Research Article A Class of New Permutation Polynomials ...

Research ArticleA Class of New Permutation Polynomials over F2n

Qian Liu 12 Ximeng Liu 12 and Jian Zou 12

1College of Computer and Data Science Fuzhou University Fuzhou 350116 China2Key Laboratory of Information Security of Network Systems Fuzhou University Fuzhou 350116 China

Correspondence should be addressed to Qian Liu lqmovafoxmailcom

Received 14 August 2021 Accepted 15 October 2021 Published 15 November 2021

Academic Editor Li Guo

Copyright copy 2021Qian Liu et al)is is an open access article distributed under theCreative CommonsAttribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

In this paper according to the known results of some normalized permutation polynomials with degree 5 over F2n we determinesufficient and necessary conditions on the coefficients (b1 b2) isin F22n such that f(x) x3x2 + b1x

2x + b2x permutes F2n Meanwhile we obtain a class of complete permutation binomials over F2n

1 Introduction

Denote Fq as a finite field with q elements where q is a primepower then Flowastq is its multiplicative group)e bijectivity of theassociated polynomialmapping(s)f x⟼f(x) from Fq intoitself (and f(x) + x) makes a polynomial f(x) isin Fq[x] as a(complete) permutation polynomial [1] Research interests in(complete) permutation polynomials over finite fields havebeen aroused due to their widespread applications in cryp-tography [2 3] combinatorial designs [4] design theory [1 5]coding theory [6] and other areas of mathematics and engi-neering [1 7] Readers could refer to [1 8] for a comprehensivesurvey about (complete) permutation polynomials So it isimportant to realize that there is significance in discoveringnew methods to construct permutation polynomials Readerscan find recent progress in [9 10]

Few-term permutation polynomials especially bino-mials and trinomials have not only simple algebraic formbut excellent properties )e following form over F2n fromNiho exponents has drawn much attention

f(x) xr 1 + b1x

s 2mminus 1( )+ b2x

t 2mminus 1( )1113872 1113873 (1)

where n 2m r s t isin Z b1 b2 isin F2n Note that (s t) can beviewed as modulo 2m + 1 It is a hard problem to findnecessary and sufficient conditions on (b1 b2) for (1) being apermutation polynomial over F2n with given (r s t) In mostknown cases the coefficients are assumed to be trivial suchas (b1 b2) (1 1) some pairs (r s t) in (1) were made in

[10ndash15] For (r s t) (1 1 2) Hou [16] completely char-acterized all non-trivial (b1 b2) over Fq2 through the Her-mite criterion After that Tu et al [17] studied the case overFq2 where (r s t) (1 2m 2) by solving low-degree equa-tions with variable in the unit circle )e latter only obtainedthe sufficient conditions but conjectured their necessitybased on numerical experiments which were then proved byBartoli [18] and Hou [19] respectively using algebraiccurves over finite fields and the HassendashWeil bound In 2018Tu and Zeng [20] determined all (b1 b2) with (r s t)

(1 2mminus 1 2nminus 1) and gave sufficient conditions for (r s t)

(1 3 middot 22mminus 2 22mminus 2) over F2n the necessity of which was laterproved by Hou [21] using a similar method described in[19] Recently Zheng et al [22] claimed sufficient conditionsfor both (r s t) (1 1 (2k(2mk minus 1)((2k minus 1)(2m minus 1))))

and (r s t) (1 (2k(2k minus 1)) (minus1(2k minus 1))) over F2n Furthermore they conjectured the necessity of the formerbased on numerical experiments Very recently by makinguse of the similar approach in [18] Bartoli and Timpanella[23] provided necessary conditions for(r s t) (n + m m n) over finite fields with characteristic 2

To our knowledge only above seven different pairs(r s t) had been characterized completely However theexponents of f(x) in (1) are also Niho exponents )ismotivates us to explore new permutation polynomials withgeneral coefficients (b1 b2) from non-Niho exponents witheven characteristics By transforming the problem intostudying some normalized permutation polynomials with

HindawiJournal of MathematicsVolume 2021 Article ID 5872429 7 pageshttpsdoiorg10115520215872429

degree 5 over even characteristics we determine the coef-ficients (b1 b2) for f(x) x3x2 + b1x

2x + b2x being apermutation over F2n

)e rest of this paper is arranged as follows Somenotations and useful results are presented in Section 2 InSection 3 the sufficient and necessary conditions are shownto determine the coefficients of a class of permutationpolynomials over F2n Finally Section 4 provides someconcluding remarks

2 Preliminaries

Let m and n be two positive integers with m|n and F2n be afinite field with 2n elements the trace function from F2n toF2m is denoted by Trn

m(middot) where

Trnm(x) x + x

2m

+ x22m

+ middot middot middot + x2(n(mminus1))m

(2)

If the usual complex conjugation of any x isin F22m isdefined as x x2m the following relations hold

(i) For all x isin F22m x + x isin F2m xx isin F2m (ii) For all x y isin F22m x + y x + y xy xy

U2m+1 denotes the unit circle of F22m as

U2m+1 x isin F22m x2m+1

xx 11113966 1113967 (3)

Lemma 1 (see [24]) f(x) isin Fq[x] with f(0) 0 is apermutation polynomial over Fq iff f x⟼f(x) from Flowastqinto itself is a bijection

Hereinafter we claim that f(x) is a permutationpolynomial over Flowastq upon the bijectivity of f(x) from Flowastq toFlowastq which is the only case we consider

Lemma 2 (see [1]) -e irreducibility of f(x) isin Fq[x] withdegree n remains over Fpm iff gcd(m n) 1

Definition 1 (see [25]) A permutation polynomialf(x) isin Fq[x] of degree n is said to be normalized if thefollowing properties hold

(i) f(x) is monic and the value of f(x) at 0 is equal to 0(ii) )e coefficient of xnminus 1 equals 0 if p∤n where p is the

characteristic of Fq

Remark 1 Let b c isin Fq and a isin Flowastq for any permutationpolynomial f(x) isin Fq[x] there exists a unique normalizedform provided by g(x) af(x + b) + c

Theorem 1 (Hermitersquos criterion) (see [1]) Let p be thecharacteristic of Fq -us f(x) isin Fq[x] is a permutationpolynomial over Fq iff

(i) f(x) has exactly one solution over Fq(ii) forallt isin Z where 1le tle q minus 2 tne 0(modp) the degree

of the reduction of f(x)t(modxq minus x) is no greaterthan q minus 2

From Hermitersquos criterion the characterization of allnormalized permutation polynomials with degree no greaterthan 5 in Fq is due to ([1] Section 72) and they are listed inTable 1

3 A Class of Permutation Polynomials over F2n

In this section we consider the coefficients b1 b2 isin F2n suchthat the polynomial f(x) x3x2 + b1x

2x + b2x permutesover F2n

)e main results in this paper are given in the followingtheorems

Theorem 2 For two positive integers m and n with n 2mlet b1 b2 isin F2n Define g(z) z5 + z3(b1b1 + b2 + b2)

+z2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2]+

z[(b1 + b1)4 + (b1 + b1)

2(b1b1 + b2 + b2) + b2b2] isin F2m [z]-en the polynomial

f(x) x3x2

+ b1x2x + b2x (4)

permutes F2n iff one of the following two cases holds

(1) If m equiv 2(mod4) g(x) x5 permutes F2m (2) If m is odd g(x) x5 + ax3 + 5minus 1a2x permutes F2m

Proof Based on Lemma 1 in order to prove that f(x) is apermutation polynomial in Flowast2n we only consider that for anyc isin Flowast2n the equation

x3x2

+ b1x2x + b2x c (5)

has exactly one root in Flowast2n Let x λy where λ isin U2m+1 and y isin F2m )en equation

(5) becomes

λy5

+ b1λy3

+ b2λy c (6)

Raising both sides of equation (6) to the power 2m leadsto

y5

+ b1y3

+ b2y λc (7)

Since c isin Flowast2n we can let c de where d isin Flowast2m ande isin U2m+1 )en equation (7) can be rewritten as

y5

+ b1y3

+ b2y λ deminus 1

(8)

Raising both sides of equation (8) to the power 2m yields

y5

+ b1y3

+ b2y λminus 1de (9)

Multiplying equations (8) and (9) we can obtain

y10

+ y8

b1 + b11113872 1113873 + y6

b1b1 + b2 + b21113872 1113873

+ y4

b1b2 + b1b21113872 1113873 + y2b2b2 d

2

(10)

)e substitution of y2 with y in equation (10) leads to

y5

+ y4

b1 + b11113872 1113873 + y3

b1b1 + b2 + b21113872 1113873

+ y2

b1b2 + b1b21113872 1113873 + yb2b2 d2

(11)

Let y z + (b1 + b1) and we can get

2 Journal of Mathematics

z5

+ z3

b1b1 + b2 + b21113872 1113873

+ z2

b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b21113960 1113961

+ z b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b21113876 1113877

constant(12)

where constant b1b1(b31 + b13) + (b1 + b1)

2(b1b2 + b1b2) +

(b1 + b1)(b2b2 + b21b12) + d2 and constant isin F2m

From the above discussion we can deducef(x) x3x2 + b1x

2x + b2x permutes F2n iffg(z) z5 + z3(b1b1 + b2 + b2) + z2[(b1 + b1)

(b1b1 + b2 + b2) +b1b2 + b1b2] +z[(b1 + b1)4

+(b1 + b1)2(b1b1 + b2 + b2) + b2b2] permutes F2m

Based on Table 1 we know that if g(z) is a normalizedpermutation polynomial of degree 5 over Fq (q 2m) then itmust have the following forms

(1) z5 q≢1(mod5)(2) z5 + az3 + 5minus 1a2z (a arbitrary) q equiv plusmn 2(mod5)

When q≢1(mod5) we have 4∤m which includes thatm equiv 2(mod4) and m is odd When m is odd that isq plusmn 2(mod5) we mainly study the permutation poly-nomial of z5 + az3 + 5minus 1a2z (a arbitrary) Whenm equiv 2(mod4) we consider the permutation polynomial ofz5 Hence g(z) permutes F2m iff one of the following twocases holds

(1) If m equiv 2(mod4) g(z) z5 permutes F2m (2) If m is odd g(z) z5 + az3 + 5minus 1a2z permutes F2m

In conclusion we deduce that the polynomial

f(x) x3x2

+ b1x2x + b2x (13)

permutes F2n iff the following two cases are satisfied

(1) If m equiv 2(mod4) g(x) x5 permutes F2m (2) If m is odd g(x) x5 + ax3 + 5minus 1a2x permutes

F2m

Remark 2 When b1 and b2 are both 0 if 4∤m thengcd(2m+1 + 3 2n minus 1) 1 so we achieve that the monomialf(x) x3x2 permutes F2n If b1 b2 isin F2n are not both 0 thenf(x) is binomial or trinomial in F2n We will investigate thepermutation behavior of those polynomials in the sequel

Theorem 3 For two positive integers m and n with n 2m

and m being even let b1 b2 isin F2n which are not both 0 -enthe polynomial

f(x) x3x2

+ b1x2x + b2x (14)

permutes F2n iff m equiv 2(mod4) b1 θb2 and b2 is a root ofx2 + θminus 2ωx + θminus 4ω 0 where θ isin Flowast2m and ω isin Flowast2m is aprimitive third root of unity

Proof Based on )eorem 2 we deduce that f(x) x3x2 +

b1x2x + b2x permutes F2n iff

g(x) x5 + x3(b1b1 + b2 + b2) + x2[(b1 + b1)(b1b1 + b2 +

b2) + b1b2 + b1b2] + x[(b1 + b1)4 + (b1 + b1)

2(b1b1 + b2 +

b2) + b2b2] permutes F2m When m equiv 2(mod4) we get thatf(x) permutes F2n iff g(x) x5 permutes F2m

ldquorArrrdquo Comparing the coefficients of g(x) and x5 we have

b1b1 + b2 + b2 0 (15)

b1b2 + b1b2 0 (16)

b1 + b11113872 11138734 + b2b2 0 (17)

From equation (16) we can directly obtain that(b1b2) (b1b2) ie (b1b2) isin F2m Hence there existssome element θ isin Flowast2m such that b1 θb2

Plugging b1 θb2 into equation (15) we have

θ2b2b2 + b2 + b2 0 (18)

By substituting b1 with θb2 in equation (17) we obtain

θ4 b2 + b21113872 11138734

+ b2b2 0 (19)

Combining equations (18) and (19) we know thatθ12(b2b2)

3 1 )is means that b2b2 θminus 4ω andb2 + b2 θminus 2ω where ω isin Flowast2m is a primitive third root ofunity

)erefore we conclude that b2 and b2 are exactly tworoots of the equation

x2

+ θminus 2ωx + θminus 4ω 0 (20)

Hence the above analysis shows that for m equiv 2(mod4)if f(x) permutes F2n then b1 θb2 and b2 is a root ofx2 + θminus 2ωx + θminus 4ω 0 where θ isin Flowast2m and ω isin Flowast2m is aprimitive third root of unity

In what follows we will proceed to prove the necessityldquolArrrdquo For some θ isin Flowast2m let b1 θb2 and then we know

(b1b2) isin Flowast2m and this is equivalent to (b1b2) (b1b2)which implies that b1b2 + b1b2 0

Suppose that b2 and b2 are two different roots of equation(20) in F2n F2m then we have b2b2 θminus 4ω andb2 + b2 θminus 2ω Recall that b1 θb2 and we have

Table 1 Normalized permutation polynomials with degree nogreater than 5 in Fq

Normalized permutation polynomial q

x All q

x2 Even q

x3 q≢ 1(mod3)

x3 minus ax (a is not a square of Fq ) 3|q

x4 plusmn 3x q 7x4 + a1x

2 + a2x (0 is a unique solution in Fq) Even q

x5 q≢ 1(mod5)

x5 minus ax (a is not a fourth power of Fq ) 5|q

x5 + ax (a2 2) q 9x5 plusmn 2x2 q 7x5 + ax3 plusmn x2 + 3a2x (a is not a square of Fq ) q 7x5 + ax3 + 5minus 1a2x (all a) q equiv plusmn 2(mod5)

x5 + ax3 + 3a2x (a is not a square of Fq ) q 13x5 minus 2ax3 + a2x (a is not a square of Fq ) 5|q

Journal of Mathematics 3

b1b1 θ2b2b2 θminus 2ω

b1 + b1 θ b2 + b21113872 1113873 θminus 1ω(21)

which implies that

b1b1 + b2 + b2 θminus 2ω + θminus 2ω 0

b1 + b11113872 11138734

+ b2b2 θminus 4ω4+ θminus 4

0(22)

where ω isin Flowast2m is a primitive third root of unity)erefore equations (15)ndash(17) are satisfied Further-

more we know that if m equiv 2(mod4) b1 θb2 and b2 is aroot of x2 + θminus 2ωx + θminus 4ω 0 where θ isin Flowast2m and ω isin Flowast2m isa primitive third root of unity and thus f(x) is a permu-tation polynomial over F2n

To summarize we conclude that the polynomial

f(x) x3x2

+ b1x2x + b2x (23)



and m being odd let b1 b2 isin F2n which are not both 0 -enthe polynomial

f(x) x3x2

+ b1x2x + b2x (24)


(1) b1 0 b2 θω or b2 θω2 where θ isin Flowast2m andω isin F2n is a primitive third root of unity

(2) b1 θω or b1 θω2 b2 θ2 + θηω2 orb2 θ2 + θηω where θ isin Flowast2m η isin F2m and ω isin F2n isa primitive third root of unity

Proof Based on)eorem 2 if m is odd then f(x) permutesF2n iff g(x) x5 + ax3 + 5minus 1a2x permutes F2m whereg(x) x5 + x3(b1b1 + b2 + b2)+

x2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2] + x[(b1 + b1)4 +

(b1 + b1)2 (b1b1 + b2 + b2) + b2b2]

Since gcd(2 m) 1 x2 + x + 1 is a irreducible polyno-mial in F2 and we deduce that x2 + x + 1 also remains ir-reducible over F2m by Lemma 2 Let ω be one of the roots ofh(x) x2 + x + 1 in F2n then ω2 + ω + 1 0 and ω is aprimitive third root of unity Furthermore we can view 1and ω as a basis of vector space F2n upon F2m

ldquorArrrdquo Comparing the coefficients of g(x) andx5 + ax3 + 5minus 1a2x we have

b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b2 0 (25)

5minus 1b1b1 + b2 + b21113872 11138732 b1 + b11113872 11138734

+ b1 + b11113872 11138732 b1b1 + b2 + b21113872 1113873

+ b2b2

(26)

From equation (26) we know that

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873

+ b2b2 0

(27)

)en we can discuss the solutions of g(x) 0 asfollows

Case 1 b1 0 )en equation (27) turns to

b2 + b21113872 11138732

b2b2 (28)

Since 1 and ω are basis of vector space F2n upon F2m forany element b2 isin F2n we can set b2 λ + θω withλ isin F2m θ isin Flowast2m

By plugging b2 λ + θω into equation (28) we obtain

θ2(ω + ω)2

(λ + θω)(λ + θω) (29)

Since m is odd we know that ω ω2m

ω2 )enequation (29) can be written as

λ2 + λθ 0 (30)

which implies that λ 0 or λ θ Consequently we con-clude that b2 θω or b2 θω2

)e above analysis indicates that if f(x) permutes F2n

and m is odd then b1 0 b2 θω or b2 θω2 whereθ isin Flowast2m and ω isin F2n is a primitive third root of unity

Case 2 b1 ne 0 )us equation (25) becomes

b1b2 + b1b2 b1b1 b1 + b11113872 1113873 (31)

and this is equivalent to

b1 +b2

b1 b1 +

b2

b1 (32)

From equation (32) we can directly obtain thatb1 + (b2b1) isin F2m Hence there exists some element η isin F2m

such that b1 + (b2b1) η ie b2 b1(b1 + η)Plugging b2 b1(b1 + η) into equation (27) gives

b1 +b11113872 11138732η2 +b1b1 b1 +b11113872 1113873η+ b1 +b11113872 1113873

4 b1b1η

2+ b1+b11113872 1113873

3η

+b1b1 b1 +b11113872 11138732

(33)

Comparing the coefficients on both sides of equation(33) we have

b1 + b11113872 11138732

b1b1 (34)

Discussion similar to that in Case 1 shows that b1 θωor b1 θω2 Meanwhile we obtain b2 θ2 + θηω2 orb2 θ2 + θηω which follows from b2 b1(b1 + η)

)erefore the above analysis demonstrates that if f(x)

permutes F2n and m is odd then b1 θω or b1 θω2 b2

θ2 + θηω2 or b2 θ2 + θηω where θ isin Flowast2m η isin F2m andω isin F2n is a primitive third root of unity


All in all when m is odd if the polynomial f(x) per-mutes F2n then one of the following two cases is satisfied

(1) b1 0 b2 θω or b2 θω2 where θ isin Flowast2m andω isin F 2n is a primitive third root of unity


Next we will continue to prove the necessityldquolArrrdquo If b1 0 b2 θω or b2 θω2 then

(b1b1 + b2 + b2)(b1 + b1) + b1b2 + b1b2 0 which means

that equation (25) is satisfied Since ω isin F2n is a primitivethird root of unity and ω2 + ω + 1 0 we can calculate thatb2 + b2 θ and b2b2 θ2 and thus the left-hand side ofequation (27) turns to (b2 + b2)

2 + b2b2 θ2 + θ2 0 whichimplies that equation (27) is satisfied )erefore when m isodd we conclude that if b1 0 b2 θω or b2 θω2(θ isin Flowast2m and ω isin F2n is a primitive third root of unity) thepolynomial f(x) is a permutation polynomial over F2n

If b1 ne 0 recall that b1 θω or b1 θω2 b2 θ2 + θηω2

or b2 θ2 + θηω and we have

b1b1 θω(θω) θ2ω3 θ2

b1 + b1 θω + θω2 θ

b2 + b2 θ2 + θηω2+ θ2 + θηω2

θηω2+ θηω θη

b2b2 θ2 + θηω21113872 1113873 θ2 + θηω1113872 1113873 θ4 + θ3ηω + θ3ηω2

θ2η2

b1b2 θω θ2 + θηω21113872 1113873 θω2 θ2 + θηω2

1113872 1113873 θ3ω2+ θ2ηω

b1b2 θω θ2 + θηω21113872 1113873 θω θ2 + θηω1113872 1113873 θ3ω + θ2ηω2

(35)

which implies that

b1b1 + b2 + b21113872 1113873 b1 + b11113872 1113873 + b1b2 + b1b2

θ2 + θη1113872 1113873θ + θ3ω2+ θ2ηω + θ3ω + θ2ηω2

θ3 + θ2η 1 + ω + ω21113872 1113873 + θ3 ω + ω2

1113872 1113873

θ3 + θ3

0

(36)

where the third equation holds only if ω2 + ω + 1 0Consequently we conclude that equation (25) is satisfied

Next we consider

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b2

θ2 + θη1113872 11138732

+ θ4 + θ2 + θη1113872 1113873θ2 + θ4 + θ3ηω + θ3ηω2+ θ2η2

θ3η 1 + ω + ω21113872 1113873

0

(37)

where the second equation holds only if ω2 + ω + 1 0Hence we can get that equation (27) is satisfied

)erefore the above analysis shows that for m is odd ifb1 θω or b1 θω2 b2 θ2 + θηω2 or b2 θ2 + θηω where

θ isin Flowast2m η isin F2m and ω isin F2n is a primitive third root ofunity then the polynomial f(x) permutes F2n

All in all for m is odd we know thatf(x) permutes F2n ifone of the following two cases is satisfied





f(x) x3x2

+ b1x2x + b2x (38)

permutes F2n iff one of the following two cases is met



Applying )eorem 4 to b1 0 and θ 1 we have thefollowing

Corollary 1 For two positive integers m and n satisfying n

2m and m is odd the binomial f(x) x3x2 + ωx is acomplete permutation polynomial over F2n

Remark 3 Observe that a complete permutation binomialf(x) proposed in Corollary 1 is obtained by Zieve [26]However the approach we used to prove the permutationproperty is different from that in [26]

4 Concluding Remarks

In this paper by transforming the problem into studyingsome normalized permutation polynomials of degree fivewith even characteristics we investigate the coefficient pairs(b1 b2) making f(x) x3x2 + b1x

2x + b2x to be a per-mutation polynomial over F2n )e sufficient and necessaryconditions are shown in )eorems 3 and 4 Furthermore aclass of complete permutation binomials with the formf(x) x3x2 + ωx over F2n is obtained

Data Availability

No data were used to support the findings of this study

Conflicts of Interest

)e authors declare that they have no conflicts of interest

Acknowledgments

)is study was supported in part by the Educational Re-search Project of Young and Middle-Aged Teachers ofFujian Province under grant no JAT200033 the TalentFund Project of Fuzhou University under grant no GXRC-20002 and the National Natural Science Foundation ofChina under grant nos 61902073 62072109 andU1804263

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[1] R Lidl andH Niederreiter ldquoFinite fieldsrdquo EncyclopediaMathApplVol 20 Cambridge University Press Cambridge UK2nd edition 1997

[2] C Ding and Z Zhou ldquoBinary cyclic codes from explicitpolynomials overGF (2m)rdquo Discrete Mathematics vol 321pp 76ndash89 2014

[3] J Schwenk and K Huber ldquoPublic key encryption and digitalsignatures based on permutation polynomialsrdquo ElectronicsLetters vol 34 no 8 pp 759-760 1998

[4] C Ding and J Yuan ldquoA family of skew hadamard differencesetsrdquo Journal of Combinatorial -eory - Series A vol 113no 7 pp 1526ndash1535 2006

[5] H Niederreiter and K H Robinson ldquoComplete mappings offinite fieldsrdquo Journal of the Australian Mathematical SocietySeries A Pure Mathematics and Statistics vol 33 no 2pp 197ndash212 1982

[6] Y Laigle-Chapuy ldquoPermutation polynomials and applica-tions to coding theoryrdquo Finite Fields and -eir Applicationsvol 13 no 1 pp 58ndash70 2007

[7] G L Mullen ldquoPermutation polynomials over finite fieldsrdquo inFinite Fields Coding -eory and Advances in Communica-tions and Computing Lect Notes Pure Appl Math vol 141pp 131ndash151 Marcel Dekker New York NY USA 1993

[8] G L Mullen and D Pannario Handbook of Finite FieldsTaylor amp Francis Boca Raton FL USA 2013

[9] X-D Hou ldquoPermutation polynomials over finite fields - asurvey of recent advancesrdquo Finite Fields and -eir Applica-tions vol 32 pp 82ndash119 2015

[10] N Li and X Zeng ldquoA Survey on the applications of Nihoexponentsrdquo Cryptography and Communications vol 11 no 3pp 509ndash548 2019

[11] C Ding L Qu Q Wang J Yuan and P Yuan ldquoPermutationtrinomials over finite fields with even characteristicrdquo SIAMJournal on Discrete Mathematics vol 29 no 1 pp 79ndash92 2015

[12] R Gupta and R K Sharma ldquoSome new classes of permutationtrinomials over finite fields with even characteristicrdquo FiniteFields and -eir Applications vol 41 pp 89ndash96 2016

[13] N Li and T Helleseth ldquoSeveral classes of permutation tri-nomials from Niho exponentsrdquo Cryptography and Commu-nications vol 9 no 6 pp 693ndash705 2017

[14] K Li L Qu and X Chen ldquoNew classes of permutation bi-nomials and permutation trinomials over finite fieldsrdquo FiniteFields and -eir Applications vol 43 pp 69ndash85 2017

[15] Z Zha L Hu and S Fan ldquoFurther results on permutationtrinomials over finite fields with even characteristicrdquo FiniteFields and -eir Applications vol 45 pp 43ndash52 2017

[16] X-D Hou ldquoDetermination of a type of permutation trino-mials over finite fields IIrdquo Finite Fields and -eir Applica-tions vol 35 pp 16ndash35 2015

[17] Z Tu X Zeng C Li and T Helleseth ldquoA class of newpermutation trinomialsrdquo Finite Fields and -eir Applicationsvol 50 pp 178ndash195 2018

[18] D Bartoli ldquoOn a conjecture about a class of permutationtrinomialsrdquo Finite Fields and -eir Applications vol 52pp 30ndash50 2018

[19] X-D Hou ldquoOn a class of permutation trinomials in char-acteristic 2rdquo Cryptography and Communications vol 11no 6 pp 1199ndash1210 2019

[20] Z Tu and X Zeng ldquoTwo classes of permutation trinomialswith Niho exponentsrdquo Finite Fields and -eir Applicationsvol 53 pp 99ndash112 2018


[21] X-D Hou ldquoOn the Tu-Zeng permutation trinomial of type(1∕4 3∕4)rdquo Discrete Mathematics vol 344 no 3 Article ID112241 2021

[22] L Zheng H Kan and J Peng ldquoTwo classes of permutationtrinomials with Niho exponents over finite fields with evencharacteristicrdquo Finite Fields and -eir Applications vol 68Article ID 101754 2020

[23] D Bartoli and M Timpanella ldquoOn trinomials of type xn+m

(1 +Axm (qminus1) +Bxn (qminus 1)) n m odd over Fq2 q 22s + 1rdquoFinite Fields and-eir Applications vol 72 Article ID 1018162021

[24] D Wu P Yuan C Ding and Y Ma ldquoPermutation trinomialsover F2mrdquo Finite Fields and -eir Applications vol 46pp 38ndash56 2017

[25] C Malvenuto and F Pappalardi ldquoEnumerating permutationpolynomials II k-cycles with minimal degreerdquo Finite Fieldsand -eir Applications vol 10 no 1 pp 72ndash96 2004

[26] M E Zieve ldquoPermutation polynomials induced from per-mutations of subfields and some complete sets of mutuallyorthogonal Latin squaresrdquo 2013 httpsarxivorgabs13121325


degree 5 over even characteristics we determine the coef-ficients (b1 b2) for f(x) x3x2 + b1x

2x + b2x being apermutation over F2n

)e rest of this paper is arranged as follows Somenotations and useful results are presented in Section 2 InSection 3 the sufficient and necessary conditions are shownto determine the coefficients of a class of permutationpolynomials over F2n Finally Section 4 provides someconcluding remarks

2 Preliminaries

Let m and n be two positive integers with m|n and F2n be afinite field with 2n elements the trace function from F2n toF2m is denoted by Trn

m(middot) where

Trnm(x) x + x

2m

+ x22m

+ middot middot middot + x2(n(mminus1))m

(2)

If the usual complex conjugation of any x isin F22m isdefined as x x2m the following relations hold

(i) For all x isin F22m x + x isin F2m xx isin F2m (ii) For all x y isin F22m x + y x + y xy xy

U2m+1 denotes the unit circle of F22m as

U2m+1 x isin F22m x2m+1

xx 11113966 1113967 (3)

Lemma 1 (see [24]) f(x) isin Fq[x] with f(0) 0 is apermutation polynomial over Fq iff f x⟼f(x) from Flowastqinto itself is a bijection

Hereinafter we claim that f(x) is a permutationpolynomial over Flowastq upon the bijectivity of f(x) from Flowastq toFlowastq which is the only case we consider

Lemma 2 (see [1]) -e irreducibility of f(x) isin Fq[x] withdegree n remains over Fpm iff gcd(m n) 1

Definition 1 (see [25]) A permutation polynomialf(x) isin Fq[x] of degree n is said to be normalized if thefollowing properties hold

(i) f(x) is monic and the value of f(x) at 0 is equal to 0(ii) )e coefficient of xnminus 1 equals 0 if p∤n where p is the

characteristic of Fq

Remark 1 Let b c isin Fq and a isin Flowastq for any permutationpolynomial f(x) isin Fq[x] there exists a unique normalizedform provided by g(x) af(x + b) + c

Theorem 1 (Hermitersquos criterion) (see [1]) Let p be thecharacteristic of Fq -us f(x) isin Fq[x] is a permutationpolynomial over Fq iff

(i) f(x) has exactly one solution over Fq(ii) forallt isin Z where 1le tle q minus 2 tne 0(modp) the degree

of the reduction of f(x)t(modxq minus x) is no greaterthan q minus 2

From Hermitersquos criterion the characterization of allnormalized permutation polynomials with degree no greaterthan 5 in Fq is due to ([1] Section 72) and they are listed inTable 1

3 A Class of Permutation Polynomials over F2n

In this section we consider the coefficients b1 b2 isin F2n suchthat the polynomial f(x) x3x2 + b1x

2x + b2x permutesover F2n

)e main results in this paper are given in the followingtheorems

Theorem 2 For two positive integers m and n with n 2mlet b1 b2 isin F2n Define g(z) z5 + z3(b1b1 + b2 + b2)

+z2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2]+

z[(b1 + b1)4 + (b1 + b1)

2(b1b1 + b2 + b2) + b2b2] isin F2m [z]-en the polynomial

f(x) x3x2

+ b1x2x + b2x (4)


(1) If m equiv 2(mod4) g(x) x5 permutes F2m (2) If m is odd g(x) x5 + ax3 + 5minus 1a2x permutes F2m

Proof Based on Lemma 1 in order to prove that f(x) is apermutation polynomial in Flowast2n we only consider that for anyc isin Flowast2n the equation

x3x2

+ b1x2x + b2x c (5)

has exactly one root in Flowast2n Let x λy where λ isin U2m+1 and y isin F2m )en equation

(5) becomes

λy5

+ b1λy3

+ b2λy c (6)

Raising both sides of equation (6) to the power 2m leadsto

y5

+ b1y3

+ b2y λc (7)

Since c isin Flowast2n we can let c de where d isin Flowast2m ande isin U2m+1 )en equation (7) can be rewritten as

y5

+ b1y3

+ b2y λ deminus 1

(8)

Raising both sides of equation (8) to the power 2m yields

y5

+ b1y3

+ b2y λminus 1de (9)

Multiplying equations (8) and (9) we can obtain

y10

+ y8

b1 + b11113872 1113873 + y6

b1b1 + b2 + b21113872 1113873

+ y4

b1b2 + b1b21113872 1113873 + y2b2b2 d

2

(10)

)e substitution of y2 with y in equation (10) leads to

y5

+ y4

b1 + b11113872 1113873 + y3

b1b1 + b2 + b21113872 1113873

+ y2

b1b2 + b1b21113872 1113873 + yb2b2 d2

(11)

Let y z + (b1 + b1) and we can get


z5

+ z3

b1b1 + b2 + b21113872 1113873

+ z2

b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b21113960 1113961

+ z b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b21113876 1113877

constant(12)


2(b1b2 + b1b2) +




(b1b1 + b2 + b2) +b1b2 + b1b2] +z[(b1 + b1)4







f(x) x3x2

+ b1x2x + b2x (13)



F2m




f(x) x3x2

+ b1x2x + b2x (14)




g(x) x5 + x3(b1b1 + b2 + b2) + x2[(b1 + b1)(b1b1 + b2 +

b2) + b1b2 + b1b2] + x[(b1 + b1)4 + (b1 + b1)

2(b1b1 + b2 +



b1b1 + b2 + b2 0 (15)

b1b2 + b1b2 0 (16)

b1 + b11113872 11138734 + b2b2 0 (17)



θ2b2b2 + b2 + b2 0 (18)


θ4 b2 + b21113872 11138734

+ b2b2 0 (19)




x2








x All q

x2 Even q

x3 q≢ 1(mod3)




x5 q≢ 1(mod5)






b1 + b1 θ b2 + b21113872 1113873 θminus 1ω(21)

which implies that


b1 + b11113872 11138734


0(22)




f(x) x3x2

+ b1x2x + b2x (23)




f(x) x3x2

+ b1x2x + b2x (24)





x2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2] + x[(b1 + b1)4 +

(b1 + b1)2 (b1b1 + b2 + b2) + b2b2]



b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b2 0 (25)

5minus 1b1b1 + b2 + b21113872 11138732 b1 + b11113872 11138734

+ b1 + b11113872 11138732 b1b1 + b2 + b21113872 1113873

+ b2b2

(26)


b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873

+ b2b2 0

(27)



b2 + b21113872 11138732

b2b2 (28)



θ2(ω + ω)2

(λ + θω)(λ + θω) (29)



λ2 + λθ 0 (30)





b1b2 + b1b2 b1b1 b1 + b11113872 1113873 (31)


b1 +b2

b1 b1 +

b2

b1 (32)



b1 +b11113872 11138732η2 +b1b1 b1 +b11113872 1113873η+ b1 +b11113872 1113873

4 b1b1η

2+ b1+b11113872 1113873

3η

+b1b1 b1 +b11113872 11138732

(33)


b1 + b11113872 11138732

b1b1 (34)
















b1 + b1 θω + θω2 θ

b2 + b2 θ2 + θηω2+ θ2 + θηω2



θ2η2

b1b2 θω θ2 + θηω21113872 1113873 θω2 θ2 + θηω2

1113872 1113873 θ3ω2+ θ2ηω


(35)

which implies that

b1b1 + b2 + b21113872 1113873 b1 + b11113872 1113873 + b1b2 + b1b2

θ2 + θη1113872 1113873θ + θ3ω2+ θ2ηω + θ3ω + θ2ηω2

θ3 + θ2η 1 + ω + ω21113872 1113873 + θ3 ω + ω2

1113872 1113873

θ3 + θ3

0

(36)


Next we consider

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b2

θ2 + θη1113872 11138732

+ θ4 + θ2 + θη1113872 1113873θ2 + θ4 + θ3ηω + θ3ηω2+ θ2η2

θ3η 1 + ω + ω21113872 1113873

0

(37)









f(x) x3x2

+ b1x2x + b2x (38)











Data Availability




Acknowledgments


References






























z5

+ z3

b1b1 + b2 + b21113872 1113873

+ z2

b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b21113960 1113961

+ z b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b21113876 1113877

constant(12)


2(b1b2 + b1b2) +




(b1b1 + b2 + b2) +b1b2 + b1b2] +z[(b1 + b1)4







f(x) x3x2

+ b1x2x + b2x (13)



F2m




f(x) x3x2

+ b1x2x + b2x (14)




g(x) x5 + x3(b1b1 + b2 + b2) + x2[(b1 + b1)(b1b1 + b2 +

b2) + b1b2 + b1b2] + x[(b1 + b1)4 + (b1 + b1)

2(b1b1 + b2 +



b1b1 + b2 + b2 0 (15)

b1b2 + b1b2 0 (16)

b1 + b11113872 11138734 + b2b2 0 (17)



θ2b2b2 + b2 + b2 0 (18)


θ4 b2 + b21113872 11138734

+ b2b2 0 (19)




x2








x All q

x2 Even q

x3 q≢ 1(mod3)




x5 q≢ 1(mod5)






b1 + b1 θ b2 + b21113872 1113873 θminus 1ω(21)

which implies that


b1 + b11113872 11138734


0(22)




f(x) x3x2

+ b1x2x + b2x (23)




f(x) x3x2

+ b1x2x + b2x (24)





x2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2] + x[(b1 + b1)4 +

(b1 + b1)2 (b1b1 + b2 + b2) + b2b2]



b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b2 0 (25)

5minus 1b1b1 + b2 + b21113872 11138732 b1 + b11113872 11138734

+ b1 + b11113872 11138732 b1b1 + b2 + b21113872 1113873

+ b2b2

(26)


b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873

+ b2b2 0

(27)



b2 + b21113872 11138732

b2b2 (28)



θ2(ω + ω)2

(λ + θω)(λ + θω) (29)



λ2 + λθ 0 (30)





b1b2 + b1b2 b1b1 b1 + b11113872 1113873 (31)


b1 +b2

b1 b1 +

b2

b1 (32)



b1 +b11113872 11138732η2 +b1b1 b1 +b11113872 1113873η+ b1 +b11113872 1113873

4 b1b1η

2+ b1+b11113872 1113873

3η

+b1b1 b1 +b11113872 11138732

(33)


b1 + b11113872 11138732

b1b1 (34)
















b1 + b1 θω + θω2 θ

b2 + b2 θ2 + θηω2+ θ2 + θηω2



θ2η2

b1b2 θω θ2 + θηω21113872 1113873 θω2 θ2 + θηω2

1113872 1113873 θ3ω2+ θ2ηω


(35)

which implies that

b1b1 + b2 + b21113872 1113873 b1 + b11113872 1113873 + b1b2 + b1b2

θ2 + θη1113872 1113873θ + θ3ω2+ θ2ηω + θ3ω + θ2ηω2

θ3 + θ2η 1 + ω + ω21113872 1113873 + θ3 ω + ω2

1113872 1113873

θ3 + θ3

0

(36)


Next we consider

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b2

θ2 + θη1113872 11138732

+ θ4 + θ2 + θη1113872 1113873θ2 + θ4 + θ3ηω + θ3ηω2+ θ2η2

θ3η 1 + ω + ω21113872 1113873

0

(37)









f(x) x3x2

+ b1x2x + b2x (38)











Data Availability




Acknowledgments


References































b1 + b1 θ b2 + b21113872 1113873 θminus 1ω(21)

which implies that


b1 + b11113872 11138734


0(22)




f(x) x3x2

+ b1x2x + b2x (23)




f(x) x3x2

+ b1x2x + b2x (24)





x2[(b1 + b1)(b1b1 + b2 + b2) + b1b2 + b1b2] + x[(b1 + b1)4 +

(b1 + b1)2 (b1b1 + b2 + b2) + b2b2]



b1 + b11113872 1113873 b1b1 + b2 + b21113872 1113873 + b1b2 + b1b2 0 (25)

5minus 1b1b1 + b2 + b21113872 11138732 b1 + b11113872 11138734

+ b1 + b11113872 11138732 b1b1 + b2 + b21113872 1113873

+ b2b2

(26)


b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873

+ b2b2 0

(27)



b2 + b21113872 11138732

b2b2 (28)



θ2(ω + ω)2

(λ + θω)(λ + θω) (29)



λ2 + λθ 0 (30)





b1b2 + b1b2 b1b1 b1 + b11113872 1113873 (31)


b1 +b2

b1 b1 +

b2

b1 (32)



b1 +b11113872 11138732η2 +b1b1 b1 +b11113872 1113873η+ b1 +b11113872 1113873

4 b1b1η

2+ b1+b11113872 1113873

3η

+b1b1 b1 +b11113872 11138732

(33)


b1 + b11113872 11138732

b1b1 (34)
















b1 + b1 θω + θω2 θ

b2 + b2 θ2 + θηω2+ θ2 + θηω2



θ2η2

b1b2 θω θ2 + θηω21113872 1113873 θω2 θ2 + θηω2

1113872 1113873 θ3ω2+ θ2ηω


(35)

which implies that

b1b1 + b2 + b21113872 1113873 b1 + b11113872 1113873 + b1b2 + b1b2

θ2 + θη1113872 1113873θ + θ3ω2+ θ2ηω + θ3ω + θ2ηω2

θ3 + θ2η 1 + ω + ω21113872 1113873 + θ3 ω + ω2

1113872 1113873

θ3 + θ3

0

(36)


Next we consider

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b2

θ2 + θη1113872 11138732

+ θ4 + θ2 + θη1113872 1113873θ2 + θ4 + θ3ηω + θ3ηω2+ θ2η2

θ3η 1 + ω + ω21113872 1113873

0

(37)









f(x) x3x2

+ b1x2x + b2x (38)











Data Availability




Acknowledgments


References








































b1 + b1 θω + θω2 θ

b2 + b2 θ2 + θηω2+ θ2 + θηω2



θ2η2

b1b2 θω θ2 + θηω21113872 1113873 θω2 θ2 + θηω2

1113872 1113873 θ3ω2+ θ2ηω


(35)

which implies that

b1b1 + b2 + b21113872 1113873 b1 + b11113872 1113873 + b1b2 + b1b2

θ2 + θη1113872 1113873θ + θ3ω2+ θ2ηω + θ3ω + θ2ηω2

θ3 + θ2η 1 + ω + ω21113872 1113873 + θ3 ω + ω2

1113872 1113873

θ3 + θ3

0

(36)


Next we consider

b1b1 + b2 + b21113872 11138732

+ b1 + b11113872 11138734

+ b1 + b11113872 11138732

b1b1 + b2 + b21113872 1113873 + b2b2

θ2 + θη1113872 11138732

+ θ4 + θ2 + θη1113872 1113873θ2 + θ4 + θ3ηω + θ3ηω2+ θ2η2

θ3η 1 + ω + ω21113872 1113873

0

(37)









f(x) x3x2

+ b1x2x + b2x (38)











Data Availability




Acknowledgments


References

































f(x) x3x2

+ b1x2x + b2x (38)











Data Availability




Acknowledgments


References






























Research Article A Class of New Permutation Polynomials ...

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