# The Reducibility of a Special Binary Pentanomial

Authors: Ryul Kim; Yun Mi Kim
Journal: Algebra
 (2014)

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## Abstract

Swan’s theorem determines the parity of the number of irreducible factors of a binary trinomial. In this work, we study the parity of the number of irreducible factors for a special binary pentanomial with even degreexm+xn1+xn2+xn3+1, where 0<n3<n2<n1≤m/2, and exactly one of  n1,n2,  and  n3 is odd. This kind of irreducible pentanomials can be used for a fast implementation of trace and square root computations in finite fields of characteristic 2.


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## Body

## 1. Introduction

Irreducible polynomials of low weight over a finite field are frequently used in many applications such as coding theory and cryptography due to efficient arithmetic implementation in an extension field and, thus, it is important to determine the irreducibility of such polynomials. The weight of a polynomial means the number of its nonzero coefficients.Characterization of the parity of the number of irreducible factors of a given polynomial is of significance in this context. If a polynomial has an even number of irreducible factors, then it is reducible and, thus, the study on the parity of this number can give a necessary condition for irreducibility. Swan [1] gives the first result determining the parity of the number of irreducible factors of trinomials over F2. Vishne [2] extends Swan’s theorem to trinomials over an even-dimensional extension of F2. Many Swan-like results focus on determining the reducibility of higher weight polynomials over F2; see for example [3, 4]. Some researchers obtain the results on the reducibility of polynomials over a finite field of odd characteristic. We refer to [5, 6].On the other hand, Ahmadi and Menezes [7] estimate the number of trace-one elements on the trinomial and pentanomial bases for a fast and low-cost implementation of trace computation. They also present a table of irreducible pentanomials whose corresponding polynomial bases have exactly one trace-one element. Each pentanomial of even degree in this table is of the form xm+xn1+xn2+xn3+1∈F2[x], where 0<n3<n2<n1≤m/2, and exactly one of n1,n2,  and  n3 is odd. In this work, we characterize the parity of the number of irreducible factors of this pentanomial. We describe some preliminary results related to Swan-like results in Section 2 and determine the reducibility of the pentanomial mentioned above in Section 3.


## 2. Preliminaries

In this section, we recall Swan’s theorem determining the parity of the number of irreducible factors of a polynomial overF2 and some results about the discriminant and the resultant of polynomials.LetK be a field and let f(x)=a∏i=0m-1‍(x-xi)∈K[x], where x0,…,xm-1 are the roots of f(x) in an extension of K. The discriminant of f(x) is defined by
(1)D(f)=a2m-2∏0≤i<j<m‍(xi-xj)2.
From the definition, it is clear that f(x) has a repeated root if and only if D(f)=0. Since D(f) is a symmetric function with respect to the roots of f(x), it is an element of K.The following theorem, due to Swan, relates the parity of the number of irreducible factors of a polynomial overF2 with its discriminant.Theorem 1 (see [1, 8]).
Suppose that the polynomialf(x)∈F2[x] of degree m has no repeated roots and let r be a number of irreducible factors of f(x) over F2. Let F(x)∈Z[x] be any monic lift of f(x) to the integers. Then, D(F)≡1 or 5  (mod⁡8), and r≡m  (mod⁡2) if and only if D(F)≡1  (mod⁡8).Letg(x)=b∏j=0n-1‍(x-yj)∈K[x], where y0,…,yn-1 are the roots of g(x) in an extension of K. The resultant of f(x) and g(x) is defined by
(2)R(f,g)=(-1)mnbm∏j=0n-1‍f(yj)=an∏i=0m-1‍g(xi).
It is well known that
(3)D(f)=(-1)m(m-1)/2R(f,f′),
where f′(x) denotes the derivative of f(x) with respect to x. An alternate formula for the discriminant of a monic polynomial f(x) is
(4)D(f)=(-1)m(m-1)/2∏i=0m-1‍f′(xi),
see [9].Let(5)f(x)=xm+a1xm-1+⋯+am=∏i=0m-1‍(x-xi)∈K[x];
then, for all k,1≤k<m, the coefficients ak of f(x)(6)ak=(-1)k∑0≤i1<i2⋯<ik<m‍xi1xi2⋯xik
are the elementary symmetric polynomials of xi. Since each ak∈K, it follows that S(x0,…,xm-1)∈K for every symmetric polynomial S∈K[x0,…,xm-1].The following natation will be used throughout the paper. For all integersp,q and k  (0≤k<m), let
(7)S(k,p)=∑0≤i1,…,ik≤m-1ij≠ilxi1p⋯xikp,S[k,p]=∑0≤i1<i2<⋯<ik≤m-1‍xi1p⋯xikp,Sp,q=∑0≤i,j≤m-1i≠jxipxjq.
We denote S(1,p)=S[1,p] simply as Sp and put S(0,p)=S[0,p]=1. Then, the following lemma holds.Lemma 2 (see [10, 11]).
(1)S0=S(1,0)=S[1,0]=m. (2) Sp,q=Sp·Sq-Sp+q. (3) S(k,p)=k!·S[k,p].The following formula, called Newton’s identity, is often used for computation of the discriminant.Theorem 3 (see [12]).
Letf(x),Sp and x0,x1,…,xm-1 be as above. Then, for every p≥1,
(8)Sp+Sp-1a1+Sp-2a2+⋯+Sp-n+1an-1+nmSp-nan=0,
where n=min⁡⁡{p,m}.The reciprocal polynomial off(x)=a0xm+a1xm-1+⋯+am-1x+am with a0≠0 over a finite field Fq is defined by
(9)f*(x)≔xmf(1x)=amxm+am-1xm-1+⋯+a1x+a0∈Fq[x].
See Lidl and Niederreiter [12] for more details.


## 3. Main Results

In this section, we characterize the parity of the number of irreducible factors for the pentanomial(10)f(x)=xm+xn1+xn2+xn3+1∈F2[x],
where m is even; 0<n3<n2<n1≤m/2 and exactly one of n1,n2,  and  n3 is odd. For our purpose, we use Swan’s theorem and Newton’s identity. In [10, 11], Newton’s identity has also been used to solve similar problems where it is enough to determine the power sums Sk with indices k≥-2, but, for (10), one should calculate much more negative indexed power sums. We return this calculation to one of positive indexed power sums by using reciprocals.It is clear that (10) has no repeated roots because its derivative has a unique root 0. Let F(x)∈Z[x] be the monic lift of f(x) in (10) to the integers and let x0,…,xm-1 denote the roots of F(x) in some extension of the rational numbers. The derivative of F(x) is
(11)F′(x)=mxm-1+n1xn1-1+n2xn2-1+n3xn3-1.
Note that ∏i=0m-1‍xi=1. Our work is divided into three cases according to which one of n1,n2,   and  n3 is odd.Case 1 (n3 is odd).
We can write the resultant ofF and F′ as
(12)R(F,F′)=∏i=0m-1‍(mxim-1+n1xin1-1+n2xin2-1+n3xin3-1)=∏i=0m-1‍(mxim-n3+n1xin1-n3+n2xin2-n3+n3).
Since m,n1 and n2 are even, we have
(13)R(F,F′)≡n3m+mn3m-1∑i=0m-1‍xim-n3+m2n3m-2×∑i<j‍xim-n3xjm-n3+n1n3m-1×∑i=0m-1‍xin1-n3+n12n3m-2×∑i<j‍xin1-n3xjn1-n3+n2n3m-1×∑i=0m-1‍xin2-n3+n22n3m-2×∑i<j‍xin2-n3xjn2-n3+mn1n3m-2×∑i≠j‍xim-n3xjn1-n3+mn2n3m-2×∑i≠j‍xim-n3xjn2-n3+n1n2n3m-2×∑i≠j‍xin1-n3xjn2-n3(mod⁡8).
Using Lemma 2 and the fact that the square of every odd integer is congruent to 1 modulo 8, we get
(14)R(F,F′)≡1+mn3Sm-n3+12m2(Sm-n32-S2m-2n3)+n1n3Sn1-n3+12n12(Sn1-n32-S2n1-2n3)+n2n3Sn2-n3+12n22(Sn2-n32-S2n2-2n3)+mn1(Sm-n3·Sn1-n3-Sm+n1-2n3)+mn2(Sm-n3·Sn2-n3-Sm+n2-2n3)+n1n2(Sn1-n3·Sn2-n3-Sn1+n2-2n3)(mod⁡8).
Newton’s identity shows that if 0<k<m-n1, then Sk=0 and
(15)S2m-2n3=-(Sm-2n3+Sm-n3+Sm+n2-2n3+Sm+n1-2n3).
Therefore,
(16)R(F,F′)≡1+mn3Sm-n3+12m2Sm-n32+12m2Sm-2n3+12m2Sm-n3+12m2Sm+n2-2n3+12m2Sm+n1-2n3-12n12S2n1-2n3-12n22S2n2-2n3-mn1Sm+n1-2n3-mn2Sm+n2-2n3-n1n2Sn1+n2-2n3(mod⁡8).
The indices of terms in the above equation have a relation
(17)2n2-2n3<n1+n2-2n3<2n1-2n3≤m-2n3<m-n3<m+n2-2n3<m+n1-2n3.
Since m+n1-2n3<2m-n1-n3, we determine all Sk for k<2m-n1-n3 by applying Newton’s identity to get Table 1.Table 1
The values ofSk for k<2m-n1-n3.
k
S
k
k
S
k
m
-
n
1
n
1
-
m
2
m
-
2
n
1
m
-
n
1
m
-
n
2
n
2
-
m
2
m
-
n
1
-
n
2
2
m
-
n
1
-
n
2
m
-
n
3
n
3
-
m
2
m
-
2
n
2
m
-
n
2
m
-
m
Other
0Note thatS2m-2n2=m-n2 does only cover the case of 2m-2n2<2m-n1-n3; that is, 2n2>n1+n3. Since Sm+n1-2n3, Sm+n2-2n3,  and  Sn1+n2-2n3 are all even, we have
(18)-mn1Sm+n1-2n3-mn2Sm+n2-2n3-n1n2Sn1+n2-2n3≡0(mod⁡8).
Therefore,
(19)R(F,F′)≡1+m+12m2-12m3-12m2n3+12m2Sm-2n3+12m2Sm+n2-2n3+12m2Sm+n1-2n3-12n12S2n1-2n3-12n22S2n2-2n3(mod⁡8).
With reference to Table 1, we can determine all unknown terms in the above equation. We consider two subcases.Subcase 1 (m is divisible by 4).
Then, we see easily that(20)D(F)≡1+m-1-(-1)n1/24n12S2n1-2n3-1-(-1)n2/24n22S2n2-2n3(mod⁡8);
hence, the value of D(F) modulo 8 depends on a pair (k1,k2):=(n1mod⁡4,n2mod⁡4). Let S:={3n1-2n3,2n1+n2-2n3,n1+2n2-2n3,3n2-2n3}.Theorem 4.
Suppose thatn3 is odd and m is divisible by 4. Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.
Consider(1)
m≡0  (mod⁡8):
(a)
(k1,k2)=(0,0);(b)
(k1,k2)=(0,2) and m≠3n2-2n3;(c)
(k1,k2)=(2,0) and m≠3n1-2n3;(d)
(k1,k2)=(2,2) and m∉S.(2)
m≡4  (mod⁡8):
(a)
(k1,k2)=(0,2) and m=3n2-2n3;(b)
(k1,k2)=(2,0) and m=3n1-2n3;(c)
(k1,k2)=(2,2) and m∈S.Proof.
If(k1,k2)=(0,0), then D(F)≡1+m  (mod⁡8) and, therefore, we have
(21)D(F)≡{1(mod⁡8),m≡0(mod⁡8)5(mod⁡8),m≡4(mod⁡8).
If (k1,k2)=(0,2), then
(22)D(F)≡1+m-12n22S2n2-2n3(mod⁡8)
and 2n2-2n3≠m-n1. So if 2n2-2n3=m-n2, that is, m=3n2-2n3, then
(23)D(F)≡{5(mod⁡8),m≡0(mod⁡8)1(mod⁡8),m≡4(mod⁡8).
And if m≠3n2-2n3, then (21) holds again. Similarly, if (k1,k2)=(2,0), then
(24)D(F)≡1+m-12n12S2n1-2n3(mod⁡8)
and 2n1-2n3≠m-n2. Thus, if 2n1-2n3=m-n1, that is, m=3n1-2n3, then (23) holds and if m≠3n2-2n3, then (21) holds. If (k1,k2)=(2,2), then
(25)D(F)≡1+m-12n12S2n1-2n3-12n22S2n2-2n3(mod⁡8),
and S2n1-2n3 or S2n2-2n3 can be nonzero only when it is equal to either m-n1 or m-n2. Analyzing the possible cases shows that m∈S implies (23) and m∉S implies (21). Now, applying Swan’s theorem completes the proof.Subcase 2 (m is not divisible by 4).
Then,m≡±2 (mod⁡8) and we can write
(26)D(F)≡(-1)m(m-1)/2(A+B+C)(mod⁡8),
where
(27)A≔1+m+12m2-12m3-12m2n3,B≔12m2(Sm-2n3+Sm+n2-2n3+Sm+n1-2n3),C≔-12(n12S2n1-2n3+n22S2n2-2n3).
It is clear that if m≡2  (mod⁡8), then
(28)A≡{7(mod⁡8),n3≡1(mod⁡4)3(mod⁡8),n3≡-1(mod⁡4)
and if m≡-2  (mod⁡8), then
(29)A≡{3(mod⁡8),n3≡1(mod⁡4)7(mod⁡8),n3≡-1(mod⁡4).
Now determineB and C. First, assume that m-2n3=m-n1; that is, n1=2n3. Since m+n2-2n3<m and m+n1-2n3=m, we have
(30)B=12m2(n1-2m)≡12m2n1≡4(mod⁡8).
And then C≡0  (mod⁡8) because 2n2-2n3<2n1-2n3<m-n1. Next, assume that m-2n3=m-n2; that is, n2=2n3. Clearly, Sm+n2-2n3=Sm=-m and m+n1-2n3>m. Since m+n1-2n3≠2m-n1-n2, if m+n1-2n3=2m-2n1, that is, m=3n1-2n3, then Sm+n1-2n3=m-n1 and if m≠3n1-2n3, then Sm+n1-2n3=0. Therefore, we get
(31)B≡{0(mod⁡8),m=3n1-2n34(mod⁡8),m≠3n1-2n3
and also C≡0(mod⁡8), similarly. When n1≠2n3 and n2≠2n3, a similar consideration shows
(32)B≡{4(mod⁡8),m∈S0(mod⁡8),otherwise,C≡{-12mn12(mod⁡8),m=2n1+n2-2n3-12mn22(mod⁡8),m=n1+2n2-2n30(mod⁡8),otherwise.
Summarizing the above discussion and applying Swan’s theorem, we have the following theorem.Theorem 5.
Suppose thatn3 is odd and m≡±2  (mod⁡8). Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.
Consider(1)
(mmod⁡8,n3mod⁡4)=(2,-1) or (-2,1):
(a)
n1=2n3;(b)
n2=2n3 and m≠3n1-2n3;(c)
n1≠2n3,n2≠2n3 and either m=3n1-2n3 or m=3n2-2n3;(d)
n1≠2n3,n2≠2n3 and either m=2n1+n2-2n3,n1≡0  (mod⁡4) or m=n1+2n2-2n3,n2≡0  (mod⁡4);(2)
(mmod⁡8,n3mod⁡4)=(2,1) or (-2,-1):
(a)
n2=2n3 and m=3n1-2n3;(b)
n1≠2n3,n2≠2n3 and either m=2n1+n2-2n3,n1≡2  (mod⁡4) or m=n1+2n2-2n3,n2≡2  (mod⁡4);(c)
n1≠2n3,n2≠2n3 and m∉S.Case 2 (n2 is odd).
Similarly, we have(33)R(F,F′)=∏i=0m-1‍(mxim-n2+n1xin1-n2+n3xin3-n2+n2)≡1+mn2Sm-n2+12m2(Sm-n22-S2m-2n2)+n1n2Sn1-n2+12n12(Sn1-n22-S2n1-2n2)+n3n2Sn3-n2+12n32(Sn3-n22-S2n3-2n2)+mn1(Sm-n2·Sn1-n2-Sm+n1-2n2)+mn3(Sm-n2·Sn3-n2-Sm+n3-2n2)+n1n3(Sn1-n2·Sn3-n2-Sn1+n3-2n2)(mod⁡8).
From Newton’s identity, we see easily that Sm-n1=n1-m,  Sm-n2=n2-m,  Sm-n3=n3-m, and Sm=-m are nonzero for k≤m and if k is even with k<2m-2n2, then Sk is even. To calculate Sk for negative indices, we observe
(34)F*(x)=xm+xm-n3+xm-n2+xm-n1+1∈Z[x],
a monic lift of the reciprocal polynomial of f(x) to the integers. Denote the kth power sum of the roots of F*(x) in some extension of the rational numbers by Tk. Then, clearly S-k=Tk for every positive integer k. We can apply Newton’s identity to Tk to see that Tk is equal to 0 for odd k<n2 and is even for even k<2n2. From the above discussion, we have
(35)R(F,F′)≡1+m-m2n2+12m2-12m2S2m-2n2-12n12S2n1-2n2-12n32T2n2-2n3-n1n3Sn1+n3-2n2(mod⁡8).First consider the case whenn1+n3>2n2.Theorem 6.
Suppose thatn2 is odd and n1+n3>2n2. Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.
Consider(1)
n3∣2n2:
(a)
n1=2n2,m≠3n1-2n2 and either m≡4(mod⁡8) or m-2n2≡0  (mod⁡8);(b)
n1≠2n2,m=3n1-2n2 and either m≡0  (mod⁡8) or m-2n2≡0  (mod⁡8);(c)
n1≠2n2,m≠3n1-2n2 and either m≡4  (mod⁡8) or m-2n2≡4  (mod⁡8);(2)
n3∤2n2:
(a)
n1=2n2,m=3n1-2n2, and m≡4  (mod⁡8);(b)
n1=2n2,m≠3n1-2n2, and either m≡0  (mod⁡8) or m-2n2≡4  (mod⁡8);(c)
n1≠2n2,m=3n1-2n2, and either m≡4  (mod⁡8) or m-2n2≡4  (mod⁡8);(d)
n1≠2n2,m≠3n1-2n2, and either m≡0  (mod⁡8) or m-2n2≡0  (mod⁡8).Proof.
We determine the unknown terms in (35). Clearly, Sn1+n3-2n2=0 from 0<n1+n3-2n2<m-n1. Let again
(36)A≔1+m-m2n2+12m2,B≔-12m2S2m-2n2-12n12S2n1-2n2,C≔-12n32T2n2-2n3.
It is easy to see that
(37)A≡{1(mod⁡8),m≡0or2(mod⁡8)5(mod⁡8),m≡4or6(mod⁡8).
We also see that S2n1-2n2 is equal to n1-m if m=3n1-2n2 and equal to 0, otherwise, since 2n1-2n2<m-n2. And, by Newton’s identity, we have
(38)S2m-2n2=-Sm+n1-2n2-Sm-n2-Sm+n3-2n2-Sm-2n2=S2n1-2n2+(m+n1-2n2)am+n1-2n2+(m+n3-2n2)am+n3-2n2+(m-2n2)am-2n2+(m-n2).
With reference to the nonzero coefficients of F(x), we obtain that S2m-2n2=S2n1-2n2+3(m-n2) if n1=2n2 and S2m-2n2=S2n1-2n2+(m-n2) otherwise. Now we can determine B modulo 8. If n1=2n2 and m=3n1-2n2, then S2n1-2n2=n1-m and thus S2m-2n2=2m+n1-3n2. Clearly n1≡2  (mod⁡4) and m≡0  (mod⁡4); hence B≡4  (mod⁡8). Consideration for the other cases is similar so we describe only the results: if n1=2n2,m≠3n1-2n2, then
(39)B≡{0(mod⁡8),m≡0(mod⁡4)2(mod⁡8),m≡2(mod⁡4),n2≡1(mod⁡4)-2(mod⁡8),m≡2(mod⁡4),n2≡-1(mod⁡4);
if n1≠2n2,  m=3n1-2n2, then
(40)B≡{4(mod⁡8),m≡0(mod⁡4)2(mod⁡8),m≡2(mod⁡4),n2≡1(mod⁡4)-2(mod⁡8),m≡2(mod⁡4),n2≡-1(mod⁡4);
and if n1≠2n2,m≠3n1-2n2, then
(41)B≡{0(mod⁡8),m≡0(mod⁡4)-2(mod⁡8),m≡2(mod⁡4),n2≡1(mod⁡4)2(mod⁡8),m≡2(mod⁡4),n2≡-1(mod⁡4).
Next, we compute C modulo 8. If we denote the coefficient of xm-k in F*(x) by bk, then
(42)T2n2-2n3=-T2n2-3n3-Tn2-2n3-(2n2-2n3)b2n2-2n3.
Since n2-2n3 is odd <n2 and 2n2-3n3 is even <2n2, Tn2-2n3=0 and T2n2-3n3 is even. It follows, therefore, that T2n2-2n3≡T2n2-3n3+(2n2-2n3)b2n2-2n3(mod⁡4). Repeating this process, we get T2n2-2n3≡∑i=2l‍(2n2-in3)b2n2-in3(mod⁡4), where ln3≤2n2<(l+1)n3. Thus, we obtain
(43)C≡{4(mod⁡8),n3∣2n20(mod⁡8),otherwise.
Now, Swan’s theorem is used to complete the proof.The remaining cases whenn1+n3=n2 or n1+n3<n2 gives the following theorems, whose proofs follow a similar way and, hence, are omitted.Theorem 7.
Suppose thatn2 is odd and n1+n3=2n2. Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.(1)
n3∣2n2, and either m≡4  (mod⁡8) or m-2n2≡0  (mod⁡8);(2)
n3∤2n2, and either m≡0(mod⁡8) or m-2n2≡4  (mod⁡8).Theorem 8.
Suppose thatn2 is odd and n1+n3<2n2. Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.
Consider(1)
n3∣2n2,  n3∣2n2-n1, and either m≡0  (mod⁡8) or m-2n2≡0  (mod⁡8);(2)
n3∣2n2,  n3∤2n2-n1, and either m≡4(mod⁡8) or m-2n2≡4  (mod⁡8);(3)
n3∤2n2,  n3∣2n2-n1,n3≡0  (mod⁡4), and either m≡0  (mod⁡8) or m-2n2≡0  (mod⁡8);(4)
n3∤2n2,  n3∣2n2-n1,n3≡2  (mod⁡4) and either m≡4  (mod⁡8) or m-2n2≡4  (mod⁡8);(5)
n3∤2n2,  n3∤2n2-n1, and either m≡0  (mod⁡8) or m-2n2≡0  (mod⁡8).Case 3 (n1 is odd).
Analogously, to Case2, we can write the resultant of F(x) and its derivative as follows:
(44)R(F,F′)=∏i=0m-1‍(mxim-n1+n2xin2-n1+n3xin3-n2+n1)≡1+mn1Sm-n1+12m2(Sm-n12-S2m-2n1)+n2n1Tn1-n2+12n22(Tn1-n22-T2n1-2n2)+n3n1Tn1-n3+12n32(Tn1-n32-T2n1-2n3)+mn2(Sm-n1·Tn1-n2-Sm+n2-2n1)+mn3(Sm-n1·Tn1-n3-Sm+n3-2n1)+n2n3(Tn1-n2·Tn1-n3-T2n1-n2-n3)(mod⁡8).
Straightforward calculations show that Sm-n1=n1-m and Sm+n2-2n1 and Sm+n3-2n1 are even. In this case, we have also that Tk is equal to 0 for odd k<n1 and is even for even k<2n1. It follows, therefore, that
(45)R(F,F′)≡1+m-m2n1+12m2-12m2S2m-2n1-12n22T2n1-2n2-12n32T2n1-2n3(mod⁡8).
Now, let
(46)E≔12n22T2n1-2n2+12n32T2n1-2n3.We present the following result for the reducibility off(x) in (10) depending on the value of E modulo 8.Theorem 9.
Suppose thatn1 is odd. Then, the pentanomial f(x) in (10) has an even number of irreducible factors over F2 if and only if one of the following conditions hold.(1)
m=2n1 and E≡4  (mod⁡8);(2)
m≠2n1,  E≡0  (mod⁡8), and either m≡0  (mod⁡8) or m-2n1≡0  (mod⁡8);(3)
m≠2n1,  E≡4  (mod⁡8), and either m≡4  (mod⁡8) or m-2n1≡4  (mod⁡8).Proof.
First, we computeS2m-2n1 in (45). By Newton’s identity, we get
(47)S2m-2n1=-Sm-n1-Sm-2n1+n2-Sm-2n1+n3-Sm-2n1-(2m-2n1)a2m-2n1.
Since m-2n1<m-2n1+n3<m-2n1+n2<m-n1, we have Sm-2n1+n2=Sm-2n1+n3=Sm-2n1=0 and S2m-2n1=m-n1-(2m-2n1)a2m-2n1; hence S2m-2n1 is equal to n1-m if m=2n1 and equal to m-n1 otherwise. A simple calculation shows that if m=2n1, then R(F,F′)≡3-E(mod⁡8). If m≠2n1, then
(48)R(F,F′)≡1+m+12m2-12m2n1-12m3-E(mod⁡8)
and, thus, we have
(49)R(F,F′)≡{1-E(mod⁡8),m≡0(mod⁡8)5-E(mod⁡8),m≡4(mod⁡8)7-E(mod⁡8),m-2n1≡0(mod⁡8)3-E(mod⁡8),m-2n1≡4(mod⁡8).
Now applying Swan’s theorem completes the proof.


## 4. Conclusion

We have determined the parity of the number of irreducible factors of a pentanomial (10) under the condition that 0<n3<n2<n1≤m/2 and exactly one of n1,n2,  and  n3 is odd. Our discussion is based on Swan’s theorem. If n1 is odd, we obtained only a result which depends on E modulo 8 instead of exponents of the terms of a given pentanomial. In this case, a complete characterization of the reducibility of the given pentanomial seems to be more difficult.



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*Source: 482837-2014-09-02.xml*