On the Number of Irreducible Polynomials Over GF(2) with ......field with elements. We will denote...
Transcript of On the Number of Irreducible Polynomials Over GF(2) with ......field with elements. We will denote...
On the Number of Irreducible Polynomials Over
GF(2) with some Prescribed Coefficients
KΓΌbra AfΕar1
Ankara University
Department of Mathematics
06100, Ankara, Turkey
Necmettin Erbakan University
Department of Mathematics and Computer Science
42060, Konya, Turkey
ZΓΌlfΓΌkar SaygΔ±2
TOBB University of Economics and Technology
Department of Mathematics
06560, Ankara, Turkey
Mathematica Aeterna, Vol. 7, 2017, no. 3, 269 - 286
TOBB University of Economics and Technology
Department of Mathematics
06560, Ankara, Turkey
Erdal GΓΌner4
Ankara University
Department of Mathematics
06100, Ankara, Turkey
Abstract
In this paper we evaluate the number of monic irreducible polynomials in π½2[π₯] of
even degree π whose first four coefficients have prescribed values. This problem
first studied in [7] and some approximate results are obtained. Our results extends
the results given in [7] in some cases.
Mathematics Subject Classification: 12E05; 12E20
Keywords: Irreducible Polynomials, Finite Fields, Trace Function.
1. Introduction
Let π be a positive integer, π be a prime number and π = ππ. Let π½π be the finite
field with π elements. We will denote the number of monic irreducible polynomials
in π½π[π₯] of degree π by ππ(π) and the number of monic irreducible polynomials in
3Ernist Tilenbaev
270 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
π½π[π₯] of degree π with first π coefficients prescribed to π1, π2, β¦ , ππ β π½π by
ππ(π, π1, π2, β¦ , ππ).
In the literature ππ(π, π1) was studied by Carlitz [2] and ππ(π, π1, π2) was given
by Kuzmin [8]. Cattell et al. [3] reconsidered π2(π, π1, π2), which is a special case
of Kuzmin [8]. Results for three prescribed coefficients are given by Kuzmin [8],
Cattell et al. [3], Yucas and Mullen [16] and Fitzgerald and Yucas [4] for the case
π = 2. Moisio and Ranto [13] considered some special cases for π = 2 and π = 3.
Lalin and Larocque [10] proved Kuzminβs [8] results using elementary
combinatorial methods, together with the theory of quadratic forms over finite
fields. Ahmadi et al. [1] computed π2π(π, 0,0,0) using the number of points on
certain supersingular curves over finite fields. Granger [5] considered
ππ(π, π1, π2, β¦ , ππ) for π β€ 7 where π = 5 or π = 2 and π is odd and gave an
explicit formula for π = 3 where π = 3 and also gave an algorithm which gives
exact expressions in terms of the number of points of certain algebraic varieties
over π½π.
In this paper we obtained some result on π2(π, π1, π2, π3, π4). Our results extend
the results given in [7]. We have combined some previous results with the
properties of the extended trace functions.
2. Preliminaries and Previous Results
In this section we will present some useful notations and previous results. To obtain
our desired numbers we will use the properties of the first four traces which we
denote by ππ1, ππ2, ππ3 and ππ4. For any π β₯ 1 these trace functions are the
generalizations of the usual trace function and for any π β π½ππ they can be
expressed as
πππ(π) = β πππ1+ππ2+β―+πππ
0β€π1<π2<β―<ππβ€π
.
On the Number of Irreducible Polynomials ... 271
For π1 β π½π, πΈπ(π, π1) denote the number of elements π β π½ππ such that ππ1(π) =
π1 and in general πΈπ(π, π1, π2, β¦ , ππ) be the number of elements π β π½ππ for which
ππ1(π) = π1, ππ2(π) = π2, β¦ , πππ(π) = ππ. In literature it is shown that
ππ(π, π1, π2, β¦ , ππ) is directly related with πΈπ(π, π1, π2, β¦ , ππ) and this relations
can be described using the MΓΆbius inversion formula (see, for example [11]). By
using the MΓΆbius inversion formula π2(π, π1, π2, π3, π4) is given in terms of
πΈ2(π, π1, π2, π3, π4) in [7]. For completeness of the paper we present this theorem
in appendix.
π denotes the MΓΆbius function defined as
π(π) = {1 if π = 1,
(β1)π if π is square β free and a product of π distinct primes,0 otherwise.
πΈ2(π, π1, π2) and πΈ2(π, π1, π2, π3) is obtained in [16] and respectively. We will use
these results in our main theorem. For this reason we will present these numbers in
the following theorems.
Teorem 1: [16] Let π β₯ 2 be an integer and π = 2π. Then we have πΈ2(π, π1, π2) =
2πβ2 + πΊ2(π, π1, π2) where πΊ2(π, π1, π2) is given in Table 1.
Table 1 Values of πΊ2(π, π1, π2)
m (mod 4) (0, 0) (0, 1) (1, 0) (1, 1)
0 -2πβ1 2πβ1 0 0
1 0 0 -2πβ1 2πβ1
2 2πβ1 -2πβ1 0 0
3 0 0 2πβ1 -2πβ1
272 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
Teorem 2: [16] Let π β₯ 2 be an integer and π = 2π. Then we have
πΈ2(π, π1, π2, π3) = 2πβ3 + πΊ2(π, π1, π2, π3) where πΊ2(π, π1, π2, π3) is given in
Table 2.
Table 2 Values of πΊ2(π, π1, π2, π3)
m(mod12) 000 001 010 011 100 101 110 111
0
0 0 0 0
1 or 5
2 or 10 0
0
3
0
4 or 8
0
0 0 0 0
6
7 or 11
9
0
3. Main Results
In this section we will present our main results that extends the results given in [7].
Throughout the rest of the paper we assume that π β‘ 4 (πππ 8). We start by a
useful lemma.
Lemma 3: Let π β‘ 4 (πππ 8). Then we have πΈ2(π, π1, π2, π3, π4) =
πΈ2(π, π1, π1 + π2, π1 + π3, π4 + π3 + π2 + π1 + 1).
Proof: For any πΌ β π½2π it is enough to consider the values of πππ(πΌ) and
πππ(1 + πΌ) for all π = 1,2,3,4. By definition of the trace function we directly obtain
that ππ1(1 + πΌ) = ππ1(πΌ) since π is even. For π = 2 we have
ππ2(1 + πΌ) = β (1 + πΌ)ππ(1 + πΌ)π
π
0β€π<πβ€πβ1
= β 1ππ1π
π
0β€π<πβ€πβ1
+ β 1πππΌπ
π
0β€π<πβ€πβ1
+ β πΌππ1π
π
0β€π<πβ€πβ1
+ β πΌπππΌπ
π
0β€π<πβ€πβ1
On the Number of Irreducible Polynomials ... 273
= ππ2(1) + β 1ππ
0β€πβ€πβ1
β πΌππ
0β€πβ€πβ1
+ ππ2(πΌ)
= ππ2(1) + (π β 1)ππ1(πΌ) + ππ2(πΌ)
= ππ1(πΌ) + ππ2(πΌ)
since π β‘ 4 (πππ 8) π β 1 β‘ 1(πππ 2) and ππ2(1) = 0.
For π = 3 we have
ππ3(1 + πΌ) = β (1 + πΌ)ππ(1 + πΌ)π
π(1 + πΌ)π
π
0β€π<π<πβ€πβ1
= β 1ππ1π
π1π
π
0β€π<π<πβ€πβ1
+ β 1πππΌπ
ππΌπ
π
0β€π<π<πβ€πβ1
+ β πΌπππΌπ
π1π
π
0β€π<π<πβ€πβ1
+ β πΌππ1π
ππΌπ
π
0β€π<π<πβ€πβ1
β πΌππ1π
π1π
π
0β€π<π<πβ€πβ1
+ β 1πππΌπ
π1π
π
0β€π<π<πβ€πβ1
+ β 1ππ1π
π
0β€π<π<πβ€πβ1
πΌππ+ β πΌπ
ππΌπ
ππΌπ
π
0β€π<π<πβ€πβ1
= ππ3(1) + β 1ππ
0β€πβ€πβ1
β πΌπππΌπ
π
0β€π<πβ€πβ1
+ β πΌππ
0β€πβ€πβ1
β 1ππ1π
π
0β€π<πβ€πβ1
+ ππ3(πΌ)
= ππ3(1) + (π β 2)ππ2(πΌ) + (π β 12
)ππ1(πΌ) + ππ3(πΌ)
= ππ1(πΌ) + ππ3(πΌ)
since π β‘ 4 (πππ 8), π β 2 β‘ 0(πππ2), (π β 12
) β‘ 1 (πππ 2) and ππ3(1) = 0.
Similarly, for π = 4 by expanding
ππ4(1 + πΌ) = β (1 + πΌ)ππ(1 + πΌ)π
π(1 + πΌ)π
π(1 + πΌ)π
π
0β€π<π<π<πβ€πβ1
we obtained the desired result ππ4(1 + πΌ) = ππ1(πΌ) + ππ2(πΌ) + ππ3(πΌ) + ππ4(πΌ) + 1
which completes the proof.
Combining Lemma 3 with Theorem 2 we present values of πΈ2(π, π1, π2, π3, π4) in
some cases in the following corollary.
274 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
Corollary 4: Let π β‘ 4 (πππ8). We have
πΈ2(π, 0,0,0,0) = πΈ2(π, 0,0,0,1) = {2πβ4
π
2β‘ 2 ππ 10 (πππ 12)
2πβ4 + 3 β 2π 2β β3π
2β‘ 6 (πππ 12)
and
πΈ2(π, 0,1,1,0) = πΈ2(π, 0,1,1,1) = {2πβ4 β 2π 2β β2 π
2β‘ 2 ππ 10 (πππ 12)
2πβ4 + 2π 2β β3 π
2β‘ 6 (πππ 12)
Proof: From Lemma 3 we know that πΈ2(π, 0,0,0,0) = πΈ2(π, 0,0,0,1) and
πΈ2(π, 0,1,1,0) = πΈ2(π, 0,1,1,1). Furthermore we have
πΈ2(π, 0,0,0,0) + πΈ2(π, 0,0,0,1) = πΈ2(π, 0,0,0) and
πΈ2(π, 0,1,1,0) + πΈ2(π, 0,1,1,1) = πΈ2(π, 0,1,1)
Therefore using Theorem 3 we get
πΈ2(π, 0,0,0,0) = πΈ2(π, 0,0,0,1) =πΈ2(π, 0,0,0)
2
= {2πβ4
π
2β‘ 2 ππ 10 (πππ 12)
2πβ4 + 3 β 2π 2β β3 π
2β‘ 6(πππ 12)
and
πΈ2(π, 0,1,1,0) = πΈ2(π, 0,1,1,1) =πΈ2(π, 0,1,1)
2
= {2πβ4 β 2π 2β β2
π
2β‘ 2 ππ 10 (πππ12)
2πβ4 + 2π 2β β3π
2β‘ 6 (πππ12)
Furthermore, by combining Lemma 3 and Theorem 1 with Theorem 7 given in the
appendix we directly obtain the following result which extends Theorem 7.
On the Number of Irreducible Polynomials ... 275
Theorem 5: Let π β‘ 4 (πππ8). Then we have
ππ2(π, 1,1,1,0)
=β π(π)πΈ2(π πβ , 1,0,0,0)
π|π
πβ‘1,3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,1)π|π
πβ‘5,7 (πππ8)
ππ2(π, 1,0,0,1)
= β π(π)πΈ2(π πβ , 1,0,0,1)
π|π
πβ‘1,7 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,0)π|π
πβ‘3,5 (πππ8)
ππ2(π, 1,1,1,1)
= β π(π)πΈ2(π πβ , 1,0,0,1)
π|π
πβ‘1,3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,0)π|π
πβ‘5,7 (πππ8)
ππ2(π, 1,0,0,0)
= β π(π)πΈ2(π πβ , 1,0,0,0)
π|π
πβ‘1,7 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,1)π|π
πβ‘3,5 (πππ8)
ππ2(π, 0,0,1,0) = β π(π)πΈ2(π πβ , 0,0,1,0)
π|ππ πππ
ππ2(π, 0,0,1,1) = β π(π)πΈ2(π πβ , 0,0,1,1)
π|ππ πππ
ππ2(π, 1,1,0,0)
= β π(π)πΈ2(π πβ , 1,1,0,0) + β π(π)πΈ2(π πβ , 1,1,0,1)π|π
πβ‘3,5 (πππ8)
π|π
πβ‘1,7 (πππ8)
ππ2(π, 1,0,1,1)
276 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
= β π(π)πΈ2(π πβ , 1,1,0,0)
π|π
πβ‘1,3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,1)π|π
πβ‘5,7 (πππ8)
ππ2(π, 1,1,0,1)
= β π(π)πΈ2(π πβ , 1,1,0,1)
π|π
πβ‘1,7 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,0)π|π
πβ‘3,5 (πππ8)
ππ2(π, 1,0,1,0)
= β π(π)πΈ2(π πβ , 1,1,0,1) + β π(π)πΈ2(π πβ , 1,1,0,0)π|π
πβ‘5,7 (πππ8)
π|π
πβ‘1,3 (πππ8)
ππ2(π, 0,1,1,1) = ππ2(π, 0,1,1,0) = β π(π)πΈ2(π πβ , 0,1,1,1)π|ππ πππ
ππ2(π, 0,0,0,0) = ππ2(π, 0,0,0,1)
= β π(π)πΈ2(π πβ , 0,0,0,0) β β π(π)2π2πβ2
π|ππ πππ
π|ππ πππ
ππ2(π, 0,1,0,0)
=
(
β π(π)πΈ2(π πβ , 0,1,0,0)
π|π
πβ‘1(πππ4)
β β π(π)πΈ2(π 2πβ , 1,0)
π|π
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,0,1)
π|π
πβ‘3 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,1)π|π
πβ‘3 (πππ4) )
ππ2(π, 0,1,0,1)
=
(
β π(π)πΈ2(π πβ , 0,1,0,1)
π|π
πβ‘1 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,1)
π|π
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,0,0)
π|π
πβ‘3 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,0)π|π
πβ‘3 (πππ4) )
On the Number of Irreducible Polynomials ... 277
In Theorem 5 we see that the values of π2(π, π1, π2, π3, π4) are directly related with
the values of πΈ2(π, π1, π2, π3, π4) and πΈ2(π/π, π1, π2, π3, π4) where π is an odd
divisor of π. In this paper we are only dealing with πβs of the form π = 8π + 4,
therefore we need the following useful result.
Proposition 6: Let π be a positive integer satisfying π β‘ 4 (πππ 8) and π be a
positive odd divisor of π. Then we have π/π β‘ 4 (πππ 8).
Proof: Assume that π = 8π + 4 = 4(2π + 1) for some positive integer π. Then we
have π/π = 4π‘ where π‘ = (2π + 1)/π is an odd integer. Therefore, we have π/π =
4π‘ β‘ 4 (πππ 8).
4. Values of π΅π(π, ππ, ππ, ππ, ππ)
In this section using Corollary 4, Theorem 5 and Proposition 6 we will present the
values of π2(π, π1, π2, π3, π4) depending on the coefficients (π1, π2, π3, π4) and the
prime decomposition of π. In some cases we obtained the exact values of
π2(π, π1, π2, π3, π4).
Assume that π = 4. In this case we know that the only monic irreducible
polynomials are π₯4 + π₯ + 1, π₯4 + π₯3 + 1 and π₯4 + π₯3 + π₯2 + π₯ + 1; therefore we
have π2(π, 0,0,1,1) = π2(π, 1,0,0,1) = π2(π, 1,1,1,1) = 1. Furthermore we have
πΈ2(4,0,0,1,1) = πΈ2(4,1,0,0,1) = πΈ2(4,1,1,1,1) = 4, πΈ2(4,0,1,0,1) = 2,
πΈ2(4,0,0,0,0) = πΈ2(4,0,0,0,1) = 1 and πΈ2(4, π1, π2, π3, π4) = 0 in all other 10
cases.
Now assume that π = 4ππ for some odd prime π and positive integer π . In this
case note that the only positive odd divisors of π are ππ where 0 β€ π β€ π. But note
that by definition π(1) = 1, π(π) = β1 and π( ππ) = 0 for π β {2, β¦ , π}.
In the following four cases we have the exact values of π2(π, π1, π2, π3, π4)
278 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
π2(π, 0,1,1,1) = π2(π, 0,1,1,0) =1
π(πΈ2(π, 0,1,1,1) β πΈ2(π/π, 0,1,1,1))
= {
1
π(2πβ4β2π 2β β2 β 2π/πβ4+2π (2π)β β2) ππ π β 3
1
π(2πβ4 + 2π 2β β3 β 2π/πβ4 β 2π (2π)β β3) ππ π = 3
and
π2(π, 0,0,0,0) = π2(π, 0,0,0,1)
=1
π(πΈ2(π, 0,0,0,0) β πΈ2(π/π, 0,0,0,0) β 2
π 2β β2 + 2π (2π)β β2)
= {
1
π(2πβ4 β 2π πβ β4 β 2π 2β β2 + 2π (2π)β β2) ππ π β 3
1
π(2πβ4 + 3 β 2π 2β β3 β 2
ππβ4β 3 β 2π (2π)β β3 β 2π 2β β2 + 2π (2π)β β2) ππ π = 3
Now assume that π = 4π1π1π2
π2 for some odd primes π1, π2 (W.L.O.G assume
that π1 < π2) and positive integers π1, π2 . In this case note that the only positive
odd divisors of π are π1π , π2
π and π1ππ2
π where 0 β€ π β€ π1 and 0 β€ π β€ π2. But
note that by definition π(1) = 1, π( π1) = π( π2) = β1, π( π1π2) = 1 , π( π1π) =
π( π2π) = 0 for π β {2,β¦ , π1}, π β {2, β¦ , π2} , π( π1
ππ2π) = 0 for π β
{1,2, β¦ , π1}, π β {2,β¦ , π2} and π( π1ππ2
π) = 0 for π β {2,β¦ , π1}, π β
{1,2, β¦ , π2}.
In the following four cases we have the exact values of π2(π, π1, π2, π3, π4)
π2(π, 0,1,1,1) = π2(π, 0,1,1,0)
=1
π(πΈ2(π, 0,1,1,1) β πΈ2 (
π
π1, 0,1,1,1) β πΈ2 (
π
π2, 0,1,1,1)
+ πΈ2 (π
π1π2, 0,1,1,1)) =
{
1
π(2πβ4 + 2π 2β β3 β 2
ππ1β4+ 2π (2π1)β β2 β 2
ππ2β4β 2π (2π2)β β3 + 2
ππ1π2
β4β 2π (2π1π2)β β2) ππ π1 = 3 πππ π1 = 1
1
π(2πβ4 + 2π 2β β3 β 2
ππ1β4β 2π (2π1)β β3 β 2
ππ2β4β 2π (2π2)β β3 + 2
ππ1π2
β4+ 2π (2π1π2)β β3) ππ π1 = 3 πππ π1 > 1
1
π(2πβ4β2π 2β β2 β 2
ππ1β4+2π (2π1)β β2 β 2
ππ2β4+2π (2π2)β β2 + 2
ππ1π2
β4β2π (2π1π2)β β2) ππ π1 β 3
On the Number of Irreducible Polynomials ... 279
and
π2(π, 0,0,0,0) = π2(π, 0,0,0,1)
=1
π(πΈ2(π, 0,0,0,0) β πΈ2 (
π
π1, 0,0,0,0) β πΈ2 (
π
π2, 0,0,0,0) + πΈ2 (
π
π1π2, 0,0,0,0)
β2π 2β β2 + 2π (2π1)β β2 + 2π (2π2)β β2 β 2π (2π1π2)β β2
)
=
{
1
π(2
πβ4 + 3 β 2π 2β β3 β 2ππ1β4β 2
ππ2β4β 3. 2π 2π2β β3 + 2
ππ1π2
β4
β2π 2β β2 + 2π (2π1)β β2 + 2π (2π2)β β2 β 2π (2π1π2)β β2)
1
π( 2πβ4 + 3 β 2π 2β β3 β 2
ππ1β4β 3 β 2π (2π1)β β3 β 2
ππ2β4β 3. 2π 2π2β β3
+2π
π1π2β4+ 3. 2π 2π1π2β β3 β 2π 2β β2 + 2π (2π1)β β2 + 2π (2π2)β β2 β 2π (2π1π2)β β2
)
ππ π1 = 3 πππ π1 = 1ππ π1 = 3 πππ π1 > 1
1
π( 2πβ4 β 2π π1β β4 β 2π π2β β4 + 2π π1π2β β4
β2π 2β β2 + 2π (2π1)β β2 + 2π (2π2)β β2 β 2π (2π1π2)β β2) ππ π1 β 3
For the general case assume that π = 4π1π1π2
π2β―ππ ππ where π1, π2, β¦ , ππ are
relatively prime odd primes and π1, π2, β¦ , ππ are positive integers. Then by the
above argument using Corollary 4 and Theorem 5 we can easily evalute the values
of π2(π, 0,0,0,0) (= π2(π, 0,0,0,1)) and π2(π, 0,1,1,1) (= π2(π, 0,1,1,0)).
For the other 12 cases as we donβt know the values of πΈ2(π, π1, π2, π3, π4) we can
use the same approximation argument given in [7]. Note that using Theorem 5 one
can observe some further equalities depending on the prime factorization of π.
As an example we evaluate the values of π2(20, π1, π2, π3, π4) and presented them
in Table 3. The bold entries in Table 3 are the exact values of π2(20, π1, π2, π3, π4)
that are obtained using our results.
Table 3 Values of π2(20,π1, π2, π3, π4)
π1, π2, π3, π4 Komaβs estimate Our estimate Exact Value
0, 0, 0, 0 3264 3264 3264
0, 0, 0, 1 3264 3264 3264
0, 0, 1, 0 3276.75 3276.75 3264
0, 0, 1, 1 3276.75 3276.75 3315
0, 1, 0, 0 3264.75 3264.75 3280
0, 1, 0, 1 3263.25 3263.25 3248
0, 1, 1, 0 3276.75 3264 3264
0, 1, 1, 1 3276.75 3264 3264
280 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
1, 0, 0, 0 3276.75 3276.75 3275
1, 0, 0, 1 3276.75 3276.75 3304
1, 0, 1, 0 3276.75 3276.75 3264
1, 0, 1, 1 3276.75 3276.75 3264
1, 1, 0, 0 3276.75 3276.75 3264
1, 1, 0, 1 3276.75 3276.75 3264
1, 1, 1, 0 3276.75 3276.75 3275
1, 1, 1, 1 3276.75 3276.75 3304
References
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Applications, 2016, 42, 128-164.
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On the Number of Irreducible Polynomials ... 281
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282 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
Received: September 18, 2017
Appendix
Theorem 7 [7] Let π be even. Then we have
π2(π, 1,1,1,0)
=1
π
(
β π(π)πΈ2(π πβ , 1,1,1,0)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,0)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,1)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,1)π|π
πβ‘7 (πππ8) )
π2(π, 1,0,0,1)
=1
π
(
β π(π)πΈ2(π πβ , 1,0,0,1)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,0)
π|π
πβ‘3(πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,0)π|π
πβ‘5(πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,1)π|π
πβ‘7 (πππ8) )
π2(π, 1,1,1,1)
=1
π
(
β π(π)πΈ2(π πβ , 1,1,1,1)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,1)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,0)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,0)π|π
πβ‘7 (πππ8) )
π2(π, 1,0,0,0)
=1
π
(
β π(π)πΈ2(π πβ , 1,0,0,0)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,1)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,0,1)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,1,0)π|π
πβ‘7 (πππ8) )
On the Number of Irreducible Polynomials ... 283
π2(π, 0,0,1,0) =1
π( β π(π)πΈ2(π πβ , 0,0,1,0)
π|ππ πππ
)
π2(π, 0,0,1,1) =1
π( β π(π)πΈ2(π πβ , 0,0,1,1)
π|ππ πππ
)
π2(π, 1,1,0,0)
=1
π
(
β π(π)πΈ2(π πβ , 1,1,0,0)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,0)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,1)
π|π
πβ‘5(πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,1)π|π
πβ‘7 (πππ8) )
π2(π, 1,0,1,1)
=1
π
(
β π(π)πΈ2(π πβ , 1,0,1,1)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,0)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,0)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,1)π|π
πβ‘7 (πππ8) )
π2(π, 1,1,0,1)
=1
π
(
β π(π)πΈ2(π πβ , 1,1,0,1)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,1)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,0)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,0)π|π
πβ‘7 (πππ8) )
284 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner
π2(π, 1,0,1,0)
=1
π
(
β π(π)πΈ2(π πβ , 1,0,1,0)
π|π
πβ‘1 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,1)
π|π
πβ‘3 (πππ8)
+ β π(π)πΈ2(π πβ , 1,0,1,1)π|π
πβ‘5 (πππ8)
+ β π(π)πΈ2(π πβ , 1,1,0,0)π|π
πβ‘7 (πππ8) )
π2(π, 0,1,1,1) =1
π
(
β π(π)πΈ2(π πβ , 0,1,1,1)
π|π
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,1,0)π|π
πβ‘3 (πππ4) )
π2(π, 0,1,1,0) =1
π
(
β π(π)πΈ2(π πβ , 0,1,1,0)
π|π
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,1,1)π|π
πβ‘3 (πππ4) )
π2(π, 0,0,0,0) =1
π
(
β π(π)πΈ2(π πβ , 0,0,0,0)
π|ππ πππ
β β π(π)πΈ2(π 2πβ , 0,0)
π|π,π πβ ππ£πππ πππ )
π2(π, 0,0,0,1) =1
π
(
β π(π)πΈ2(π πβ , 0,0,0,1)
π|ππ πππ
β β π(π)πΈ2(π 2πβ , 0,1)
π|π,π πβ ππ£πππ πππ )
On the Number of Irreducible Polynomials ... 285
π2(π, 0,1,0,0)
=1
π
(
β π(π)πΈ2(π πβ , 0,1,0,0)
π|π
πβ‘1 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,0)
π|π,π πβ ππ£ππ
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,0,1)
π|π
πβ‘3 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,1)
π|π,π πβ ππ£πππβ‘3 (πππ4) )
π2(π, 0,1,0,1)
=1
π
(
β π(π)πΈ2(π πβ , 0,1,0,1)
π|π
πβ‘1 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,1)
π|π,π πβ ππ£ππ
πβ‘1 (πππ4)
+ β π(π)πΈ2(π πβ , 0,1,0,0)
π|π
πβ‘3 (πππ4)
β β π(π)πΈ2(π 2πβ , 1,0)
π|π,π πβ ππ£ππ
πβ‘3 (πππ4) )
286 Kubra Afsar, Zulfukar SaygΔ±, Ernist Tilenbaev, Erdal Guner