# On some permutation polynomials over finite fields

## Full text

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### OVER FINITE FIELDS

AMIR AKBARY AND QIANG WANG

Received 8 March 2005 and in revised form 5 May 2005

Letpbe prime,q=pm, andq1=7s. We completely describe the permutation behav-ior of the binomialP(x)=xr(1 +xes) (1e6) over a finite fieldF

q in terms of the sequence{an}defined by the recurrence relationan=an−1+ 2an−2−an−3(n≥3) with initial valuesa0=3,a1=1, anda2=5.

1. Introduction

LetFq be a finite field of q=pm elements with characteristic p. A polynomialP(x)

Fq[x] is called apermutation polynomialofFqifP(x) induces a bijective map fromFqto itself. In general, finding classes of permutation polynomials ofFqis a diﬃcult problem (see [3, Chapter 7] for a survey of some known classes). An important class of permuta-tion polynomials consists of permutapermuta-tion polynomials of the formP(x)=xrf(x(q1)/l), wherelis a positive divisor ofq−1 andf(x)Fq[x]. These polynomials were first stud-ied by Rogers and Dickson for the case f(x)=g(x)l, whereg(x)Fq[x] [3, Theorem 7.10]. A very general result regarding these polynomials is given in [8]. In recent years, several authors have considered the case that f(x) is a binomial (e.g., [2,9] and [1]).

Here we consider the binomialP(x)=xr+xuwithr < u. Lets=(ur,q1) andl= (q−1)/s. Then we can rewriteP(x) asP(x)=xr(1 +xes), wheres=(q1)/land (e,l)=1. IfP(x)=xr(1 +xes) is a permutation binomial ofF

q, thenP(x) has exactly one root inFq and thuslis odd. Whenl=3, 5, the permutation behavior ofP(x) was studied by Wang [9]. In the casel=5, the permutation binomialP(x) is determined in terms of the Lucas sequence{Ln}, where

Ln=

2 cosπ 5

n +

2 cos2π 5

n

. (1.1)

More precisely, it is proved that under certain conditions onr,s=(q−1)/5, ande, the binomialP(x)=xr(1 +xes) is a permutation binomial if and only ifL

s=2 inFp[9, The-orem 2].

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In this paper, we consider the casel=7 (see [1] for some results related to generall). Here we introduce a Lucas-type sequence{an}by

an=

2 cosπ 7

n +

2 cos2π 7

n +

2 cos3π

7 n

(1.2)

for integern≥0. It turns out that{an}∞n=0is an integer sequence satisfying the recurrence relation

an=an−1+ 2an−2−an−3 (1.3)

with initial valuesa0=3,a1=1, anda2=5 (seeLemma 2.1). This is the sequence A094648 in Sloane’s Encyclopedia [6]. Next we extend the domain of{an}∞n=0to include negative integers. For negative integer−n, we have

a−n=

4 cosπ 7cos 2π 7 n +

4 cosπ 7cos 3π 7 n + 4 cos2π

7 cos 3π

7 n

. (1.4)

Note that{an}∞n=−∞is an integer sequence, so we can consider this sequence as a se-quence inFp. Here we investigate the relation between this sequence inFp and permu-tation properties of binomialP(x)=xr(1 +xes) over a finite fieldF

q=Fpm. We have the

following Theorem.

Theorem1.1. Letq−1=7sand1≤e≤6. ThenP(x)=xr(1 +xes)is a permutation

bi-nomial ofFqif and only if(r,s)=1,2s≡1 (modp),2r+es≡0 (mod 7), and{an}satisfies one of the following:

(a)as=a−s=3inFp;

(b)a−cs−1= −1 +α,a−cs= −1−α, anda−cs+1=1inFp, wherecis the inverse ofs+ 2e5rmodulo 7andα2+α+ 2=0inF

p.

The sequence{an}is calleds-periodicoverFp ifan=an+ks inFp for integersk and n. Condition (a) in the above theorem is equivalent tos-periodicity ofan overFp (see Lemma 2.4). Equivalently we can say{an}iss-periodic overFpwhenever{an} = {a0n}in

Fp, where{a0

n}∞n=−∞is the unique sequence inFpdefined by the recursion (1.3) and initial valuesa0s−1=2,a0s =3, and a0s+1=1. Similarly condition (b) can be written as{an} =

{ac,α

n }inFp, where{anc,α}∞n=−∞is the unique sequence inFpdefined by the recursion (1.3) and initial valuesa−cs−1= −1 +α,a−cs= −1−α, anda−cs+1=1. SoTheorem 1.1states that under certain conditions onr,s=(q−1)/7, andethe binomialP(x)=xr(1 +xes) is a permutation binomial ofFpif and only if the Lucas-type sequence{an}is equal to{a0n} or{ac,α

n }inFp(for more explanation, seeExample 3.2).

It is clear that if the Legendre symbol (p7)= −1, then condition (b) in the above the-orem is never satisfied (the equation x2+x+ 2=0 does not have any solution inF

p). Moreover, in this case, we can show that condition (a) is always satisfied, and so we have the following.

Corollary1.2. Letq−1=7s,1≤e≤6, and letpbe a prime with(p7)= −1. ThenP(x)=

xr(1 +xes)is a permutation binomial ofF

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Theorem 1.1gives a complete characterization of permutation binomials of the form P(x)=xr(1 +xe(q−1)/7). Moreover, our theorem together with the above corollary can lead to an eﬃcient algorithm for constructing such permutation binomials. Note that{an}is a recursive sequence and therefore conditions (a) and (b) can be quickly verified and so by employing the above theorem it is easy to find new permutation binomials over certain

Fq. Also by an argument similar to the proof of [1, Corollary 1.3], we can show that under the conditions ofTheorem 1.1onq, there are exactly 3φ(q−1) permutation binomials P(x)=xr(1 +xe(q−1)/7) ofF

q. Here,φis the Euler totient function.

In the next section, we study certain properties of the sequence{an}that will be used in the proof of our theorem.Theorem 1.1andCorollary 1.2are proved inSection 3.

2. The sequence{an}

We first show that{an}appears in the closed expression for the lacunary sum of binomial coeﬃcients

S(2n, 7,a) :=

2n

k=0

k≡a(mod 7)

2n

k

. (2.1)

Lemma2.1. The sequence{an}∞n=0satisfies the recursionan=an−1+ 2an−2−an−3(n≥3), a0=3,a1=1,a2=5, and

S(2n, 7,a)=                       

22n+ 2a 2n

7 if2n−2a≡0 (mod 7),

22na 2n+1

7 if2n−2a≡1, 6 (mod 7), 22n+a

2n+1−a2n1

7 if2n−2a≡2, 5 (mod 7), 22na

2n+a2n1

7 if2n−2a≡3, 4 (mod 7).

(2.2)

Proof. Note that 2 cos (π/7),2 cos (2π/7), and 2 cos (3π/7) are the roots of the polyno-mialg(x)=x3x22x+ 1, soa

nsatisfies the given recursion. We know that

S(2n, 7,a)=22n

7 + 2 7

3

t=1

2 cosπt 7

2n cosπt

7 (2n−2a)

, (2.3)

(see [7, page 232, Lemma 1.3]). This together with (1.2) and (1.3) implies the result. Next we have a general formula for the productanam.

Lemma2.2. Letmandnbe integers andm≤n. Then

anam=am+n+ (1)m(a−man−m−an−2m). (2.4)

In particular,

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Proof. Letδ=2 cos (π/7),η= −2 cos (2π/7), and=2 cos (3π/7). We havean=δn+ηn+

nanda

−n=(−δη)n+ (−δ)n+ (−η)n. Considering these, a routine calculation implies

the result.

In the next two lemmas, we study the periodicity of{an}overFp.

Lemma2.3. Letp=2, 7be a prime. Then the sequence{an}∞n=−∞is7s-periodic overFp.

Proof. We know thatg(x)=x3x22x+ 1 is the characteristic polynomial of the recur-sion associated toan. Letδ,η, andbe the roots ofg(x) in a splitting fieldFofg(x) over

Fp. Since p=2, 7, we know thatanis 7s-periodic inFpif and only ifδ7s7s=7s=1 inF.

We can show thatg(x) is either irreducible inFp[x] or it splits inFp[x]. Now ifg(x) splits overFp, thenδp−1=ηp−1=p−1=1 inF

pand thereforeanhas period 7s=q−1. Ifp=7k+ 1 or 6, by [5, Theorem 7],g(x) splits overFp. Ifp=7k+ 2, 3, 4, or 5 andg(x) is irreducible overFp, then, by [3, Theorems 8.27 and 8.29],anis periodic inFpwith the least period dividing p31. Also sinceq1=pm10 (mod 7), in these cases, 3|m. Hence,anis periodic inFpwith the least period dividing 7s=q−1. We continue by describing a necessary and suﬃcient condition under which the se-quence{an}∞n=−∞will be a periodic sequence inFpwith the even periods.

Lemma2.4. Letp=2, 7be a prime and letsbe a fixed even positive integer. Then

{an}iss-periodic overFp⇐⇒as=a−s=3inFp. (2.6)

Proof. With the notation in the proof ofLemma 2.3, we know that{an}∞n=−∞iss-periodic if and only if diag(δ,η,)s=IinF. Here diag(δ,η,) is a diagonal matrix with entriesδ, η, and and I is the identity matrix. We know that a diagonal matrix is equal to the identity matrix if and only if (x−1)3 is the characteristic polynomial of the diagonal matrix. By employing this fact, together with the identitiesan=δn+ηn+nanda−n= (−δη)n+ (δ)n+ (η)ninF, we have

diag(δ,η,)s=IinF⇐⇒as=a−s=3 inFp. (2.7) The following two lemmas play important roles in the proof ofTheorem 1.1.

Lemma2.5. Let p=2, 7be a prime,s=(q−1)/7, and letc(1≤c≤6)be a fixed

inte-ger. If the sequence{an}∞n=−∞satisfiesacs+1=a2cs1−a2cs+1=a3cs−a3cs1=a4cs−a4cs1= a5cs1−a5cs+1=a6cs+1=1inFp, then

acs=a2cs=a4cs, a3cs=a5cs=a6cs (2.8)

inFp.

Proof. From the recurrence relation ofan, we geta2cs1−a2cs+1=2a2cs−a2cs+2.So, by the conditions of the lemma, we have

(A)a2 cs+1=1;

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We employ Lemmas2.2and2.3to deduce new identities from (A), (B), and (C). For simplicity of our exposition, we leta−(cs+1).

First of all (A) together withLemma 2.2implies

a2cs+2=1 + 2γ. (2.9)

From (2.9) and 2a2cs−a2cs+2=1, we have

a2cs=1 +γ. (2.10)

Next from (B), (2.9), (2.10),Lemma 2.2, andacs+1=1, we get

1=(2a2cs−a2cs+2)2

=4a22cs4a2csa2cs+2+a22cs+2

= −4(1 +γ)γ+a22cs+2

= −4(1 +γ)γ+a4cs+4+ 2a−(2cs+2)

= −4(1 +γ)γ+a4cs+4+ 2(γ2+ 2).

(2.11)

This implies

a4cs+4=2(1 +γ)25=2a22cs5. (2.12)

Note thata4cs−a4cs1=1 and the recurrence relation (1.3) imply

a4cs+2=a4cs+1+a4cs+ 1, (2.13)

and

a4cs+3=3a4cs+1+ 1. (2.14)

Now applying the recurrence relation a4cs+4=a4cs+3+ 2a4cs+2−a4cs+1 together with (2.13) and (2.14) to the left-hand side of (2.12) and applying Lemmas2.2and2.3to the right-hand side of (2.12) yield

a4cs+1=a5cs2. (2.15)

Finally, from (C), we have

a24cs2a4csa4cs1+a24cs1=1. (2.16)

Applying Lemmas2.2and2.3on this equality yields

acs+ 2a3cs2acs−12a3cs+2+acs−2=1. (2.17)

Now by employing the recurrence relation acs+1=acs+ 2acs−1−acs−2 in the previous identity andacs+1=1, we obtain

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Sincea3cs−a3cs1=1, from the recurrence relation (1.3), we have

a3cs+2=a3cs+1+a3cs+ 1. (2.19)

Applying this identity in (2.18) yields

acs=a3cs+1+ 2. (2.20)

Now we are ready to finish the proof. Note that by changing sto−s all the above equations remain true, so, by changing sto−s in (2.15) and applyingLemma 2.3, we have

a3cs+1=a2cs2. (2.21)

This together with (2.20) impliesacs=a2cs. Changingsto−sin this equality yields a6cs=a5cs. These identities together with Lemmas2.2and2.3imply that

acs=a2cs=a4cs, a3cs=a5cs=a6cs. (2.22)

Lemma2.6. Letp=2, 7be a prime,s=(q−1)/7, and letc(1≤c≤6)be a fixed integer. If

the sequence{an}∞n=−∞satisfies

a6cs1= −1 +α, a6cs= −1−α, a6cs+1=1, (2.23) whereαis a root of equationx2+x+ 2=0inF

p, then we haveacs=a2cs=a4cs=α,a3cs= a5cs=a6cs= −1−α,acs−1= −2−α,acs+1=1,a5cs1=12α, anda5cs+1= −2αinFp. Proof. From Lemmas2.2and2.3, we have the following six identities:

a26cs1=a5cs22acs+1, a6cs1a6cs=a5cs1−a1acs+1+acs+2, a6cs1a6cs+1=a5cs−a2acs+1+acs+3,

a2

6cs=a5cs+ 2acs, a6csa6cs+1=a5cs+1+acs−acs+1,

a26cs+1=a5cs+22acs−1.

(2.24)

Replacing the known values of the variables in the above identities, writinga5cs2 and a5cs+2in terms ofa5cs1,a5cs, anda5cs+1, and writingacs+2andacs+3in terms ofacs−1,acs, andacs+1yield

(1 +α)2=2a5cs1+a5cs−a5cs+12acs+1, 1−α2=a5cs1−acs−1+ 2acs,

1 +α=a5cs−acs−1+acs−2acs+1, (1 +α)2=a

5cs+ 2acs,

1−α=a5cs+1+acs−acs+1, 1= −a5cs1+ 2a5cs+a5cs+12acs−1.

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Solving this system of linear equations and noting thatα2+α+ 2=0 imply the desired values foracs−1,acs,acs+1,a5cs1,a5cs, anda5cs+1. By setting up two similar systems of linear equations, one can derive the desired values fora2cs,a3cs, anda4cs.

3. Permutation binomials and the sequence{an}

The main tool in the proof ofTheorem 1.1is the following well-known theorem of Her-mite [3, Theorem 7.4].

Theorem3.1 (Hermite’s criterion). P(x)is a permutation polynomial ofFqif and only if

(i)P(x)has exactly one root inFq;

(ii)for each integertwith1≤t≤q−2andt≡0 (modp), the reduction of[P(x)]t mod(xqx)has degree less than or equal toq2.

Finally, we are ready to prove the main result of this paper.

Proof ofTheorem 1.1. First we assume thatP(x) is a permutation binomial. Then p=2, since otherwiseP(0)=P(1)=0. Also, in this case, it is known that (r,s)=1 [8, Theorem 1.2] and 2s1 (modp) [4, Theorem 4.7]. Next we note that the coecient ofxq−1 in the expansion of [P(x)]ksisS(ks, 7,ke5r), so ifP(x) is a permutation binomial, then by Hermite’s criterionS(ks, 7,−ke5r)=0 inF

pfork=1,. . ., 6.

We next show that 2r+es≡0 (mod 7). Otherwise, 2r+es≡0 (mod 7) andLemma 2.1 yields that

S(ks, 7,−ke5r)=2ks+ 2aks

7 inFp (3.1)

fork=1,. . ., 6. From here ifP(x) is a permutation binomial, we have

as=a2s= ··· =a6s= −1

2 inFp. (3.2)

Using Lemmas 2.3 and 2.2, we have 1/4=a2

s=a2s+ 2a6s=3as= −3/2. Hence, (1/2) ((1/2) + 3)=0 inFpwhich is a contradiction since 7|(q−1). Hence, 2r+es≡0 (mod 7). It remains to show that ifP(x) is a permutation binomial, then either (a) or (b) holds. Letcbe the inverse ofs+ 2e5r modulo 7. Hermite’s criterion together withLemma 2.1 implies that

acs+1=1, a2cs1−a2cs+1=1, a3cs−a3cs1=1,

a4cs−a4cs1=1, a5cs1−a5cs+1=1, a6cs+1=1, (3.3)

inFp. So, byLemma 2.5, we have

acs=a2cs=a4cs, a3cs=a5cs=a6cs, (3.4)

inFp. From Lemmas2.2and2.3, we have

a2

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By subtracting these two equations and employing (3.4), we get

(acs−a6cs)(acs+a6cs+ 1)=0 inFp. (3.6)

Ifα=βinFp, then byLemma 2.2and (3.4) we havea7cs=acsa6cs−a6csa5cs+a4cs=a4cs. Since byLemma 2.3a7cs=a0=3 inFp, we havea4cs=3 inFp. This together with (3.4) andacs=a6csimplies condition (a).

Ifα=β, then from (3.6) we haveacs+a6cs+ 1=0. This together with (3.5) implies thatαandβare roots of the equationx2+x+ 2=0 inF

pand thereforeβ= −1−α. FromLemma 2.2, we have

acsacs+1=a2cs+1+a6csa1−a6cs+1. (3.7)

This together withacs=α,a6cs= −1−α, andacs+1=a6cs+1=1 implies thata2cs+1=2α+ 2. Note thata2cs1=1 +a2cs+1, and soa2cs1=2α+ 3 and thusa2cs+2=a2cs+1+ 2a2cs a2cs1=2α−1. Finally, by Lemma 2.2, we have a2cs+1=a2cs+22a6cs1 which implies a6cs1=α−1. Hence, in this case,ansatisfies condition (b).

Conversely we assume that the conditions inTheorem 1.1are satisfied and we show thatP(x) is a permutation binomial. First note that 2s1 (modp) follows thatpis odd. Hence, it is obvious thatP(x) has only one root inFq. Since, (r,s)=1, the possible coeﬃ -cient ofxq−1in the expansion of [P(x)]tcan only happen ift=ksfor somek=1,. . ., 6. So by Hermite’s criterion, it is suﬃcient to show thatS(ks, 7,−ke5r)=0 inF

pfork=1,. . ., 6. Now ifansatisfies condition (a), then byLemma 2.4aniss-periodic overFp. Using the initial values ofan, 2r+es≡0 (mod 7), andLemma 2.1, we haveS(ks, 7,−ke5r)=0 inF

p and thusP(x) is a permutation binomial overFq.

Next we assume thatansatisfies condition (b). Then, byLemma 2.6, we also have

acs=a2cs=a4cs, a3cs=a5cs=a6cs= −1−α,

acs−1= −2−α, acs+1=1, a5cs1=12α, a5cs+1= −2α. (3.8)

By using 2s=1,a

cs+1=a6cs+1=1, andLemma 2.1, we have

S(kcs, 7,−kce5r)=0 fork=1, 6. (3.9)

To demonstrateS(kcs, 7,−kce5r)=0 for otherk’s, it is sucient to show that

a2cs1−a2cs+1=1, a3cs−a3cs1=1,

a4cs−a4cs1=1, a5cs1−a5cs+1=1. (3.10)

From the values for a5cs1 and a5cs+1, it is clear that a5cs1−a5cs+1=1. Next note that by considering appropriate systems of linear equations as described in the proof of Lemma 2.6, we can deduce that

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Table 3.1

Type IV Type III Type II Type I

2731 4999 7309 874 651

3389 18 439 20 063 941 879 15 583 20 441 33 587 1 018 879 62 791 33 503 37 199 1 036 267 65 899 55 609 37 339 1 074 277

..

. ... ... ...

Soa2cs1−a2cs+1=a3cs−a3cs1=a4cs−a4cs1=1. These relations show thatS(ks, 7,

−ke5r)=0 inF

pfork=1,. . ., 6. Hence,P(x) is a permutation binomial ofFq. Next we prove that if (p7)= −1 then the sequence an is always s-periodic. That is, as=a−s=3.

Proof ofCorollary 1.2. Following the notation in the proof ofLemma 2.3, letbe a root of g(x)=x3x22x+ 1 in an extension ofFp. We need to prove thats=1. If p 6 (mod 7), then by [5, Theorem 7] we haveFp. Since (p−1, 7)=1, in this case,is a 7th power inFpand therefores=1 inFp. To prove the result forp≡3 or 5 (mod 7), first of all note thatg(x) is either irreducible inFp[x] or it splits inFp[x]. If it splits overFp, thenis a 7th power inFpand sos=1 inFp. Otherwise,g(x) splits overFp3. Now since

p≡1, 2 or 4 (mod 7), we have (p31, 7)=1, so is a 7th power inF

p3 and therefore

(p31)/7

=1 in Fp3. Also since 7|(q−1), we have 6|m. This and (p 31)/7

=1 inFp3 implies thats=1 inFq. Hence,{a

n}iss-periodic and so byLemma 2.4,as=a−s=3.

NowTheorem 1.1implies the result.

Example 3.2. An algorithm for finding permutation binomialsP(x)=xr(1 +xe(q−1)/7) of a given fieldFq can be easily implemented by usingTheorem 1.1andCorollary 1.2. Moreover, our theorem together with Lemmas2.4 and 2.6 implies that under certain conditions onr,s, andethe binomialxr(1 +xes) is a permutation polynomial overF

qif and only if the Lucas-type sequence{an}becomes one of the following four sequences overFp:

(I)a−s−1=2,a−s=3,a−s+1=1,as−1=2,as=3, andas+1=1;

(II)a−s−1= −1 +α,a−s= −1−α,a−s+1=1,as−1= −2−α,as=α, andas+1=1; (III)a−2s1= −1 +α,a−2s= −1−α,a−2s+1=1,a2s1= −2−α,a2s, anda2s+1=1; (IV)a−3s1= −1 +α,a−3s= −1−α,a−3s+1=1,a3s1= −2−α,a3s, anda3s+1=1. Note that the sequence (I) iss-periodic and in (II), (III), and (IV),αis a root of equa-tionx2+x+ 2=0 inFp.

Table 3.1gives some prime numbers pwith p≡1 (mod 7) and 2(p1)/71 (modp) whose corresponding sequence{an}overFpis in the form (I) (resp., (II), (III), (IV)).

Here p=2731 (resp., 4999, 7309, 874 651) is the smallest prime p≡1 (mod 7) with 2(p1)/71 (modp) whose corresponding sequence {a

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Table 3.2

p=2731 p=4999 p=7309 p=874651

an a−3s−1=1001 a−2s−1=760 a−s−1=3858 a−s−1=2

a−3s=1728 a−2s=4237 a−s=3449 a−s=3

a−3s+1=1 a−2s+1=1 a−s+1=1 a−s+1=1

a3s−1=1727 a2s−1=4236 as−1=3448 as−1=2

a3s=1002 a2s=761 as=3859 as=3

a3s+1=1 a2s+1=1 as+1=1 as+1=1

(r,e,s) (7, 1, 390) (5, 1, 714) (7, 1, 1044) (1, 1, 124 950) (23, 1, 390) (19, 1, 714) (13, 1, 1044) (11, 1, 124 950) (37, 1, 390) (23, 1, 714) (35, 1, 1044) (13, 1, 124 950) (49, 1, 390) (37, 1, 714) (41, 1, 1044) (19, 1, 124 950) (77, 1, 390) (47, 1, 714) (49, 1, 1044) (23, 1, 124 950)

..

. ... ... ...

Acknowledgments

Research of both authors was partially supported by NSERC.

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Amir Akbary: Department of Mathematics and Computer Science, University of Lethbridge, 4401 University Drive West, Lethbridge, AB, Canada T1K 3M4