Therefore, 𝑓(𝑖) (𝑥) has a root in F2 for all 0 ≤ 𝑖 ≤ deg(𝑔(𝑥)). As desired, 𝑔(𝑥) ∈ 𝑇(2,𝑛) . On the other hand, let 𝑔(𝑥) ∈ 𝑇(2,𝑛) . Write 𝑔(𝑥) = deg(𝑔(𝑥)) 𝑔𝑖 𝑥𝑖 and consider the following two cases. ∑𝑖=0

2

𝑇(2,2) = {0, 1, 𝑥 , 𝑥 + 𝑥, 𝑥 + 𝑥 + 1} ,

good punctured polynomials over the binary field F2 are given in Section 2. In Section 3, good punctured polynomials of degree 𝑛 over F𝑞 , where 𝑞 > 2, are studied. The good punctured polynomials of degree less than or equal to 2 over fields F𝑞 are completely determined. Lower bounds of the size of the set of good punctured polynomials of degree greater than 2 are provided as well. Conclusion and some discussions about future researches on punctured polynomials are provided in Section 4.

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̂ (2,2) | = 3, respectively. Hence, |𝑇(2,2) | = 5 and |𝑇 In this paper, we focus on the characterization and enumeration of the good punctured polynomials of degree 𝑛 over F𝑞 . The complete characterization and enumeration of

Case 1 (𝑔0 = 0) Case 1.1 (𝑔1 = 0). Then 𝑔(𝑥) ∈ 𝐶. Case 1.2 (𝑔1 = 1). Then deg(𝑔(𝑥)) ≥ 1. Since 𝑔(0) (𝑥) = deg(𝑔(𝑥))−1 ∑𝑖=0 𝑔𝑖+1 𝑥𝑖 , we have 𝑔(0) (0) = 𝑔1 = 1. It follows that deg(𝑔(𝑥))−1 0 = 𝑔(0) (1) = ∑𝑖=0 𝑔𝑖+1 = 𝑔(1). Hence, 𝑔(𝑥) ∈ 𝐵.

International Journal of Mathematics and Mathematical Sciences Case 2 (𝑔0 = 1). Since 𝑔(0) (0) = 𝑔0 = 1, we have 0 = 𝑔(0) (1) = deg(𝑔(𝑥)) ∑𝑖=1 𝑔𝑖 . Suppose that there exists 1 ≤ 𝑗 < deg(𝑔(𝑥)) such that 𝑔𝑗 = 0. Since 𝑔(𝑗) (0) = 𝑔0 = 1 and 𝑔(deg(𝑔(𝑥))) (0) = 𝑔0 = 1, we have 𝑗−1

deg(𝑔(𝑥))

0 = 𝑔(𝑗) (1) = ∑𝑔𝑖 +

𝑖=𝑗+1

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deg(𝑔(𝑥))−1

0 = 𝑔(deg(𝑔(𝑥))) (1) =

∑ 𝑖=0

𝑔𝑖 .

It follows that 0 = 𝑔𝑗 = 𝑔deg(𝑔(𝑥)) = 1, a contradiction. Hence, 𝑔𝑖 = 1 for all 0 ≤ 𝑖 ≤ deg(𝑔(𝑥)). Since 0 = 𝑔(0) (1) = deg(𝑔(𝑥)) ∑𝑖=1 𝑔𝑖 , the degree of 𝑔(𝑥) must be even. We conclude that 𝑔(𝑥) ∈ 𝐴. From the two cases, we have 𝑔(𝑥) ∈ 𝐴 ∪ 𝐵 ∪ 𝐶, and, hence, 𝑇(2,𝑛) ⊆ 𝐴 ∪ 𝐵 ∪ 𝐶. Therefore, 𝑇(2,𝑛) = 𝐴 ∪ 𝐵 ∪ 𝐶 as desired. Corollary 5. If 𝑛 is a positive integer, then {2 󵄨 { 󵄨󵄨 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = { {3 ⋅ 2𝑛−2 + ⌊ 𝑛 ⌋ + 1 2 {

𝑖𝑓 𝑛 = 1, 𝑖𝑓 𝑛 ≥ 2.

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𝑖 where 𝐴 = {∑2𝑘 𝑖=0 𝑥 | 0 ≤ 𝑘 ≤ 𝑛/2}, 𝐵 = {𝑥𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−1 and 𝑓(1) = 0}, and 𝐶 = {𝑥2 𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−2 }. Since 𝐴 and 𝐵 ∪ 𝐶 are disjoint, by the inclusion-exclusion principle, we have 󵄨 󵄨󵄨 (10) 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = |𝐴 ∪ 𝐵 ∪ 𝐶| = |𝐴| + |𝐵| + |𝐶| − |𝐵 ∩ 𝐶| .

Clearly, |𝐴| = ⌊𝑛/2⌋ + 1 and |𝐶| = 2𝑛−1 . Observe that 𝑥𝑓(𝑥) = 𝑖 𝑥 ∑𝑛−1 𝑖=0 𝑓𝑖 𝑥 ∈ 𝐵 if and only if |{0 ≤ 𝑖 ≤ 𝑛 − 1 | 𝑓𝑖 = 1}| is even. Hence, |𝐵| = 2

.

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It is not difficult to see that 2

𝐵 ∩ 𝐶 = {𝑥 𝑓 (𝑥) | 𝑓 (𝑥) ∈ F2 [𝑥]𝑛−2 , 𝑓 (1) = 0} ,

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and, hence, |𝐵 ∩ 𝐶| = 2𝑛−2 .

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Therefore, 𝑛 𝑇(2,𝑛) = ⌊ ⌋ + 1 + 2𝑛−1 + 2𝑛−1 − 2𝑛−2 2 𝑛−2

=3⋅2 as desired.

𝑛 +⌊ ⌋+1 2

̂∪𝐵 ̂∪𝐶 ̂ 𝑖𝑓 𝑛 𝑖𝑠 𝑒V𝑒𝑛, {𝐴 ̂ (2,𝑛) = 𝑇 {̂ ̂ 𝐵∪𝐶 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑, {

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̂ = {𝑥(𝑓(𝑥) + 𝑥𝑛−1 ) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−2 ̂ = {∑𝑛 𝑥𝑖 }, 𝐵 where 𝐴 𝑖=0 ̂ = {𝑥2 (𝑓(𝑥) + 𝑥𝑛−2 ) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−3 }. 𝑎𝑛𝑑 𝑓(1) = 1}, and 𝐶 Proof. We prove the statement by determining the elements 𝑖 in 𝑇(2,𝑛) of degree 𝑛. Let 𝐴 = {∑2𝑘 𝑖=0 𝑥 | 0 ≤ 𝑘 ≤ 𝑛/2}, 𝐵 = {𝑥𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−1 and 𝑓(1) = 0}, and 𝐶 = {𝑥2 𝑓(𝑥) | 𝑓(𝑥) ∈ F2 [𝑥]𝑛−2 } be defined as in Theorem 4. It is not difficult to see that the set of elements in 𝐵 (resp., ̂ (resp., 𝐶). ̂ 𝐶) of degree 𝑛 is 𝐵 ̂ If 𝑛 is even, then the set of elements in 𝐴 of degree 𝑛 is 𝐴. In the case where 𝑛 is odd, the set of elements in 𝐴 of degree 𝑛 is empty. By Theorem 4, the result, therefore, follows. Corollary 7. If 𝑛 is a positive integer, then

Proof. By direct calculation, we have 𝑇(2,1) = {0, 1} and |𝑇(2,1) | = 2. Next, assume that 𝑛 ≥ 2. By Theorem 4, we have 󵄨󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = |𝐴 ∪ 𝐵 ∪ 𝐶| , (9) 󵄨 󵄨

𝑛−1

̂ (2,𝑛) of good punctured polyNext, we determine the set 𝑇 nomials of degree 𝑛 over the binary field F2 . Since 𝑇(2,1) = F2 , ̂ (2,𝑛) can be determined ̂ (2,1) = 0. For 𝑛 ≥ 2, the set 𝑇 we have 𝑇 as follows. Theorem 6. If 𝑛 ≥ 2 is a positive integer, then

∑ 𝑔𝑖 ,

𝑖=0

3

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0 { { { { { {3 󵄨󵄨 ̂ 󵄨 { 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = { 󵄨 󵄨 { { 3 ⋅ 2𝑛−3 { { { { 𝑛−3 {3 ⋅ 2 + 1

𝑖𝑓 𝑛 = 1, 𝑖𝑓 𝑛 = 2, 𝑖𝑓 𝑛 ≥ 3 𝑖𝑠 𝑜𝑑𝑑,

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𝑖𝑓 𝑛 ≥ 4 𝑖𝑠 𝑒V𝑒𝑛.

̂ (2,1) = 0 and Proof. By direct calculation, we have 𝑇 ̂ (2,2) = {𝑥2 , 𝑥2 + 𝑥, 𝑥2 + 𝑥 + 1} . 𝑇

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̂ (2,2) = 3. ̂ (2,1) | = 0 and 𝑇 Hence, we have |𝑇 Next, assume that 𝑛 ≥ 3. By Theorem 6, we have 󵄨󵄨̂ ̂ ̂󵄨󵄨 󵄨󵄨 {󵄨󵄨󵄨𝐴 ∪ 𝐵 ∪ 𝐶󵄨󵄨󵄨 󵄨󵄨 ̂ 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨 = {󵄨 󵄨 󵄨 󵄨󵄨𝐵 ̂∪𝐶 ̂󵄨󵄨󵄨󵄨 󵄨 {󵄨󵄨

if 𝑛 is even, if 𝑛 is odd,

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̂ 𝐵, ̂ and 𝐶 ̂ are defined as in Theorem 6. Since 𝐴 ̂ and where 𝐴, ̂∪𝐶 ̂ are disjoint, by the inclusion-exclusion principle, we 𝐵 have 󵄨󵄨 ̂ 󵄨 󵄨̂ ̂ ̂󵄨󵄨 󵄨󵄨̂󵄨󵄨 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨̂󵄨󵄨 󵄨󵄨 ̂ ̂󵄨󵄨 󵄨 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = 󵄨󵄨󵄨𝐴 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 (19) 󵄨 󵄨 󵄨 ∪ 𝐵 ∪ 𝐶󵄨󵄨 = 󵄨󵄨𝐴󵄨󵄨 + 󵄨󵄨𝐵󵄨󵄨 + 󵄨󵄨𝐶󵄨󵄨 − 󵄨󵄨𝐵 ∩ 𝐶󵄨󵄨 . ̂ = 1 and |𝐶| ̂ = 2𝑛−2 . We note that |𝐴| 𝑛−1 𝑖 𝑛−1 ̂ if and only )∈𝐵 Since 𝑥(𝑓(𝑥)+𝑥 ) = 𝑥(∑𝑛−2 𝑖=0 𝑓𝑖 𝑥 +𝑥 if 󵄨󵄨 󵄨 (20) 󵄨󵄨{𝑖 | 0 ≤ 𝑖 ≤ 𝑛 − 2, 𝑓𝑖 = 1}󵄨󵄨󵄨 is odd, we have 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨𝐵󵄨󵄨 = 2𝑛−2 . 󵄨 󵄨

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4

International Journal of Mathematics and Mathematical Sciences Theorem 10. Let 𝑞 > 2 be a prime power. Then

Table 1: Punctured polynomials over F2 . 𝑛 󵄨󵄨 󵄨 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨 󵄨 󵄨

1 2 3 4

5

6

7

8

9

10

11

12

13

2 5 8 15 27 52 100 197 389 774 1542 3079 6151 0 3 3 7 12 25 48 97 192 385 768 1537 3072

= {𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 | (𝑎2 , 𝑎1 , 𝑎0 ) ∈ F𝑞∗ × F𝑞∗ × F𝑞 }

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∪ {𝑎2 𝑥2 | 𝑎2 ∈ F𝑞∗ } ∪ F𝑞 .

It is not difficult to see that ̂∩𝐶 ̂ = {𝑥2 (𝑓 (𝑥) + 𝑥𝑛−2 ) | 𝑓 (𝑥) ∈ F2 [𝑥]𝑛−3 , 𝑓 (1) 𝐵 = 1} ,

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and, hence, 󵄨󵄨 ̂ ̂󵄨󵄨 󵄨󵄨𝐵 ∩ 𝐶󵄨󵄨 = 2𝑛−3 . 󵄨 󵄨

(23)

Therefore, by (18), we have 𝑛−2 𝑛−2 𝑛−3 󵄨 {1 + 2 + 2 − 2 󵄨󵄨 ̂ 󵄨󵄨𝑇(2,𝑛) 󵄨󵄨󵄨 = { 󵄨 󵄨 2𝑛−2 + 2𝑛−2 − 2𝑛−3 { 𝑛−3 {3 ⋅ 2 + 1 ={ 3 ⋅ 2𝑛−3 {

𝑇(𝑞,2)

if 𝑛 is even, if 𝑛 is odd (24)

if 𝑛 is even, if 𝑛 is odd

as desired. ̂ (2,𝑛) | for 𝑛 = Table 1 presents the numbers |𝑇(2,𝑛) | and |𝑇 ̂ (2,𝑛) | in 1, 2, . . . , 13. The relation |𝑇(2,𝑛) | = |𝑇(2,𝑛−1) | + |𝑇 Remark 2 can be easily seen.

3. Punctured Polynomials over Nonbinary Finite Fields In this section, we focus on punctured polynomials over nonbinary finite fields. Given a prime power 𝑞 > 2, the characterization and enumeration of good punctured polynomials of degree less than or equal to 2 over F𝑞 are completely ̂ (𝑞,𝑛) determined. For 𝑛 ≥ 3, we construct subsets of 𝑇(𝑞,𝑛) and 𝑇 which lead to lower bounds of the cardinalities of 𝑇(𝑞,𝑛) and ̂ (𝑞,𝑛) , respectively. 𝑇

Proof. Let 𝐴 = {𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 | (𝑎2 , 𝑎1 , 𝑎0 ) ∈ F𝑞∗ × F𝑞∗ × F𝑞 } and 𝐵 = {𝑎2 𝑥2 | 𝑎2 ∈ F𝑞∗ }. Let 𝑓(𝑥) ∈ 𝐴 ∪ 𝐵 ∪ F𝑞 . We write 𝑓(𝑥) = 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 and consider the proof as two cases. Case 1 (𝑓(𝑥) ∈ 𝐴). We have 𝑓(0) (𝑥) = 𝑎2 𝑥 + 𝑎1 , 𝑓(1) (𝑥) = 𝑎2 𝑥 + 𝑎0 , and 𝑓(2) (𝑥) = 𝑎1 𝑥 + 𝑎0 . Since 𝑎2 and 𝑎1 are nonzero, it follows that −𝑎1 𝑎2−1 , −𝑎0 𝑎2−1 , and −𝑎0 𝑎1−1 are roots of 𝑓(0) (𝑥), 𝑓(1) (𝑥), and 𝑓(2) (𝑥), respectively. Hence, 𝑓(𝑥) ∈ 𝑇(𝑞,2) . Case 2 (𝑓(𝑥) ∈ 𝐵). Then 𝑓(𝑥) = 𝑎2 𝑥2 for some 𝑎2 ∈ F𝑞∗ . It follows that 0 is a root of 𝑓(0) (𝑥) = 𝑎2 𝑥, 𝑓(1) (𝑥) = 𝑎2 𝑥, and 𝑓(2) (𝑥) = 0. Therefore, 𝑓(𝑥) ∈ 𝑇(𝑞,2) . Case 3 (𝑓(𝑥) ∈ F𝑞 ). Then, by the definition, 𝑓(𝑥) ∈ 𝑇(𝑞,2) . On the other hand, let 𝑓(𝑥) = 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 ∈ 𝑇(𝑞,2) . If 𝑎2 = 0, then 𝑓(𝑥) = 𝑎1 𝑥 + 𝑎0 ∈ 𝑇(𝑞,1) , and, hence, 𝑓(𝑥) = 𝑎0 ∈ F𝑞 by Theorem 8. Assume that 𝑎2 ≠ 0. Then 𝑓(0) (𝑥) = 𝑎2 𝑥+𝑎1 , 𝑓(1) (𝑥) = 𝑎2 𝑥 + 𝑎0 , and 𝑓(2) (𝑥) = 𝑎1 𝑥 + 𝑎0 have a root in F𝑞 . Case 1 (𝑎1 = 0). We have that 𝑓(2) (𝑥) = 𝑎0 has a root in F𝑞 which implies that 𝑎0 = 0. Therefore, 𝑓(𝑥) = 𝑎2 𝑥2 ∈ 𝐵. Case 2 (𝑎1 ≠ 0). Since 𝑓(0) (𝑥) = 𝑎2 𝑥 + 𝑎1 has a root in F𝑞 , we have 𝑎2 ≠ 0. Hence, 𝑓(𝑥) ∈ 𝐴. From the two cases, it can be concluded that 𝑓(𝑥) ∈ 𝐴 ∪ 𝐵 ∪ F𝑞 . As desired, we have 𝑇(𝑞,2) = 𝐴 ∪ 𝐵 ∪ F𝑞 . Corollary 11. If 𝑞 > 2 is a prime power, then 󵄨󵄨 󵄨 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 = 𝑞 (𝑞2 − 2𝑞 + 3) − 1. 󵄨 󵄨

Theorem 8. If 𝑞 > 2 is a prime power, then 𝑇(𝑞,1) = F𝑞 . Proof. By the definition, F𝑞 ⊆ 𝑇(𝑞,1) . Let 𝑓(𝑥) = 𝑎𝑥+𝑏 ∈ 𝑇(𝑞,1) . Since 𝑓(0) (𝑥) = 𝑎 has a root in F𝑞 , we have 𝑎 = 0. Hence, 𝑓(𝑥) = 𝑏 ∈ F𝑞 as desired.

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Proof. Let 𝐴 = {𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 | (𝑎2 , 𝑎1 , 𝑎0 ) ∈ F𝑞∗ × F𝑞∗ × F𝑞 } and 𝐵 = {𝑎2 𝑥2 | 𝑎2 ∈ F𝑞∗ } be defined as in the proof of Theorem 10. It is not difficult to see that 𝐴, 𝐵, and F𝑞 are disjoint. By Theorem 10, we have

The next corollary follows immediately from Theorem 8. 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 = 󵄨󵄨󵄨𝐴 ∪ 𝐵 ∪ F𝑞 󵄨󵄨󵄨 = |𝐴| + |𝐵| + 󵄨󵄨󵄨F𝑞 󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨

Corollary 9. If 𝑞 > 2 is a prime power, then the following statements hold:

= (𝑞 − 1) (𝑞 − 1) 𝑞 + (𝑞 − 1) + 𝑞

(i) |𝑇(𝑞,1) | = 𝑞.

= 𝑞 (𝑞2 − 2𝑞 + 3) − 1

̂ (𝑞,1) = 0. (ii) 𝑇 ̂ (𝑞,1) | = 0. (iii) |𝑇

as desired.

(27)


̂ (𝑞,3) . Let 𝑔(𝑥) = 𝑥(𝑎2 𝑥2 + 𝑎1 𝑥 + First, we show that 𝑆 ⊆ 𝑇 ∗ 𝑎0 ) ∈ 𝑆, where 𝑎2 ∈ F𝑞 and 𝑎1 , 𝑎0 ∈ F𝑞 . Then 𝑔(0) (𝑥) = 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 has a root in F𝑞 and 0 is a root of

Corollary 12. If 𝑞 is a prime power, then ̂ (𝑞,2) 𝑇 = {𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 | (𝑎2 , 𝑎1 , 𝑎0 ) ∈ F𝑞∗ × F𝑞∗ × F𝑞 }

5

𝑔(1) (𝑥) = 𝑎2 𝑥2 + 𝑎1 𝑥,

(28)

∪ {𝑎2 𝑥2 | 𝑎2 ∈ F𝑞∗ } .

𝑔(2) (𝑥) = 𝑎2 𝑥2 + 𝑎0 𝑥,

Proof. From Theorem 10, it is not difficult to see that the polynomials of degree less than 2 in 𝑇(𝑞,2) are 𝑓(𝑥) = 𝑎 ∈ F𝑞 . Hence, the result follows. Corollary 13. If 𝑞 is a prime power, then 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨 = (𝑞 − 1) (𝑞2 − 𝑞 + 1) . 󵄨 󵄨

(29)

𝑔(3) (𝑥) = 𝑎1 𝑥2 + 𝑎0 𝑥. ̂ (𝑞,3) is good punctured. Hence, 𝑔(𝑥) ∈ 𝑇 By Theorem 14, the number of monic irreducible polynomials of degree 2 over F𝑞 is 1 1 𝐿 (𝑞, 2) = ∑𝜇 (𝑑) 𝑞2/𝑑 = (𝜇 (1) 𝑞2 + 𝜇 (2) 𝑞1 ) 2 𝑑|2 2

Proof. From Corollaries 11 and 12, it follows that 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨 = 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 − 𝑞 = 𝑞 (𝑞2 − 2𝑞 + 3) − 1 − 𝑞 󵄨 󵄨 󵄨 󵄨 = (𝑞 − 1) (𝑞2 − 𝑞 + 1) .

1 = (𝑞2 − 𝑞) . 2 (30)

In the case where 𝑛 ≥ 3, determining the sets 𝑇(𝑞,𝑛) and ̂ 𝑇(𝑞,𝑛) is more tedious and complicated. For these cases, we ̂ (𝑞,𝑛) |. give constructive lower bounds of |𝑇(𝑞,𝑛) | and |𝑇 The following results are important tools in constructing ̂ (𝑞,𝑛) |. lower bounds of |𝑇(𝑞,𝑛) | and |𝑇 Theorem 14 (see [14, Page 588]). Let 𝑛 be a positive integer and let 𝑞 be a prime power. Then the number of monic irreducible polynomials of degree 𝑛 in F𝑞 [𝑥] is 1 𝐿 (𝑛, 𝑞) = ∑𝜇 (𝑑) 𝑞𝑛/𝑑 , 𝑛 𝑑|𝑛

(35)

Hence, the number of irreducible polynomials of degree 2 over F𝑞 is 1 (𝑞 − 1) (𝑞2 − 𝑞) . 2

(36)

By Theorem 15, the number of polynomials of degree 2 over F𝑞 having a root in F𝑞 is |𝑆| = (𝑞 − 1) 𝑞2 −

1 (𝑞 − 1) (𝑞2 − 𝑞) 2

1 = (𝑞 − 1) 𝑞 (𝑞 + 1) . 2

(37)

̂ (𝑞,3) , we have Since 𝑆 ⊆ 𝑇 1 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨 ≥ |𝑆| = (𝑞 − 1) 𝑞 (𝑞 + 1) 󵄨 󵄨 2

(31)

(38)

as desired.

where 𝜇 (𝑛) 1 { { { { = {0 { { { 𝑟 {(−1)

(34)

Corollary 17. If 𝑞 > 2 is a prime power, then 𝑖𝑓 𝑛 = 1, 𝑖𝑓 𝑛 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑎 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑 𝑝𝑟𝑖𝑚𝑒 𝑓𝑎𝑐𝑡𝑜𝑟,

(32)

Proof. By Corollary 11 and Theorem 16, we have 󵄨󵄨󵄨𝑇 󵄨󵄨󵄨 = 𝑞 (𝑞2 − 2𝑞 + 3) − 1, 󵄨󵄨 (𝑞,2) 󵄨󵄨

𝑖𝑓 𝑛 𝑖𝑠 𝑎 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑝𝑟𝑖𝑚𝑒𝑠

is the M¨obius function. Theorem 15 (see [3, Section 4.2, Theorem 1]). Let 𝑓(𝑥) be a polynomial of degree 2 or 3 in F𝑞 [𝑥]. Then 𝑓(𝑥) is reducible if and only if 𝑓(𝑥) has a root in F𝑞 [𝑥]. Theorem 16. If 𝑞 > 2 is a prime power, then 󵄨󵄨 ̂ 󵄨󵄨 1 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨 ≥ (𝑞 − 1) 𝑞 (𝑞 + 1) . 󵄨 󵄨 2

󵄨󵄨 󵄨 3𝑞3 − 4𝑞2 + 5𝑞 − 2 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨󵄨 ≥ . 󵄨 󵄨 2

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Proof. Let 𝑆 fl {𝑥𝑓(𝑥) | 𝑓(𝑥) ∈ ̂F 𝑞 [𝑥]2 and 𝑓(𝑥) has a root in F𝑞 }.

󵄨󵄨 ̂ 󵄨󵄨 1 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨 ≥ (𝑞 − 1) 𝑞 (𝑞 + 1) . 󵄨 󵄨 2

(39)

(40)

Hence, by Remark 2, we have the relation 󵄨 󵄨 󵄨 󵄨 ̂ 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨󵄨 = 󵄨󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑇 󵄨 󵄨 󵄨 󵄨 (𝑞,3) 󵄨󵄨 󵄨 ≥ 𝑞 (𝑞2 − 2𝑞 + 3) − 1 + =

3𝑞3 − 4𝑞2 + 5𝑞 − 2 . 2

1 (𝑞 − 1) 𝑞 (𝑞 + 1) 2

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6


Theorem 18. If 𝑞 > 2 is a prime power, then 󵄨󵄨 ̂ 󵄨󵄨 1 󵄨󵄨𝑇(𝑞,4) 󵄨󵄨 ≥ (𝑞 − 1) 𝑞 (𝑞2 + 1) . 󵄨 2 󵄨

From Theorem 18, we have

Proof. Let 𝑆 fl {𝑥𝑓(𝑥) | 𝑓(𝑥) ∈ ̂F 𝑞 [𝑥]3 and 𝑓(𝑥) has a root in F𝑞 }. ̂ (𝑞,4) . Let 𝑔(𝑥) = 𝑥(𝑎3 𝑥3 + 𝑎2 𝑥2 + First, we show that 𝑆 ⊆ 𝑇 ∗ 𝑎1 𝑥+𝑎0 ) ∈ 𝑆, where 𝑎3 ∈ F𝑞 and 𝑎2 , 𝑎1 , 𝑎0 ∈ F𝑞 . Then 𝑔(0) (𝑥) = 𝑎3 𝑥3 + 𝑎2 𝑥2 + 𝑎1 𝑥 + 𝑎0 has a root in F𝑞 and 0 is a root of 𝑔(1) (𝑥) = 𝑎3 𝑥3 + 𝑎2 𝑥2 + 𝑎1 𝑥, 𝑔(2) (𝑥) = 𝑎3 𝑥3 + 𝑎2 𝑥2 + 𝑎0 𝑥, 𝑔(3) (𝑥) = 𝑎3 𝑥3 + 𝑎1 𝑥2 + 𝑎0 𝑥,

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1 (𝑞 − 1) (𝑞3 − 𝑞) . 2

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By Theorem 15, the number of polynomials of degree 3 over F𝑞 having a root in F𝑞 is

=

󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ̂ 󵄨󵄨 󵄨󵄨𝑇(𝑞,4) 󵄨󵄨󵄨 = 󵄨󵄨󵄨𝑇(𝑞,3) 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑇 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 (𝑞,4) 󵄨󵄨 ≥

3𝑞3 − 4𝑞2 + 5𝑞 − 2 1 + (𝑞 − 1) 𝑞 (𝑞2 + 1) 2 2

=

𝑞4 + 2𝑞3 − 3𝑞2 + 4𝑞 − 2 . 2

󵄨󵄨 󵄨 󵄨󵄨𝑇(𝑞,𝑛) 󵄨󵄨󵄨 ≥ 𝑞𝑛−1 + 𝑞3 − 2𝑞2 + 2𝑞. 󵄨 󵄨

1 (𝑞 − 1) (𝑞3 − 𝑞) 2

1 (𝑞 − 1) 𝑞 (𝑞2 + 1) . 2

(46)

(52)

(53)

Therefore, consider 󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨𝑇(𝑞,𝑛) 󵄨󵄨󵄨 ≥ |𝐴| + 󵄨󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 − 󵄨󵄨󵄨𝐴 ∩ 𝑇(𝑞,2) 󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 = 𝑞𝑛−1 + 𝑞 (𝑞2 − 2𝑞 + 3) − 1 − (𝑞 − 1)

(54)

= 𝑞𝑛−1 + 𝑞3 − 2𝑞2 + 2𝑞. Corollary 21. Let 𝑞 > 2 be a prime power and let 𝑛 ≥ 5 be an ̂ (𝑞,𝑛) | ≥ (𝑞 − 1)𝑞𝑛−2 . integer. Then |𝑇 Proof. The set of elements in 𝐴 of degree 𝑛 in the proof ̂ = {𝑥2 𝑓(𝑥) | 𝑓(𝑥) ∈ ̂F 𝑞 [𝑥]𝑛−2 }. By of Theorem 20 is 𝐴 Theorem 20, we have

̂ (𝑞,4) , we have Since 𝑆 ⊆ 𝑇 1 󵄨󵄨 ̂ 󵄨󵄨 󵄨󵄨𝑇(𝑞,4) 󵄨󵄨 ≥ |𝑆| = (𝑞 − 1) 𝑞 (𝑞2 + 1) 󵄨 󵄨 2

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Proof. Let 𝐴 = {𝑥2 𝑓(𝑥) | 𝑓(𝑥) ∈ F𝑞 [𝑥]𝑛−2 }. First, we show that 𝐴 ∪ 𝑇(𝑞,2) ⊆ 𝑇(𝑞,𝑛) . Clearly, 𝑇(𝑞,2) ⊆ 𝑇(𝑞,𝑛) . Let 𝑔(𝑥) ∈ 𝐴. Then 0 is a root of 𝑔(𝑖) (𝑥) for all 𝑖 = 0, 1, . . . , deg(𝑔(𝑥)), and, hence, 𝑔(𝑥) ∈ 𝑇(𝑞,𝑛) . Note that 𝐴 ∩ 𝑇(𝑞,2) = {𝑎𝑥2 | 𝑎 ∈ F𝑞∗ }. By Corollary 11, we have 󵄨 󵄨󵄨 󵄨󵄨𝑇(𝑞,2) 󵄨󵄨󵄨 = 𝑞 (𝑞2 − 2𝑞 + 2) . 󵄨 󵄨

Hence, the number of irreducible polynomials of degree 3 over F𝑞 is

|𝑆| = (𝑞 − 1) 𝑞3 −

Hence, by Remark 2, we have the relation

Theorem 20. Let 𝑞 > 2 be a prime power and let 𝑛 ≥ 5 be an integer. Then

̂ (𝑞,4) is good punctured as desired. Therefore, 𝑔(𝑥) ∈ 𝑇 By Theorem 14, the number of monic irreducible polynomials of degree 3 over F𝑞 is

1 = (𝑞3 − 𝑞) . 2

(50)

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𝑔(4) (𝑥) = 𝑎2 𝑥3 + 𝑎1 𝑥2 + 𝑎0 𝑥.

1 1 𝐿 (𝑞, 3) = ∑𝜇 (𝑑) 𝑞3/𝑑 = (𝜇 (1) 𝑞3 + 𝜇 (3) 𝑞1 ) 3 𝑑|3 2

󵄨󵄨 ̂ 󵄨󵄨 1 󵄨󵄨𝑇(𝑞,4) 󵄨󵄨 ≥ (𝑞 − 1) 𝑞 (𝑞2 + 1) . 󵄨 2 󵄨

(42)

(47)

󵄨󵄨 ̂ 󵄨 󵄨̂󵄨󵄨 𝑛−2 󵄨 󵄨󵄨𝑇(𝑞,𝑛) 󵄨󵄨󵄨 ≥ 󵄨󵄨󵄨𝐴 󵄨 󵄨 󵄨 󵄨󵄨 ≥ (𝑞 − 1) 𝑞 .

(55)

as desired. Corollary 19. If 𝑞 > 2 is a prime power, then 󵄨󵄨 󵄨 𝑞4 + 2𝑞3 − 3𝑞2 + 4𝑞 − 2 󵄨󵄨𝑇(𝑞,4) 󵄨󵄨󵄨 ≥ . 󵄨 󵄨 2

4. Conclusion and Open Problems (48)

Proof. By Corollary 17, we have 󵄨󵄨 󵄨 3𝑞3 − 4𝑞2 + 5𝑞 − 2 󵄨󵄨𝑇(𝑞,3) 󵄨󵄨󵄨 ≥ . 󵄨 󵄨 2

(49)

The concepts of punctured polynomials and good punctured polynomials are introduced. Over the finite field F2 , the complete characterization and enumeration of such polynomials are given. Over nonbinary finite fields, the good punctured polynomials of degree less than or equal to 2 are completely determined. For 𝑛 ≥ 3, constructive lower bounds of the number of good punctured polynomials of degree 𝑛 are given.

International Journal of Mathematics and Mathematical Sciences In general, the following related problems are also interesting: ̂ (𝑞,𝑛) , where 𝑞 > 2 is a ̂ (𝑞,𝑛) and 𝑇 (1) Determine the sets 𝑇 prime power and 𝑛 ≥ 3 is an integer. ̂ (𝑞,𝑛) |, ̂ (𝑞,𝑛) | and |𝑇 (2) Determine the exact values of |𝑇 where 𝑞 > 2 is a prime power and 𝑛 ≥ 3 is an integer. ̂ (𝑞,𝑛) |, where ̂ (𝑞,𝑛) | and |𝑇 (3) Improve lower bounds of |𝑇 𝑞 > 2 is a prime power and 𝑛 ≥ 3 is an integer. (4) Characterize and enumerate the good punctured polynomials of degree 𝑛 over the real numbers R or over the rational numbers Q.

Competing Interests The authors declare that there are no competing interests regarding the publication of this paper.

Acknowledgments This research is supported by the Thailand Research Fund under Research Grant TRG5780065.

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[13] A. Knopfmacher and J. Knopfmacher, “Counting polynomials with a given number of zeros in a finite field,” Linear and Multilinear Algebra, vol. 26, no. 4, pp. 287–292, 1990. [14] D. S. Dummit and R. M. Foote, Abstract Algebra, John Wiley & Sons, New York, NY, USA, 3rd edition, 2003.

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