Two types of permutation polynomials with special forms
aa r X i v : . [ c s . I T ] M a y Two types of permutation polynomialswith special forms
Dabin Zheng a , Mu Yuan a , Long Yu b a Hubei Province Key Laboratory of Applied Mathematics,Faculty of Mathematics and Statistics, Hubei University, Wuhan 430062, China b School of Mathematics and Physics, Hubei Polytechnic University,Huangshi 435003, China
Abstract
Let q be a power of a prime and F q be a finite field with q elements. In thispaper, we propose four families of infinite classes of permutation trinomialshaving the form cx − x s + x qs over F q , and investigate the relationshipbetween this type of permutation polynomials with that of the form ( x q − x + δ ) s + cx . Based on this relation, many classes of permutation trinomialshaving the form ( x q − x + δ ) s + cx without restriction on δ over F q arederived from known permutation trinomials having the form cx − x s + x qs . Keywords:
Finite field, Permutation polynomial, Symbolic computation
1. Introduction
Let q be a power of a prime. Let F q be a finite field with q elements and F ∗ q denote its multiplicative group. A polynomial f ( x ) ∈ F q [ x ] is called a per-mutation polynomial (PP) if its associated polynomial mapping f : c f ( c )from F q to itself is a bijection. Permutation polynomials over finite fieldshave important applications in cryptography, coding theory and combina-torial design theory. So, finding new permutation polynomials is of greatinterest in both theoretical and applied aspects. The study of permutationpolynomials has a long history [6, 12] and many recent results are surveyedin [14].Permutation polynomials with few terms attracts many researchers’ at-tention due to their simple algebraic representation and wide applications in Email addresses: [email protected] (Dabin Zheng), [email protected] (Mu Yuan), [email protected] (Long Yu)
Preprint submitted to Elsevier May 29, 2018 oding theory, combinatorial designs and cryptography. The recent progresson construction of permutation binomials and trinomials can be seen in [2,7, 11, 19, 18, 20, 21, 22, 23, 25, 38] and the references therein. Permuta-tion polynomials of the form x + γ Tr q n q ( x k ) were investigated in [3, 4] forsome special γ ∈ F q n . Very recently, G.Kyureghyan and M.E. Zieve [15]found almost all permutation polynomials over F q n of this form for γ ∈ F ∗ q n and q n ≤ q is odd. And Li et al. [18] studied permutationbehavior of this type of polynomials over the finite fields with even char-acteristic. In both [15] and [18], most permutation polynomials having theform x + γ Tr q n q ( x k ) occur in case of n = 2, i.e., they are trinomial permu-tations. Inspired by this, we construct four families of infinite classes ofpermutation trinomials over F q with the following form cx − x s + x qs , (1)where s is a positive integer and c ∈ F q . We use the well-known result(Lemma 1) provided by Park, Lee, Wang and Zieve to prove the first threeclasses of permutation trinomials. For the proof of permutation behavior ofthe last class of trinomials, symbolic computation method related to Gr¨ o bnerbases and resultants is used.On the other hand, in order to derive new Kloosterman sums identitiesover F q n , Helleseth and Zinoviev [13] found a class of permutation polyno-mials related to the form ( x q i − x + δ ) s + cx (2)over F q n , and Ding and Yuan [9] first studied this type of permutationpolynomials. Following line of work in [9], many permutation polynomialswith this form have been constructed [5, 21, 26, 27, 29, 31, 32, 33, 34,35, 36, 37]. Most known permutation polynomials having this form arerelated to δ . In this paper, we find a close relationship between the twotypes of permutation polynomials with the forms (1) and (2) respectively.Based on this relationship, many classes of permutation polynomials havingthe form (2) without restriction on δ are derived from known permutationpolynomials of the form (1).The remainder of this paper is organized as follows. In section 2, weintroduce some preliminary results and investigate the relationship betweenthe two types of permutation polynomials with the forms (1) and (2) re-spectively. Section 3 presents four families of infinite classes of permutationtrinomials with the form (1). In section 4, many new classes of permutationpolynomials with the form (2) are derived from known permutation polyno-mials with the form (1). Finally, concluding remark is given in section 5.2 . Preliminaries Let q be a power of a prime, and F q be a finite field with q elements. Let k be a divisor of n . The trace function from F q n to F q k is defined byTr q n q k ( x ) = n/k − X i =0 x q ik , where x ∈ F q n . Let d be a positive integer with d | ( q − µ d denotethe set of d th roots of unity in F ∗ q , i.e., µ d = n x ∈ F ∗ q | x d = 1 o . Many people investigate the permutation behavior of the polynomialswith the form x r h ( x s ) for s | ( q −
1) over F q . The following lemma is well-known which transforms the problem of proving x r h ( x s ) permutes F q intothe problem of investigating whether it permutes µ ( q − /d . Lemma 1. [24][28][39] Let d, r be positive integers with d | ( q − , and let h ( x ) ∈ F q [ x ] . Then x r h ( x ( q − /d ) permutes F q if and only if both (1) gcd ( r, ( q − /d ) = 1 and (2) x r h ( x ) ( q − /d permutes µ d := (cid:8) x ∈ F ∗ q | x d = 1 (cid:9) . The following proposition shows a close relationship between the twotypes of permutation polynomials with the forms (1) and (2).
Proposition 2.
Let m, k be integers with < k < m and ℓ = gcd( k, m ) .Let F q m be a finite field and g ( x ) ∈ F q m [ x ] . For δ ∈ F q m and c ∈ F ∗ q ℓ , if h ( x ) = g ( x ) q k − g ( x ) + cx permutes F q m then f ( x ) = g ( x q k − x + δ ) + cx permutes F q m . Proof.
For α ∈ F q m , it is sufficient to show that the equation f ( x ) = g ( x q k − x + δ ) + cx = α (3)has only one solution in F q m . Denote by y = x q k − x + δ . Taking q k -th poweron both sides of (3) and subtracting (3) we have h ( y ) = g ( y ) q k − g ( y ) + cy = α q k − α + cδ. Since h permutes F q m we obtain y = h − ( α q k − α + cδ ). So, the only onesolution of (3) in F q m is x = c − ( α − g ◦ h − ( α q k − α + cδ )).3 emark 3. In Proposition 2, let m = 2 , k = 1 and g ( x ) = x s and c ∈ F ∗ q .By Proposition 2, if cx − x s + x qs permutes F q then ( x q − x + δ ) s + cx permutes F q . Moreover, if there are no restrictions on δ and c = 1 then theinverse of Proposition 2 also holds. Proposition 4.
Let F q m be the degree m extension of the finite field F q . Let g ( x ) ∈ F q [ x ] be a polynomial over F q . For any δ ∈ F q m , f ( x ) = g ( x q − x + δ ) + x permutes F q m if and only if h ( x ) = g ( x ) q − g ( x ) + x permutes F q m . Proof.
By Proposition 2 we only prove the necessity . Denote by ϕ ( x ) = x q − x + δ , α = Tr q m q ( δ ), and S α = { x ∈ F q m | Tr q m q ( x ) = α } . By Theo-rem 2.25 in [16] we have that S α = { x q − x + δ | x ∈ F q m } . For α, α ′ ∈ F q with α = α ′ we have that S α ∩ S α ′ = ∅ and ∪ α ∈ F q S α = F q m .It is easy to verify that ϕ ◦ f ( x ) = h ◦ ϕ ( x ) for any x ∈ F q m , i.e., thefollowing diagram is commutative, F q m f −−−−→ F q m y ϕ y ϕ S α h −−−−→ S α If f ( x ) is a bijection on F q m , then from above diagram we know that h ( x ) is a surjective mapping from S α to itself. So, h ( x ) is a bijection on S α for any α ∈ F q , moreover F q m = ∪ α ∈ F q S α . Thus, h ( x ) permutes F q m . (cid:3)
3. Four classes of permutation polynomials with the form cx − x s + x sq over F q In this section we present four classes of permutation polynomials withthe form cx − x s + x sq for some proper index s and parameter c ∈ F ∗ q . Theorem 5.
Let s = q +2 q − and c ∈ F ∗ q . The polynomial f ( x ) = cx − x s + x qs permutes F q in each of the following cases: (1) q ≡ and ( − c ) q +12 = 1 ; (2) q ≡ and ( c ) q +12 = 1 . Proof.
We only prove the case (1). The proof of the case (2) is similar andthe details are omitted. 4ince q ≡ | (3 q + 5). The polynomial f ( x ) isrewritten as f ( x ) = cx − x q +54 ( q − + x ( q +54 q +1 ) ( q − . Let u = q +54 . By Lemma 1, f ( x ) permutes F q if and only if g ( x ) = x ( c + x − u − x u ) q − permutes µ q +1 .Let µ q +12 be the cyclic subgroup of µ q +1 with order q +12 . It is easy toverify that µ q +12 is exactly all square elements of µ q +1 . Since q ≡ − µ q +1 . Then − µ q +12 is the set of allnon-square elements of µ q +1 . Next we show that g ( x ) permutes µ q +12 and − µ q +12 respectively.Since q ≡ µ q +12 . For x ∈ µ q +12 , there exists an unique y ∈ µ q +12 such that x = y . Then x u = ( y ) u = y q +52 = y, x − u = x · x − u = y · y − = y. In this case, g ( x ) = c q − x , and it permutes µ q +12 .For x ∈ − µ q +12 , there exists an unique y ∈ µ q +12 such that x = − y withthe same reason above. Due to ( − c ) q +12 = 1, i.e., c ∈ − µ q +12 and y ∈ µ q +12 we have that c − y = 0. So, g ( x ) = g ( − y ) = − y ( c + ( − − u y − ( − u y ) q − = − y ( c − y ) q c − y = − y c q − y − c − y = 2 c y. Since c ∈ − µ q +12 and y ∈ µ q +12 we have c y ∈ − µ q +12 . So, g ( x ) permutes − µ q +12 . Combining these two cases, we have that g ( x ) permutes µ q +1 . (cid:3) Similarly, we obtain another family of infinite class of permutation poly-nomials over F q as follows. Theorem 6.
Let s = ( q +1) and c ∈ F ∗ q . The polynomial f ( x ) = cx − x s + x qs permutes F q in each of the following cases: q ≡ and ( − c ) q +12 = 1 ; (2) q ≡ and ( c ) q +12 = 1 . By direct verification and simplification we get a family of infinite classof permutation polynomials over F q as follows. Theorem 7.
Let q be a power of a prime with q ≡ , and s = q + q +13 . Then f ( x ) = x − x s + x qs is a permutation polynomial over F q . Proof.
Since q ≡ | ( q + 2). The polynomial f ( x )is rewritten as f ( x ) = x − x q +23 ( q − + x (cid:16) q ( q +2)3 +1 (cid:17) ( q − . Let u = q +23 . By Lemma 1, f ( x ) permutes F q if and only if g ( x ) = x (1 + x − u − x u ) q − permutes µ q +1 .First, we show that 1 + x − u − x u = 0 has no roots in µ q +1 . Otherwise,assume that there is x ∈ µ q +1 satisfying this equation, i.e.,1 + x − u − x u = 0 , x q +1 = 1 . (4)These imply x u − − x u − = x q +1 , i.e., x u − − x u − = x q . This leads to x − u − x − u = x , then x − u − x − u = 1, i.e., x u − + x u = 1. This equalitytogether with (4) implies x u − + x u = 0, i.e., x = −
1. On the other hand, itis easy to verify that x = − ∤ q . So, we get a contradiction.For any x ∈ µ q +1 , we have x u = x since u = q +23 , and g ( x ) = x x u − − x − u x − u − x u = x + x u − x − u x − u − x u = x u + x u − x u x u − x u = x u . So, g ( x ) permutes µ q +1 since gcd( u, q + 1) = 1. (cid:3) The main technique in the proof of above theorems is provided in Lemma 1.This method doesn’t work for the following theorem. Using the method pro-vided by Dobbertin [8] and Kyureghyan and Zieve [15] we present a familyof infinite class of permutation polynomials over F q . To this end, we firstgive two preliminary lemmas. 6 emma 8. There exists x ∈ F ∗ q such that x q + x q +1 + x = 0 if and onlyif | q and x q − = 1 . Proof.
Assume that x ∈ F ∗ q satisfies x q + x q +1 + x = 0 . (5)Let y = x q − , then y ∈ µ q +1 and y + y + 1 = 0. It follows that y = 1.Since gcd(3 , q + 1) = 1, we must have y = 1 and 3 | q . Conversely, it isobvious. (cid:3) Similarly, we have the following lemma.
Lemma 9.
There exists x ∈ F ∗ q such that x q − x q +1 + x = 0 if and onlyif | q and x q − = − . Theorem 10.
Let q be a power of an odd prime and s = q + q − q . Thenthe polynomial f ( x ) = x − x s + x q s permutes F q . Proof.
We prove that f ( x ) = α has only one root in F q for any α ∈ F q .To this end, we discuss the proof according to the following four situations. Case I: α = 0 and f ( x ) = 0. It is clear that this equation has a solution x = 0. Next we show that there is no x ∈ F ∗ q satisfying f ( x ) = 0. Otherwise,we have 1 − x q + q − q − + x − q + q = 0 . Denote by y = x q − , then y ∈ µ q +1 . From above equation we get y q +1 − y − q = 1 . (6)Calculation of (6) q − (6) q together with y q = y − , we obtain0 = y q + q − y − q − y q +1 + y − = y q − − y − y − q +1 + y − =( y q − − y )( y − q + 1) . It follows that y q − = 1 since y = −
1. This fact together with y q +1 = 1implies that y gcd( q − ,q +1) = y gcd( q − , = 1 . So, we have y = 1. On the other hand, substituting y q − = 1 into (6) weget y = y + 1. This together with y = 1 implies that y = 0. This is acontradiction. So, f ( x ) = 0 has only one solution x = 0 in F q .7o discuss case of α = 0, we define two sets as follows, S ± = n x ∈ F ∗ q | x q ± x q +1 + x = 0 o . Case II: α ∈ S + and f ( x ) = α . By Lemma 8, we have 3 | q and α q − α = 0. In this case we have0 = α q − α = f ( x ) q − f ( x )= x q − x − q + q +1 + x q + q − q − ( x − x q + q − q + x − q + q +1 )= x q − x − x q + q − q + x − q + q +1 = x − q − q ( x − x q ) q ( x q + q + x q ) . If x − x q = 0 then f ( x ) = x = α . If x q + q = − x q then x ( q +1)( q − = − x ∈ F ∗ q , we have x gcd(2( q +1)( q − ,q − = x q +1) = 1. It implies that x (1+ q )( q − = 1. This leads to a contradiction. Thus f ( x ) = α has only oneroot x = α . Case III: α ∈ S − and f ( x ) = α . By similar proof to Case II, we knowthat f ( x ) = α has only one solution x = α .Denote by T = F ∗ q \ ( S + ∪ S − ). From the proving process of Cases IIand III, we know that f ( x ) is a permutation over S + and S − respectively.Moreover, f ( T ) ⊆ T . It remains to show that f ( x ) is also a permutationover the set T . Case IV: α ∈ T . We will show that f ( x ) = α has only one x ∈ T satisfying this equation. Denote by y = x q , z = x q , w = x q , β = α q , γ = α q , δ = α q . The equation f ( x ) = α is reduced to x + xyw − zwy = α. (7)Taking q th, q th and q th power on both sides of (7) respectively, we obtain y + yzx − wxz = β, (8) z + zwy − yxw = γ, (9) w + wxz − yzx = δ, (10)8espectively. Adding (7) and (9) we have z = α + γ − x . Similarly, we get w = β + δ − y from (8) and (10). Substituting the two equalities into (7)and (8) respectively, we obtain ( ( − γ + x ) y + (( α + 2 γ − x )( β + δ )) y + ( − α − γ + x )( β + δ ) = 0 (cid:0) x − ( α + γ ) x + ( α + γ ) (cid:1) y − δx − β ( α + γ ) x = 0 . (11)Since x ∈ T , from Lemma 8 we have x − ( α + γ ) x + ( α + γ ) = x + xz + z = 0 . From the second equation in (11) we get y = x ( βα + βγ + δx ) α + 2 αγ + γ − ( α + γ ) x + x , H ( x ) . (12)Substituting this y into the first equation in (11), we obtain a equality A ( x ) = 0, where A ( x ) ∈ F ∗ q [ α, β, γ, δ ][ x ] and its degree on x is 5.Suppose that f ( x ) = α have another root X ∈ T different from x ,then A ( X ) = 0. Replacing α, β, γ, δ in coefficients of A ( X ) with x, y, z, w respectively, and multiplying wx yz on both sides of A ( X ) = 0, we get( X − x ) B ( X ) C ( X ) = 0 , (13)where B ( X ) = yw ( x + xz + z ) X + wxz ( z + x ) ( w + y ) − ( z + x )( w xz + wx y + wxyz + wyz − xy z ) X,C ( X ) =( w x − wxyz + y z ) X + yz ( z + x ) ( w + y )+( z + x )( w xz + wx y + wxyz − wyz − xy z − y z ) X. Hence B ( X ) = 0 or C ( X ) = 0. It has been verified that yw ( x + xz + z ) = 0above. Assume that w x − wxyz + y z = 0. Then0 = w x − wxyz + y z = ( yz ) q − ( yz ) q +1 + ( yz ) . (14)By Lemma 9 we have ( yz ) q − = −
1, i.e., y ( q − q +1) = −
1. Since gcd(2( q − q + 1) , q −
1) = 2( q − y q − = 1. This is a contradic-tion with y ( q − q +1) = −
1. So, the degree of B ( X ) and C ( X ) on X is 2,respectively.(i) When B ( X ) = 0 in (13). Taking q th power on both sides of thisequality we get B ′ ( x q ) = 0, where the coefficients of B ′ ( X ) are q th power of9orresponding coefficients of B ( X ). Replacing Y = X q with H ( X ) in (12)we obtain 0 = B ′ ( H ( X )) = C ( X ) D ( X ) , (15)where D ( X ) =( w x + w xz + zxwy + xy z + y z ) X + wx ( z + x ) ( w + y ) − ( z + x )(2 w x + w xz + wx y + zxwy − wyz + xy z ) X. If C ( X ) = 0 from (15) then X is a common root of B ( X ) and C ( X ).Then the resultant of these two polynomials should be 0. So,0 = Res ( B, C, X ) = zx ( z + x ) ( w + y ) ( w x − zxwy + y z ) × ( w xz + w z + wxyz + x y + xy z ) q +1 . (16)It has been verified that w x − zxwy + y z = 0 above. Moreover, we havethat ( z + x )( w + y ) = 0. In fact, if x + z = 0 or w + y = 0 we can derivethat B ( X ) = D ( X ) = x y X = 0. From (16) we have w xz + w z + wxyz + x y + xy z = 0 . (17)Combining (17) and B ( X ) = 0 we obtain that yw ( x + xz + z )( X − x ) = 0 . This is a contradiction with that x + xz + z = 0 and X = x .So, from (15) we must have D ( X ) = 0. If the leading coefficient of D ( X )is nonzero, i.e., w x + w xz + zxwy + xy z + y z = 0 then0 = Res ( B, D, X ) = xwyz ( z + x ) ( w + y ) (cid:0) w x − zxwy + y z (cid:1) q +1 . This is impossible from the discussion in previous paragraph. Thus, theleading coefficient of D ( X ) is zero, i.e., L ( x ) = w x + w xz + zxwy + xy z + y z = 0 . (18)The fact that wx ( z + x ) ( w + y ) = 0 implies that the degree of D ( X ) is 1.The resultant of B ( X ) and D ( X ) is zero. Then0 = Res ( B ( X ) , D ( X ) , X ) = ( z + x ) xw ( w + y ) U ( x ) , (19)where U ( x ) = − yw x + (2 zw + yzw + 2 y zw x ) x + (3 z w + 3 yz w +6 y z w + 2 y z w + y z ) x + ( z w − yz w + y z w + y z ) x − (2 yz w + y z w + 2 y z w ) x + y z w . U ( x ) = 0. By calculation of the resultant of L ( x ) and U ( x ) we get0 = Res ( L ( x ) , U ( x ) , x ) = ( w + y ) ( w + y ) w z y . (20)In this case, it is verified that y + w = 0 above. From (20) we have w = − y ,i.e., y q − = − y q − = 1. Then y gcd(4( q − ,q − = y q − = 1.This is a contradiction. Therefore, we prove that B ( X ) can not be zero.(ii) Similarly, we can prove that C ( x ) in (13) is not zero. Thus, f ( x ) = α has only one root in T , and f ( x ) permutes T . (cid:3)
4. Permutation polynomials with the form ( x q − x + δ ) s + cx over F q In this section we give many classes of permutation polynomials havingthe form ( x q − x + δ ) s + cx without restriction on δ over F q . These per-mutation polynomials are derived from known permutation trinomials withthe form (1) by using Proposition 2. First, four families of infinite classes of permutation polynomials of theform ( x q − x + δ ) s + cx over F q with odd characteristic are derived frompermutation trinomials introduced in Section 3. Theorem 11.
Let δ ∈ F q and c ∈ F ∗ q . The polynomial f ( x ) = ( x q − x + δ ) q q − + cx is a permutation over F q in each of the following cases: (1) q ≡ and c = − ; (2) q ≡ and c = 2 . Proof.
Let s = q +2 q − and g ( x ) = x s . From Theorem 5 we know that h ( x ) = cx + g ( x ) s − g ( x ) qs permutes F q under the case of q ≡ − /c ) q +12 = 1 or the case of q ≡ /c ) q +12 = 1 respectively.These are exact cases listed in (1) and (2) respectively since c ∈ F q . ByProposition 2 we obtain that f ( x ) permutes F q under one of conditionslisted above. (cid:3) Similarly, one easily verifies the following three theorems.11 heorem 12.
Let δ ∈ F q and c ∈ F ∗ q . The polynomial f ( x ) = ( x q − x + δ ) ( q +1)24 + cx is a permutation over F q in each of the following cases: (1) q ≡ and c = 2 ; (2) q ≡ and c = − . Theorem 13.
Let q be a power of a prime with q ≡ . For δ ∈ F q ,the polynomial f ( x ) = ( x q − x + δ ) q q +13 + x is a permutation over F q . Theorem 14.
Let q be a power of an odd prime. For any δ ∈ F q , thepolynomial f ( x ) = ( x q − x + δ ) q + q − q + x is a permutation over F q .4.2. Permutation polynomials over finite fields with even characteristic By Proposition 2 many classes of permutation polynomials having theform ( x q + x + δ ) s + cx without restriction on δ are derived from knownpermutation trinomials having the form cx + x s + x qs over F q with evencharacteristic. To this end, we first recall known permutation trinomialswith the form cx + x s + x qs over finite fields with even characteristic in thefollowing lemmas. Lemma 15. [17] Let k be a positive integer and q = 2 k . The trinomials f ( x ) = cx + x s + x qs are permutation polynomials over F q in each of thefollowing cases. (1) s = 2 q − . The positive integer k and c ∈ F q satisfy one of thefollowing conditions: i) k is even and c = 1 ; ii) k is odd and c = 1 . (2) s = (3 q − q + q +1)3 , and k is even and c ∈ F q satisfies c = 1 . (3) s = q +45 , and k is odd and c ∈ F q satisfies c = 1 . (4) s = q +14 , and c ∈ F q satisfies that x + x + c = 0 having no solutionin F q . s = q +67 and c = 1 . (6) s = q +3 q +26 , and k is odd and c ∈ F q satisfies c q +13 = 1 . (7) s = q − q +43 , and k is even and c = 1 . (8) s = Q + Q − Q +12 and Q = 2 k for an even k , and c ∈ F ∗ Q . Lemma 16. [19][1] Let k be a positive integer and q = 2 k . The trinomials f ( x ) = cx + x s + x qs are permutation polynomials over F q in each of thefollowing cases. (1) s = − k ′ − (2 k −
1) + 1 , where k ′ is a positive integer with gcd(2 k ′ − , k + 1) = 1 and c ∈ F ∗ k ′ ∩ F q . (2) s = k ′ +1 (2 k −
1) + 1 , where k ′ is a positive integer with gcd(2 k ′ +1 , k + 1) = 1 and c ∈ F k ′ ∩ F q . Remark 17.
It is easy to see that each s in Lemmas 15 and 16 can berewritten as the form s = i ( q −
1) + 1 , here i is interpreted as modulo q + 1 when it is negative or proper fraction. By Proposition 2, the coefficient c of x in polynomials of the form (2)should be in F ∗ q , and the permutation polynomials in the following theoremare derived directly from permutation trinomials in Lemmas 15 and 16. Theorem 18.
Let k be a positive integer and q = 2 k . For δ ∈ F q , f ( x ) =( x q + x + δ ) s + cx permutes F q in each of the following cases: (1) s = 2 q − and c = 1 . (2) s = (3 q − q + q +1)3 and k is even, and c ∈ F q satisfies c = 1 . (3) s = q +45 , c = 1 and k is odd. (4) s = q +14 , and c ∈ F q satisfies that x + x + c = 0 having no solutionin F q . (5) s = q +67 and c = 1 . (6) s = q +3 q +26 , c = 1 and k is odd. (7) s = q − q +43 , c = 1 and k is even. (8) s = Q + Q − Q +12 and Q = 2 k for an even k , and c ∈ F ∗ Q . (9) s = − k ′ − (2 k −
1) + 1 , and k ′ is a positive integer with gcd(2 k ′ − , k +1) = 1 , and c ∈ F k ′ ∩ F q . (10) s = k ′ +1 (2 k −
1) + 1 , and k ′ is a positive integer with gcd(2 k ′ + 1 , k +1) = 1 , and c ∈ F k ′ ∩ F q . emark 19. The permutation polynomial in case (1) has been proposed inTheorem 1 of [26] and Theorem 3.2 of [29]. The permutation polynomial incase (3) has been constructed in [10]. The permutation polynomial in case (4) has been constructed in Theorem 2.1 of [37] and Theorem 3.1 of [29]. Eventhough the representation of permutation polynomials having the form (2) inthose papers may be different, they have the same corresponding trinomialsof the form (1) if s is written as i ( q −
1) + 1 and i is interpreted as modulo q + 1 when it is negative or proper fraction. As far as we know, Table 1presents all permutation polynomials having the form ( x q + x + δ ) s + cx without restriction on δ over F q with even characteristic, where Ω = { c ∈ F q | x + x + c = 0 has no roots in F q } . Table 1: PPs of the form ( x k + x + δ ) s + cx without restriction on δ over F k No. k s = i (2 k −
1) + 1 c Reference1 k is even i = 2 or − c = 1 [26][29]2 all k i = 0 or 1 c = 1 [29][31]3 all k i = c = 1 [29]4 k is even i = or c = 1 [30]5 k is odd i = or c = 1 [10]6 a positive integer i = or c ∈ Ω [37][29]7 k is even i = k − or − k − c = 1 Theorem 188 a positive integer i = or c = 1 Theorem 189 k is odd i = k +46 or − k c q +13 = 1 Theorem 1810 k is even i = k − or − k c = 1 Theorem 1811 k is even i = k/ +12 or − k/ c ∈ F ∗ k/ Theorem 1812 gcd(2 k ′ − , k + 1) = 1 i = − k ′ − or k ′ k ′ − c ∈ F ∗ k T F k ′ Theorem 1813 gcd(2 k ′ + 1 , k + 1) = 1 i = k ′ +1 or k ′ k ′ +1 c ∈ F ∗ k T F k ′ Theorem 18
5. Concluding remark
In this paper, we proposed four families of infinite classes of permutationtrinomials with the form (1). From known permutation trinomials with theform (1), many classes of permutation polynomials with the form (2) arederived. As part of the future work, observe that the relationship betweenthe two types of polynomials can be generalized to the case of multiple14onomials. It would be interesting to find more permutation polynomialswith the form (1) or (2) by using this relation.
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