On the Primitivity of Trinomials over Small Finite Fields

On the Primitivity of Trinomials over Small Finite Fields

Li Yujuan

, Zhao Jinhua

, Wang Huaifu

, Ma Jing

. Science and Technology on Information Assurance Laboratory, Beijing, 100072, P.R. China

Abstract:In this paper, we explore the primitivity of trinomials over small finite fields. We extend the results of the primitivity of trinomialsxn+ax+boverF4[1] to the general formxn+axk+b. We prove that for

givennandk, one of all the trinomialsxn+axk+bwithbbeing the primitive element ofF4anda+b6= 1

is primitive overF4if and only if all the others are primitive overF4. And we can deduce that if we find one

primitive trinomial overF4, in fact there are at least four primitive trinomials with the same degree. We give the

necessary conditions if there exist primitive trinomials overF4. We study the trinomials with degreesn= 4m+ 1

andn= 21·4m_{+ 29}

, wheremis a positive integer. For these two cases, we prove that the trinomialsxn_+ax+b with degreesn= 4m_{+ 1andn= 21·4}m_{+ 29are always reducible ifm >1. If some results are obviously} true overF3, we also give it.

Keywords:finite fields¶primitive polynomials¶trinomials

1

Introduction

As usual, denoteFqthe finite field ofqelements and letFq[x]be the ring of polynomials in one variablexwith

coefficients inFq. Trinomials inFq[x]are polynomials of the formxn+axk+b(n > k >0, ab6= 0). Irreducible

and primitive trinomials have many important applications in the theory of finite fields, cryptography, and coding theory [2-4]. Hence, there are many results on the factorizations of trinomials and existence or non-existence of irreducible or primitive trinomials. For detail results one can see [5] [6] [7] [8] . However, these results on trinomials are mainly related to binary field and many basic questions concerning trinomials remain unanswered. For example, to this day no one has proved that there are infinite primitive trinomials over finite fieldFq.

In this paper, we mainly explore the primitivity of trinomialsxn_+axk_{+bover finite field}

F4. First we remark

that we only consider the trinomials of odd degrees. This is because the only primitive trinomials of even degree overF4are of the formx2+ax+b[6] . In next section, we extend the results of the primitivity of trinomials

xn+ax+boverF4[1] to the general formxn+axk+b. As a consequence, We prove that for givennandk,

one of all the trinomialsxn+axk+bwithbbeing the primitive element ofF4anda+b6= 1is primitive overF4

if and only if all the others are primitive overF4. And we can deduce that if we find one primitive trinomial over

F4, in fact there are at least four trinomials with the same degree. We give a table of necessary conditions for the

existence of primitive trinomials and other interesting results. In section 3, we discuss the primitivity of trinomials with special degrees overF4and if some results are obviously true overF3, we also give it.

2

The general form

In [1], we have given some results on the primitivity of trinomials of the special formxn_+ax+bover

F4. In this

section, we extend these results to the general formxn_+axk_{+b. These new results are mainly included in the}

following theorem 1 and theorem 2. To prove theorem 1 and theorem 2, we first give some lemmas.

Lemma 1 [3]. Iff(x) ∈ Fq[x]is a polynomial of positive degree with f(0) 6= 0, and ifris a prime not

dividingq, thenord(f(xr_{)) =rord(f(x)).}

Lemma 2. Iff(x)∈Fq[x]is a polynomial of positive degree withf(0)6= 0andf(x)has no multiple roots,

then for eacha∈F∗q,ord(f(x))dividesord(a)·ord(f(ax)).

Proof. Letβbe a root off(x), thena−1βis a root off(ax)andord(β) =ord(a·a−1β). It is obvious that

ord(a·a−1β)can divideord(a)·ord(a−1β), soord(β)can divideord(a)·ord(a−1β). Sincef(0)6= 0andf(x)

has no multiple roots, it is well known that the order off(x)is equal to the least common multiple of the orders of all its roots, hence according to the basic knowledge of the least common multiple, we haveord(f(x))can divide

ord(a)·ord(f(ax)).

Lemma 3. Letωbe a primitive element ofF4. Ifk≡0 mod 3orn≡0 mod 3, then for anya∈F∗4, the

trinomialxn_+axk_{+ωcan not be primitive over}

F4.

Proof. Letg(x) = xn+axk+ω. Ifn ≡0 mod 3andk ≡ 1 mod 3ork ≡2 mod 3, one can easily check thatg(x)has at least one root inF4. So it is reducible overF4. And of course not primitive.

Ifn≡0 mod 3andk≡0 mod 3, theng(x) = (x3₎n

3 +a(x3) k

3 +ω.According to lemma 1, we have

ord(g(x)) = 3ord(xn3 _+axk3 _+ω) ≤ 3(4n3 −1)

< 4n−1.

Sog(x)can not be primitive overF4.

Ifk≡0 mod 3andn≡1 mod 3, sincea∈F∗4,a= 1, ωorω2. Fora= 1anda=ω2, one can check that

for each caseg(x) =xn_+axk_{+ωhas a root in}

F4. Fora=ω, note thatg(ωx) =ω(xn+xk+1),g(0)6= 0andn

is odd, sog(x)has no multiple roots, then by lemma 2,ord(g(x))can divideord(ω)·ord(g(ωx)), which is equal to3ord(g(ωx)). Sinceω−1g(ωx) =xn+xk+ 1∈F2[x], we haveord(g(ωx)) =ord(ω−1g(ωx))≤2n−1, thenord(g(x))≤3ord(g(ωx))≤3(2n−1)<4n−1. Sog(x)can not be primitive overF4too.

If k ≡ 0 mod 3andn ≡ 2 mod 3, for a = 1 anda = ω2, one can check that for each caseg(x) =

xn+axk +ωhas a root inF4. Fora= ω, the proof is similar to the casek ≡0 mod 3,n≡1 mod 3and

a=ω, we omit it.

Corollary 1.Letωbe a primitive element ofF4.

1. Ifxn_+ωxk_{+ωis primitive, thenn≡1 mod 3,k≡2 mod 3orn≡2 mod 3,k≡1 mod 3.}

2. Ifxn_+xk_+ω−1_{is primitive, thenn≡1 mod 3,k≡2 mod 3orn≡2 mod 3,k≡1 mod 3.} Proof. LetT1(x) =xn+ωxk+ωandT2(x) =xn+xk+ω−1. Suppose thatT1(x)is primitive. Ifk≡0

mod 3orn≡0 mod 3, according to lemma 3,T1(x)can not be primitive over_F4. Ifn≡1 mod 3,k ≡1

mod 3and ifn≡2 mod 3,k≡2 mod 3, one can check thatT1(x)has a root inF4for each case. So ifT1(x)

is primitive, thenn≡1 mod 3,k≡2 mod 3orn≡2 mod 3,k≡1 mod 3.

Ifk ≡0 mod 3orn ≡0 mod 3, the same according to lemma 3,T2(x)can not be primitive overF4. If

n≡1 mod 3,k≡1 mod 3, thenT2(ω−1_{x) =ω}−1_(xn_+xk_{+ 1). So according to lemma 2,ord(T2(x))can}

divideord(ω−1)·ord(T2(ω−1x)), which is equal to3ord(T2(ω−1x)). SinceωT2(ω−1x) =xn+xk+ 1∈F2[x], we haveord(T2(ω−1x)) = ord(ωg(ω−1x))≤ 2n−1, thenord(T2(x)) ≤ 3ord(T2(ω−1x))≤ 3(2n−1) <

4n−1. Ifn≡2 mod 3,k≡2 mod 3, the proof is the same as the casen≡1 mod 3,k≡1 mod 3. So if

T2(x)is primitive, then alson≡1 mod 3,k≡2 mod 3orn≡2 mod 3,k≡1 mod 3. This completes the proof.

Letωbe a primitive element ofF4. In [1] we have proved that the trinomials of the special formxn+ωx+ω

was primitive overF4if and only ifxn+x+ω−1was primitive overF4. In fact, it is also true for the trinomials

of the general form.

Theorem 1.The trinomialxn_+ωxk_{+ωis primitive over}

F4if and only ifxn+xk+ω−1is primitive over

F4.

Proof.LetT1(x) =xn_+ωxk_{+ωandT2(x) =x}n_+xk_+ω−1_{. SinceT1(x)is primitive, then by corollary}

1,n≡1 mod 3,k≡2 mod 3orn≡2 mod 3,k≡1 mod 3. Letξbe a root ofT1(x), then

ω=ξ·ξ4· · ·ξ4n−1=ξ4

n₋₁

3 . (1)

Ifn≡1 mod 3andk≡2 mod 3, then

T2(ωξ) = (ωξ)n+ (ωξ)k+ω−1=ω(ξn+ωξk+ω) = 0.

Hence,ωξis a root ofxn_+xk_+ω−1_{. By (1), we haveωξ=ξ}1+4n₃−1_{. Note that}

1 + 4

n₋₁

3 = 1 +n+

n−2 X

k=0

C_nk3n−k−1.

So 3 is not a divisor of4n−1and1 +4n₃−1 ifn ≡ 1 mod 3. If pis a prime divisor of4n−1 andp 6= 3, obviously, it can not be a divisor of1 +4n₃−1. So4n_{−1and1 +} 4n₋₁

3 are relatively prime. Then the order of

ωξis also4n−1. Thus,ωξis a primitive element ofF4n. Since the degree ofxn+xk+ω−1isn, then it is the

minimal polynomial ofωξ, hencexn+xk+ω−1is primitive overF4. Ifn≡2 mod 3andk≡1 mod 3, the

proof is similar, we omit it.

Conversely, suppose thatT2(x)is primitive, then by corollary 1, we also have thatn ≡ 1 mod 3,k ≡ 2

mod 3orn ≡2 mod 3,k ≡1 mod 3. And the proof of the left is similar to the proof of necessity, we omit

Lemma 4 [3]. The monic polynomialf(x) ∈ Fq[x]of degreen > 1 is a primitive polynomial overFq if

and only if(−1)nf(0)is a primitive element ofFq and the least positive integerrfor whichxris congruent

modf(x)to some element ofFq isr = q

n₋₁

q−1. In case f(x)is primitive overFq, we havex

r _{≡ (−1)}n_f(0)

modf(x).

By lemma 4, when we check the primitivity of all trinomialsxn_+axk_+bover

F4. We only need to consider

the trinomials withb being the primitive element ofF4. Letω be a primitive element ofF4. Then b = ω or

b = ω2_{. Ifb =ω,a =ω}2_{orb = ω}2_{, a =ω, note thatω}2_{+ω+ 1 = 0, then 1 is the root ofx}n_+axk_+b.

Hence ifxn_+axk_{+bis primitive over}

F4, thena=b =ωora=b=ω2ora= 1, b=ωora= 1, b=ω2,

i.e., only the following four casesg1(x) = xn+ωxk +ω, g2(x) = xn +ω2xk +ω2, g3(x) = xn +xk+ω

andg4(x) =xn+xk +ω2 maybe primitive overF4. By theorem 1,g1(x)is primitive overF4if and only if

g4(x)is primitive over_F4 andg2(x)is primitive overF4 if and only ifg3(x)is primitive overF4. Note that

g2

1(x) =g2(x2), then the squares of all the roots ofg1(x)are just all the roots ofg2(x)and the converse is true.

Sog1(x)is primitive overF4if and only ifg2(x)is primitive overF4. The above discussion means that for given

nandk, one of all the trinomialsxn_+axk_{+bwithbbeing the primitive element of}

F4anda+b6= 1is primitive

overF4if and only if all the others are primitive overF4. And we can deduce that if we find one primitive trinomial

overF4, in fact there are at least four trinomials with the same degree.

Lemma 5 [6]. Suppose[K : F2]is even. Only two types of odd-degree trinomials have an even number of factors, namely,

1.g(x) =xn+axk+b,2|k|2n, ift2+t+a2nkb2− 2n

k has no roots in K.

2.g(x) =xn_+axk_{+b, n−k|n, ift}2_+t+a2n

kb−2has no roots in K.

Lemma 6.Letωbe a primitive element ofF4and letT1(x) =xn+ωxk+ω. Ifn >2, thenT1(x)is reducible

overF4for the cases in table 1.

Table1

k mod 15 n mod 15 k mod 15 n mod 15

1 2,8,14 2 1,4,13

4 2,8,11 8 1,4,7

7 8,11,14 11 4,7,13 13 2,11,14 14 1,7,13

Proof. For the casek mod 15 = 1andn mod 15 = 2, letβbe a root ofx2_{+ωx+ω. Thenβ}15 _{= 1and}

one can check thatβ is also the root ofT1(x). Sincex2+ωx+ω is irreducible overF4andn >2, we have

x2+ωx+ωis a factor ofT1(x). The proofs of other cases are similar, we omit them and list an irreducible factor of degree 2 for each case in the following table 2.

Table2

k mod 15 n mod 15 a factor ofT1(x) k mod 15 n mod 15 a factor ofT1(x)

2 x2_{+ωx+ω 1 x}2_+ω2_{x+ 1}

1 8 x2_{+ωx+ 1 2 4 x}2_+ω2_x+ω2

14 x2_+x+ω2 _{13 x}2_+x+ω

2 x2+ωx+ 1 1 x2+ω2x+ω2

4 8 x2+ωx+ω 8 4 x2+ω2x+ 1

11 x2+x+ω2 7 x2+x+ω

8 x2_+ω2_x+ω2 _{4 x}2_+ωx+ω

7 11 x2_+ω2_{x+ 1 11 7 x}2_+x+ω2

14 x2_{+x+ω 13 x}2_{+ωx+ 1}

2 x2+ω2x+ω2 1 x2+ωx+ω

13 11 x2_{+x+ω 14 7 x}2_{+ωx+ 1}

14 x2_+ω2_{x+ 1 13 x}2_+x+ω2

Now we can prove the following further results.

Theorem 2.If trinomialxn_+axk_{+bwithbbeing the primitive element of}

F4is primitive overF4andn >2,

thenn, ksatisfy the conditions in the table below.

Proof.By theorem 1 and the arguments after lemma 4, we only need to suppose that the trinomialxn+ωxk+ω

Table3

k mod 15 1 2 4 5 7 8 10 11 13 14

n mod 15 5,11 7,10 5,14 1,4,7,13 2,5 10,13 2,8,11,14 1,10 5,8 4,10

Fork≡1 mod 3andn≡2 mod 3, note thatk≡1 mod 3is equivalent tok≡1,4,7,10,13 mod 15.

Fork≡1,4,7,13 mod 15, sincen >2, then by lemma 6, for the corresponding values ofn mod 15in table 1,

the trinomialxn_+ωxk_{+ωis reducible over}

F4. Checking all possible values ofn mod 15and make suren≡2

mod 3for eachk≡1,4,7,13 mod 15, only for the values ofn mod 15in table 3, we can not decide whether

xn+ωxk+ωis primitive or not and for all other cases,xn+ωxk+ωcan not be primitive. Fork≡10 mod 15,

ifn≡5 mod 15, then 5 can dividenandk, by lemma 1,ord(xn+ωxk+ω) = 5ord(xn5+ωx

k

5+ω)<4n−1.

Soxn +ωxk+ω is not primitive overF4. Checking all possible values ofn mod 15and make suren ≡2

mod 3,xn+ωxk+ωcan not be primitive except forn≡2,8,11,14 mod 15. Fork≡2 mod 3andn≡1

mod 3, the proof is similar to the casek≡1 mod 3andn≡2 mod 3. We omit the proof.

Following from theorem 2, we can have

Corollary 2.For anyn(n >2)there is no primitive trinomialsxn_+ax2_+bover

F4.

Proof. Letωbe a primitive element of_F4. By theorem 1 and the arguments after lemma 4, we only need to

consider the casea= b = ω. According to corollary 1,xn_+ωx2_{+ωcan not be primitive over}

F4ifn ≡0

mod 3andn≡2 mod 3. In fact, one can directly check thatxn_+ax2_{+ωhas a root in}

F4. Ifn≡1 mod 3,

since[F4:F2] = 2,t2+t+ω2n2 ω2−2n2 =t2+t+ω2has no roots in_F4. Then by lemma 5, we havexn+ωx2+ω

is reducible overF4.

3

The special form

In this section, we go on to consider the primitivity of trinomials of the special formxn_{+ax+b. By theorem 2}

we know that there is no primitive trinomials overF4except forn≡5 mod 15andn≡11 mod 15. Hence we

consider the primitivity of trinomialxn+ax+bwhenn≡5 mod 15andn≡11 mod 15. We mainly study the trinomials with degreesn= 4m+ 1andn= 21·4m+ 29, wheremis a positive integer. It is easy to check that ifn= 4m+ 1, thenn≡5 mod 15formis odd and ifn= 21·4m+ 29, thenn≡5 mod 15formis even. For these two cases, we prove that the trinomialsxn_{+ax+bwith degreesn= 4}m_{+ 1andn= 21·4}m_{+ 29are}

always reducible ifm >1. If some results are obviously true overF3, we also give them. Theorem 3.Letmbe a positive integer. Suppose thata, b∈F∗qand

(λi, λi+1) = (1,−a)·

0 −b

1 −a

i

, i= 0,1,· · ·, q−2. (2)

Ifλi6= 0for1≤i≤q−1andλq−2b+λq−1a= 0, then all irreducible factors ofxq m₊₁

+ax+bhave degrees dividing(q+ 1)m, and therefore, periods dividingq(q+1)m_−1.

Proof.Letϕ(x) =xqm+1_+ax+band

Φ(x) =λq−1ϕ(x) +

q−2 X

i=0

λixq

(q−2−i)m_+···+qm₊₁

ϕq(q−1−i)m(x). (3)

By (2), we have that

(λi+1, λi+2) = (λi, λi+1)·

0 −b

1 −a

, i= 0,1,· · ·, q−3.

Thenλi+2+λib+λi+1a= 0,0≤i≤q−3. So by simple calculation and the condition thatλi6= 0,1≤i≤

q−1, λq−2b+λq−1a= 0, we can deduce that

Φ(x) =xqqm+···+qm+1+λq−1b.

Note that ϕ(x) can divide Φ(x). So ϕ(x) can divide Φ(x) = xqqm_+···+qm₊₁

+λq−1b, which can divide

x(q−1)(qqm_+···+qm₊₁₎

−1.Notice that for any positive integerm,q−1dividesqm_{−1, and (q}m_−1)(qqm₊ · · ·+qm_{+ 1) = q}(q+1)m_{−1. So(q−1)(q}qm_{+· · ·+q}m_{+ 1)is a divisor ofq}(q+1)m_{−1. Soϕ(x)divides}

xq(q+1)m−1_{−1. Hence all irreducible factors ofϕ(x)have degrees dividing(q+ 1)m, and therefore, periods}

Corollary 3. Letq = 3,4and letm be a positive integer. Suppose that a, b ∈ F∗q andb 6= a2, then all

irreducible factors ofxqm+1_{+ax+bhave degrees dividing(q+ 1)m.}

Proof. For the caseq = 3, letg(x) =x3m+1_{+ax+b. According to the condition thatb 6=a}2_{and since}

F3={0,1,−1}, we haveb=−1, a= 1orb=−1, a=−1. Letλ0= 1, λ1=−a. Thenλ2=a2−baccording

to theorem 3 and one can check thatλ1b+λ2a= 0whateverb=−1, a= 1orb=−1, a=−1. So by theorem

3, all irreducible factors ofx3m+1+ax+bhave degrees dividing4m.

For the caseq = 4, the proof is similar. Let ωbe a primitive element ofF4. ThenF4 ={0,1, ω, ω2}and 1 +ω+ω2 = 0. Sinceb6=a2, ifb= 1, thena6= 1, i.e.,a=ωora=ω2, ifb =ω, thena= 1ora=ω, if

b =ω2_{, thena= 1ora =ω}2_{. Letλ}

0 = 1, λ1 =a. Thenλ2 =a2−b, λ3 = 1. According to theorem 3 and

one can check that forb = 1, a=ω ora=ω2_{,b =ω, a= 1ora= ωandb =ω}2_{, a= 1ora=ω}2_{, we all}

have thata2_b+b2_{+a= 0and this is equivalent toλ}

2b+λ3a= 0. Hence by theorem 3, all irreducible factors

ofx4m₊₁

+ax+bhave degrees dividing5m.

By corollary 3, we can deduce that trinomialxqm+1_{+ax+bis reducible ifm >1. Before we give corollary}

4, we first need a lemma.

Lemma 7 [2]. Let_Fq be a finite field and letmbe a positive integer. Then the degree of every irreducible

factor ofxqm+1_{+x+ 1over}

Fqdivides3m.

Corollary 4.Suppose thatq= 3,4andmis a positive integer. Then all irreducible factors ofxqm+1_+ax+b

have degrees dividing(q+ 1)mor dividing3m.

Proof. For the caseq = 3, letg(x) = x3m+1+ax+b. According to theorem 3, all irreducible factors of

g(x) = x3m+1 +ax+bdividing4mwhenb = −1. Ifb = 1, a = 1, then by lemma 7, the degree of every irreducible factor ofxqm+1 +x+ 1over Fq divides3m. Ifb = 1, a = −1, then g(x) = x3

m₊₁

−x+ 1. Letg1(x) = g(−x), theng1(x) =x3m+1+x+ 1, it is well known that the transformation fromg(x)tog1(x)

preserves degrees of factors. So the degree of every irreducible factor ofg(x)is the same as the degree of some irreducible factor ofg1(x), hence divides3m.

For the caseq = 4, leth(x) =x4m₊₁

+ax+b. According to theorem 3, all irreducible factors ofh(x) =

x4m₊₁

+ax+bdividing5mwhenb 6=a2_{. Ifb =a}2_{, the first case ifb =a = 1, then by lemma 7, we know}

that the degree of every irreducible factor ofxqm₊₁

+x+ 1overFq divides3m. Letωbe the primitive element

of _Fq. The second case ifa = ω, then b = ω2 andh(x) = x4

m₊₁

+ωx+ω2_{. Let h}

1(x) = h(ωx), then

h1(x) =ω(x4m+1_{+x+ 1), of course the transformation fromh(x)toh1(x)also preserves degrees of factors. So}

the degree of every irreducible factor ofh(x)is the same as the degree of some irreducible factor ofh1(x), hence divides3m. The third case ifa=ω2_{, thenb=ωandh(x) =x}4m+1_+ω2_{x+ω. The proof of this case is similar}

to the second case, we omit it.

Theorem 4.The trinomialx21·4m+29_{+ax+balways has a root in}

F43 for any positive integerm.

Proof. First one can verify thatx50+ax+balways has a root inF43. Form≡0 mod 3, ifβis a root of

x50+ax+binF43, then it is also the root ofx21·4 m₊₂₉

+ax+b. Form≡1 mod 3andm≡1 mod 3, note

that21·4 + 29 = 63 + 50and21·16 + 29 = 63·5 + 50, so the roots ofx50+ax+binF43 are also the roots of

x21·4m+29+ax+b. This completes the proof.

Reference

[1] Li Yujuan, Wang Huaifu,Zhao Jinhua. On the primitivity of some trinomials over finite fields. Advances in mathemat-ics(China), accepted, 2013.

[2] S.W.Golomb. Shift register sequence. Revised edition, Aegean Park Press,1982.

[3] Lidl R., Neiderreiter, Finite Fields, Cambridge University Press, Cambridge, 1997.

[4] Savas,E., Koc,C.K.: Finite field arithmetic for cryptography. IEEE Circuits and Systems Magazine, 10(2), 40-56, 2010.

[5] Richard G. Swan . Factorization of polynomials over Finite Fields. Pacific Journal of Mathematics 12(2), 1099-1106, 1962.

[6] Uzi Vishne . Factorization of Trinomials over Galois Fields of Characteristic 2. Finite Fields and Their Applications 3(4), 370-377, 1997.

[7] J. von zur Gathen, Irreducible trinomials over Finite Fields, Mathematics of Computation, 72, 1987-2000, 2003.

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