arXiv:1606.03768v1 [cs.IT] 12 Jun 2016 inL@iubn;[email protected]. [email protected]; Let Introduction 1 rnmasfo ioepnnsoe nt ed ihee character even with fields finite over exponents Niho from trinomials fields. p finite recent over four trinomials and and binomials su of the in constructions found the cla be of can specific which of literature, works number the in small nice described a are some only trinomials currently and However, form, 23]. algebraic 18, 10, simple [3, their to trinomia due t and years binomials referred recent Permutation is example. reader for and the therein 23] property, 19, references 18, desired 13, certain 10, 8, with [3, to polynomials referred construction is recent reader some the For form, simple problem. research interesting an is mathe of designs. areas combinatorial the and in cryptography applications theory, wide coding have as [2], Dickson and [6] Hermite of permutation a is emtto oyoilof permutation a e emtto rnmasFo ioEpnnsoe Finite over Exponents Niho From Trinomials Permutation New ∗ nti ae,floigteln ftewrsddi 1,1] efurther we 11], [18, in did works the of line the following paper, this In form simple a either with polynomials permutation of construction The h uhr r ihteDprmn fIfrais Univers Informatics, of Department the with are authors The p h eeaiain fsm ale works. earlier it some and of fields, generalizations finite the over polynomials fractional linear from ceitci ute netgtd e emtto trino permutation New investigated. further is acteristic eapieand prime a be nti ae,acaso emtto rnmaso iotype Niho of trinomials permutation of class a paper, this In F F p p m m 1] emtto oyoil vrfiiefils hc eefis studie first were which fields, finite over polynomials Permutation [15]. eoetefiiefil with field finite the denote ilswt vnCharacteristic Even with Fields F p m fteascae oyoilfunction polynomial associated the if inL n o Helleseth Tor and Li Nian Abstract 1 p m lmns polynomial A elements. il rmNh xoet r obtained are exponents Niho from mials t fBre,N52 egn owy Email: Norway. Bergen, N-5020 Bergen, of ity ssonta h rsne eut are results presented the that shown is o oecntutoso permutation of constructions some for fpruainplnmaswt a with polynomials permutation of s pr 3 1 3 2 npermutation on 22] 13, 11, [3, apers vrfiiefilswt vnchar- even with fields finite over ∗ 1 ,5 2 0 1 2 n the and 22] 21, 20, 12, 5, 4, [1, o satatrsaces teto in attention researchers’ attract ls vyppr[]o h developments the on [9] paper rvey sso emtto ioil and binomials permutation of sses si.Peiey let Precisely, istic. nte a enahee in achieved been had them on f aisadegneigsuch engineering and matics : rcrandsrdproperty desired certain or c 7→ td h permutation the study f f ( ( x c ) from ) ∈ F p n m F [ 2 = x p m scalled is ] m to by d ea be F p m positive integer and p = 2, we consider the trinomials over the finite field F2n of the form

m m f(x)= x + xs(2 −1)+1 + xt(2 −1)+1, (1) where s,t are integers satisfying 1 ≤ s,t ≤ 2m. For simplicity, if the integers s,t are written as fractions m 1 3 or negative integers, then they should be interpreted as modulo 2 + 1. For instance, (s,t) = ( 2 , 4 )= (2m−1 +1, 2m−2 + 1). The objective of this paper is then to find new pairs (s,t) such that the trinomial f(x) defined by (1) is a permutation polynomial. Based on the analysis of certain equations over the

finite field F2n , some new classes of permutation trinomials over F2n with the form (1) are constructed. The presented new classes of permutation trinomials of the form (1) are determined by making use of the property of the linear fractional polynomials over the finite field F2n , which is a completely different approach, and it will be seen that many permutation trinomials of the form (1) can be obtained from them since their parameters on (s,t) are flexible.

2 Preliminaries

n Throughout this paper, let n =2m and F2n denote the finite field with 2 elements. A positive integer d j m is called a Niho exponent with respect to the finite field F2n if d ≡ 2 (mod 2 −1) for some nonnegative integer j. When j = 0, the integer d is then called a normalized Niho exponent. The Niho exponents were originally introduced by Niho [16] who investigated the cross-correlation between an m-sequence and its d-decimation. 2m For simplicity, denote the conjugate of x ∈ F2n over F2m by x, i.e., x = x . The unit circle of F2n is defined as follows: 2m+1 U = {x ∈ F2n : xx = x =1}.

Notice that the trinomial f(x) defined by (1) can be written as the form xrg(xs) for some integers r, s and g(x) ∈ F2n [x]. For this kind of polynomials, its permutation property had been characterized by the following lemma which was proved by Park and Lee in 2001 and reproved by Zieve in 2009.

Lemma 1. ([17, 24]) Let p be a prime, n a positive integer and g(x) ∈ Fpn [x]. If d,s,r > 0 such that n r s p − 1= ds, then x g(x ) is a permutation over Fpn if and only if

1) gcd(r, s) =1, and

r s 2) h(x)= x g(x) permutes the d-th roots of unity in Fpn .

In the sequel, we will use the property of the linear fractional polynomial to prove our main results. ax+b F n For the linear fractional polynomial φ(x)= cx+d with a,b,c,d ∈ 2 and ad − bc 6= 0, it can be readily

2 verified that φ(x) defines a permutation on F2n ∪ {∞}, where

a a∞ + b a a∞ + b = ∞ if a 6= 0; = if c 6= 0; = ∞ if c = 0 and ad 6=0. 0 c∞ + d c c∞ + d

Then, by utilizing the fact that the composition of any two linear fractional polynomial is still a linear fractional polynomial, we can prove

3 ax+(a+1) 2k Lemma 2. Let a ∈ U with a 6= 1 and φ(x)= (a+1)x+1 . If φ(x)= x holds for some positive integer k. Then for any i ≥ 2 we have

2k 2ik x, if ei−1 + e1 +1=0, x = k eix+1 , if e2 + e +1 6=0, ( x+ei+1 i−1 1

a 2k 2k where e1 = a+1 , ei = ∞ if ei−1 + e1 +1 = 0 and otherwise ei = φ(ei−1) for any i ≥ 2.

Proof. Note that φ(x) = ax+(a+1) = e1x+1 , where e = a . Taking 2k-th power on both sides of (a+1)x+1 x+e1+1 1 a+1

k x2 = φ(x) gives

2k 2k 2k e x +1 x2 = 1 . 2k 2k x + e1 +1

k Then, substituting x2 by φ(x)= e1x+1 , we can further obtain x+e1+1

2k+1 2k 2k (e + 1)x + e + e +1 e x +1 x2 = 1 1 1 = 2 2k 2k+1 2k x + e +1 (e1 + e1 + 1)x + e1 + e1 + e1 2

k k 2 +1 k k 2k 2 2 e1 +1 2 2 2 (e1+e1+1)x if e + e1 +1 6= 0, where e2 = k = φ(e ). If e + e1 + 1 = 0, then x = 2 = x 1 2 1 1 e +e1+1 (e1 +e1+1) 1 2 3 a 3 2 3 since e1 + e1 +1 6= 0. Otherwise we have e1 = ( a+1 ) = 1, i.e., a + a + 1 = 0 which implies that a = 1, a contradiction with a3 6= 1. By using this, we then can show by induction on i that

2k 2ik x, if ei−1 + e1 +1=0, x = k (2) eix+1 , if e2 + e +1 6=0, ( x+ei+1 i−1 1

2k 2k where ei = ∞ if ei−1 + e1 +1 = 0 and otherwise ei = φ(ei−1) for any i ≥ 2.

2(i+1)k 2k Suppose that (2) holds for i. Then for the case of i + 1 we have x = x = φ(x) if ei = ∞ and

k k 2 2 e1x+1 ei+1 = φ(e )= φ(∞)= e1 which coincides with (2). If ei 6= ∞, then by x = φ(x)= and (2) i x+e1+1 we have

2k 2k 2k 2k (i+1)k e x +1 (e e + 1)x + (e + e + 1) x2 = i = 1 i i 1 . 2k 2k 2k 2k 2k x + ei +1 (ei + e1 + 1)x + e1ei + ei + e1

3 2k 2k 2k 2k 2 If ei + e1 + 1 = 0, then e1ei + ei + e1 = e1ei +1= e1 + e1 +1 6= 0 as shown before. This means 2(i+1)k 2k that x = x and ei+1 = ∞. If ei + e1 +1 6= 0, then it can be readily verified that

2k 2k 2k e e +1 k e e + e + e e = 1 i = φ(e2 ), 1 i i 1 = e +1. i+1 2k i 2k i+1 ei + e1 +1 ei + e1 +1

Therefore, (2) holds for any i ≥ 2. This completes the proof. 

3 New Permutation Trinomials From Niho Exponents

In this section, we present two new classes of permutation polynomials over F2n of the form (1), namely

m m f(x)= x + xs(2 −1)+1 + xt(2 −1)+1, where n =2m and 1 ≤ s,t ≤ 2m.

By using the property of the linear fractional polynomials over F2n , we can obtain new classes of permutation trinomials of the form (1) as below.

Theorem 1. Let n = 2m and gcd(2k − 1, 2m +1) = 1, where m, k are positive integers with k

2k −1 Then the trinomial f(x) defined by (1) is a permutation if (s,t) = ( 2k−1 , 2k−1 ).

Proof. According to Lemma 1, to complete the proof, it is sufficient to prove that h(x) = x(1 +

m− 2k 1 s t 2 1 F n x + x ) with (s,t) = ( 2k −1 , − 2k−1 ) permutes the unit circle U of 2 , which is equivalent to

k k k k m k k m showing that h(x2 −1) = x2 −1(1 + xs(2 −1) + xt(2 −1))2 −1 = x2 −1(1 + x2 + x−1)2 −1 permutes

k U since gcd(2k − 1, 2m + 1) = 1. Notice that x2 +1 + x +1 6= 0 for any x ∈ U. Otherwise, from

k m k m m k x2 +1 + x +1=0 and x2 +1 + 1 = 0 we have x2 +1 + x + x2 +1 = 0, i.e., x2 = x2 + 1 which leads to

k m m k k k 2k k x = (x2 + 1)2 = (x2 + 1)2 = (x2 )2 = x2 . Taking 2k-th power on both sides of x2 +1 + x +1=0

2k k k 2k k k gives x2 +2 + x2 +1 = 0. Then by x2 = x and x2 +1 + x + 1 = 0 we have x2 = x which implies

k m that x = 1 due to x2 −1 = 1, x2 +1 = 1 and gcd(2k − 1, 2m +1) = 1, a contradiction. Thus, we arrive

k k k k m at x2 +1 + x +1 6= 0 for any x ∈ U, and then, h(x2 −1)= x2 −1(1 + x2 + x−1)2 −1 can be rewritten

k 2k+1 2k as h(x2 −1)= x +x +1 . x2k +1+x+1

k k 2k −1 x2 +1+x2 +1 F We next prove that h(x )= k permutes the unit circle of 2n . For any given a ∈ U, x2 +1+x+1

k k we show that h(x2 −1)= a has at most one solution in U. Note that h(x2 −1)= a can be expressed as

k ax + (a + 1) x2 = . (3) (a + 1)x +1

We then can discuss it as follows:

4 k 1) a =1. If a = 1, then (3) is reduced to x2 = x which has exactly one solution x = 1 in U since gcd(2k − 1, 2m + 1) = 1.

2) a = (a + 1)2, i.e., a2 + a + 1 = 0. This case happens only if m is odd since a ∈ U. Then, (3) can

k be reduced to x2 = a +1= a2 which has exactly one solution in U when this case happens.

3) a3 6= 1. For this case, if (3) holds for some x ∈ U, then by Lemma 2, we have that (2) holds for

2ik 1 the same x for any i ≥ 2. If k is odd, then x = x if one takes i = m. Thus, by (2) we have either x = 1 or emx+1 = 1 , it then can be readily seen that no matter which case there exists x x+em+1 x j at most one such x. If k is even, assume that k = 2 k1 for some positive j and odd k1, then we j k m k m can conclude that 2 |m since gcd(2 − 1, 2 + 1) = 1 if and only if gcd(m,k) is odd. Take i = 2j

2ik 2mk1 1 in (2), we have x = x = x since k1 is odd. Therefore, the proof can be completed as the case of k is odd.

This completes the proof.  If one takes k = 1, then Theorem 1 generalizes Theorem 3.2 in [3].

Corollary 1. Let n = 2m for a positive integer m. Then the trinomial f(x) defined by (1) is a permutation if (s,t)=(2, −1) = (2, 2m).

If one takes k = 2, then Theorem 1 generalizes Theorem 3 in [11].

Corollary 2. Let n =2m for an even integer m. Then the trinomial f(x) defined by (1) is a permutation

4 −1 2m+5 2m+1+1 if (s,t) = ( 3 , 3 ) = ( 3 , 3 ).

Theorem 2. Let n = 2m and gcd(2k +1, 2m +1) = 1, where m, k are positive integers. Then the

1 2k trinomial f(x) defined by (1) is a permutation if (s,t) = ( 2k+1 , 2k +1 ).

m Proof. According to Lemma 1, we need to show that h(x) = x(1 + xs + xt)2 −1 permutes the unit

1 2k F n k m circle U of 2 if (s,t) = ( 2k +1 , 2k +1 ). Since gcd(2 +1, 2 + 1) = 1, then it suffices to prove that

k k k k m k k m h(x2 +1)= x2 +1(1 + xs(2 +1) + xt(2 +1))2 −1 = x2 +1(1 + x + x2 )2 −1 permutes U. First, we show

k k m that x2 + x +1 6= 0 for any x ∈ U. Suppose that x2 + x +1=0 and x2 +1 +1 = 0, then one gets

2k 2m 1 2m+k 2m 1 2k 2k x = x + 1 and x = x which leads to x = (x + 1) = ( x ) . This together with x + x +1=0 2m 1 2m+1 2 implies that (x + 1) = x+1 , i.e., (x + 1) = 1, and then, we obtain that x + x + 1 = 0 due to m x2 +1 + 1 = 0. However, this is impossible if m is even since x2 + x + 1 = 0 has no solution in U for even m. If m is odd, then x2 + x + 1 = 0 has two solutions in U satisfying x3 = 1. Note that k m m k gcd(2 +1, 2 + 1) = 1 if and only if one of gcd(m,k) and gcd(m,k) is even which means that k is even

5 k k and 2k ≡ 1 (mod 3). This leads to x2 + x +1= x + x + 1 = 1, a contradiction. Hence, x2 + x +1 6=0

k k k m k 2k+1 2k for any x ∈ U, and h(x2 +1)= x2 +1(1 + x + x2 )2 −1 can be expressed as h(x2 +1)= x +x +x . x2k +x+1

k k 2k +1 x2 +1+x2 +x F We next prove that h(x )= k permutes the unit circle of 2n . For any given a ∈ U, x2 +x+1

k k we show that h(x2 +1)= a has at most one solution in U. Note that h(x2 +1)= a can be written as

k (a + 1)x + a x2 = . x + (a + 1)

Taking 2m-th power on both sides of the above equation gives

m+k (a + 1)x + a ax + (a + 1) x2 = = . (4) x + (a + 1) (a + 1)x +1

Let k′ = m + k and a′ = a. Thus, by Lemma 2, if a′3 6= 1, then for any i ≥ 2 we have

′ k ′ ′2 ′ ik x, if e i−1 + e1 +1=0, 2 ′ x = ′ k (5)  eix+1 ′2 ′ ′ , if e i−1 + e1 +1 6=0,  x+ei+1

′ ′ k  k ′ ′2 ′ ′ ′2 where ei = ∞ if e i−1 + e1 +1 = 0 and otherwise ei = φ(e i−1) for any i ≥ 2. Thus, similar as the proof of Theorem 1, we can discuss it as follows:

m+k k 1) a′ =1. If a′ = 1, then (4) is reduced to x2 = x, i.e., x2 +1 = 1, which has exactly one solution x = 1 in U since gcd(2k +1, 2m + 1) = 1.

2) a′ = (a′ +1)2, i.e., a′2 +a′ +1 = 0. This case happens only if m is odd since a′ ∈ U. Then, (4) can

m+k be reduced to x2 = a′ +1= a′2 which has exactly one solution in U when this case happens.

3) a′3 6= 1. For this case, if (4) holds for some x ∈ U, then by Lemma 2, we have that (5) holds for ′ ′ 2ik 1 the same x for any i ≥ 2. If k is odd, then x = x if one takes i = m. Thus, by (5) we have ′ 1 emx+1 1 ′ either x = or ′ = , which implies that there exists at most one such x. If k is even, x x+em+1 x ′ ′ j ′ ′ ′ then we can claim that m is even. Moreover, let k = 2 k1 for some positive j and odd k1, we ′ have 2j |m. This can be verified by the fact that gcd(2k +1, 2m + 1) = 1 if and only if one of ′ m k m k m+k k gcd(m,k) and gcd(m,k) is even, i.e., gcd(m,k) + gcd(m,k) = gcd(m,k) = gcd(m,k) is odd. Therefore, for

′ ′ ′ ′ j ′ m 2ik 2mk1 1 ′ even k =2 k , we can take i = ′ in (5) and then x = x = since k is odd. Then, the 1 2j x 1 proof can be completed as the case of k′ is odd.

6 This completes the proof.  If one takes k = 1, then Theorem 2 shows that the trinomial f(x) defined by (1) is a permutation

1 2k 1 2 for even m if (s,t) = ( 2k +1 , 2k+1 ) = ( 3 , 3 ). However, this result is covered by Theorem 3.4 in [3] up to equivalence (see Table 1). If one takes k = 2, then Theorem 2 generalizes Theorem 6 in [11].

Corollary 3. Let n = 2m satisfy gcd(5, 2m + 1) = 1. Then the trinomial f(x) defined by (1) is a 1 4 permutation if (s,t) = ( 5 , 5 ).

Table 1: Known pairs (s,t) such that f(x) defined by (1) are permutation polynomials

No. (s,t) h(x) Conditions Equivalent Pairs Proven in ±k ±2k 1 (k, −k) x see Thm. 3.4 in [3] ( 2k∓1 , 2k∓1 ) [3] x3+x2+1 1 2 2 (2, −1) x3+x+1 positive m (1, 3 ), (1, 3 ) [3] 1 x(x3+x2+1) 3 1 3 3 (1, − 2 ) x3+x+1 gcd(3,m)=1 (1, 2 ), ( 4 , 4 ) [13, 14, 7] 1 4 x5+x4+1 1 4 4 (− 3 , 3 ) x5+x+1 m even (1, 5 ), (1, 5 ) [11, 14] x4+x3+1 3 4 1 4 5 (3, −1) x(x4+x+1) m even ( 5 , 5 ), ( 3 , 3 ) [11, 14] 2 5 x7+x5+1 2 5 6 (− 3 , 3 ) x7+x2+1 m even (1, 7 ), (1, 7 ) [11] 4 3 1 4 x(x +x +1) m 1 4 7 ( 5 , 5 ) x4+x+1 gcd(5, 2 +1)=1 (1, − 3 ), (1, 3 ) [11, 14, 7] 1 x(x5+x+1) 2 5 1 5 8 (2, − 2 ) x5+x4+1 m ≡ 2, 4 (mod 6) ( 3 , 6 ), ( 4 , 4 ) [14] x6+x4+1 2 5 1 5 9 (4, −2) x(x6+x2+1) gcd(3,m)=1 ( 3 , 6 ), ( 4 , 4 ) [14] k 2k+1 2k k 10 ( 2 , −1 ) x +x +1 gcd(2k − 1, 2m +1)=1 (1, 1 ), (1, 2 ) Theorem 1 2k −1 2k −1 x2k +1+x+1 2k+1 2k+1 k 2k+1 2k k 11 ( 1 , 2 ) x +x +x gcd(2k +1, 2m +1)=1 (1, 2 ), (1, −1 ) Theorem 2 2k +1 2k +1 x2k +x+1 2k−1 2k−1

To end this section, we list all the known pairs (s,t) such that the polynomials of the form (1) are in Table 1 and we claim that all the permutation trinomials obtained in this paper are multiplicative inequivalent with the known ones. Notice that the inverse of a normalized Niho exponent, if exists, is still a normalized Niho exponent and the product of two normalized Niho exponents is also a normalized Niho exponent. With this fact, it can be readily checked that the permutation trinomials obtained in this paper are multiplicative inequivalent with the known permutation trinomials listed in Theorem 1 and Table 1 in [11]. Theorem 1 in [11] listed all the known permutation trinomials over

F2n for even n and Table 1 in [11] presented all the known permutation trinomials of the form (1). The “Equivalent Pairs” column in Table 1 are obtained based on Lemma 4 in [11], which also leads to permutation trinomials of the form (1) if they exist. Note that the fractional polynomial in No. 10 covers those of in Nos. 2 and 4 as special cases and the fractional polynomial in No. 11 covers that of

7 in No. 7 as a special case.

Further, it also should be noted that trinomial permutations over F2n with a more general form r s(2m−1)+r t(2m−1)+r fr,s,t(x) = x + x + x can also be obtained by using the same techniques as in [11] and this paper if the parameters r,s,t are suitably chosen based on Lemma 1. Lemma 1 implies that the determination of the permutation property of fr,s,t(x) over F2n is equivalent to that of the permutation

− − m− xr+xr s+xr t r s t 2 1 F n property of hr,s,t(x) = x (1 + x + x ) = 1+xs+xt over the unit circle of 2 . From the viewpoint of Lemma 1, the trinomials fr1,s1,t1 (x) and fr2,s2,t2 (x) are viewed as the same in Table 1 for different integer tuples (r1,s1,t1) and (r2,s2,t2) if hr1,s1,t1 (x) = hr2,s2,t2 (x) although fr1,s1,t1 (x) and fr2,s2,t2 (x) are multiplicative inequivalent (anyway one of them can be easily obtained from the other). With this observation, the two conjectures proposed in [7] can be settled by using the fifth and fourth fractional permutation polynomials in Table 1 over the unit circle of F2n proved in [11, 14] .

4 Conclusion Remarks

In this paper new classes of permutation trinomials over F2n with the form (1) were obtained from Niho exponents by using the property of linear fractional polynomials and some techniques in solving equations over finite fields. It was shown that the presented results generalized some earlier works in [3, 7, 11, 14]. It is interesting to generalize our method in general or find new ideas to prove the permutation property of fractional polynomials over finite fields in order to obtain more permutation trinomials.

References

[1] C. Bracken, C.H. Tan and Y. Tan, Binomial differentially 4 uniform permutations with high non- linearity, Finite Fields Appl. 18(3)(2012), pp. 537-546.

[2] L.E. Dickson, The analytic representation of substitutions on a power of a prime number of letters with a discussion of the linear , Ann. of Math., 11 (1896), pp. 65-120.

[3] C. Ding, L. Qu, Q. Wang, J. Yuan and P. Yuan, Permutation trinomials over finite fields with even , SIAM J. Dis. Math, 29 (2015), pp. 79-92.

[4] H. Dobbertin, Almost perfect nonlinear power functions on GF (2n): The Welch case, IEEE Trans Inform. Theory, 45(1999), pp. 1271-1275.

[5] H. Dobbertin, Almost perfect nonlinear power functions on GF (2n): the Niho case, Information and Computation, 151(1-2)(1999), pp. 57-72.

[6] Ch. Hermite, Sur les fonctions de sept lettres, C. R. Acad. Sci. Paris, 57 (1863), pp. 750-757.

8 [7] Rohit Gupta, R.K. Sharma, Some new classes of permutation trinomials over finite fields with even characteristic, Finite Fields Appl. 41 (2016), pp. 89-96.

[8] X. Hou, A class of permutation trinomials over finite fields, Acta Arith. 162 (2014), pp. 51-64.

[9] X. Hou, Permutation polynomials over finite fields–A survey of recent advances, Finite Fields Appl. 32 (2015), pp. 82-119.

[10] X. Hou, Determination of a type of permutation trinomials over finite fields, II, Finite Fields Appl. 35 (2015), pp. 16-35.

[11] N. Li and T. Helleseth, Several classes of permutation trinomials from Niho exponents, submitted.

[12] N. Li, T. Helleseth, and X. Tang, Further results on a class of permutation polynomials over finite fields, Finite Fields Appl. 22(2013), pp. 16-23.

[13] K. Li, L. Qu and X. Chen, New classes of permutation binomials and permutation trinomials over finite fields, available online: http://arxiv.org/pdf/1508.07590.pdf.

[14] K. Li, L. Qu, C. Li and S. Fu, New permutation trinomials constructed from fractional polynomials, available online: https://arxiv.org/pdf/1605.06216v1.pdf

[15] R. Lidl, H. Niederreiter, Finite Fields, 2nd ed. Cambridge Univ. Press, Cambridge, 1997.

[16] Y. Niho, Multivalued cross-correlation functions between two maximal linear recursive sequence, Ph.D. dissertation, Univ. Southern Calif., Los Angeles, 1972.

[17] Y. H. Park and J. B. Lee, Permutation polynomials and group permutation polynomials, Bull. Austral. Math. Soc. 63 (2001), pp. 67-74.

[18] Z. Tu, X. Zeng, L. Hu and C. Li, A class of binomial permutation polynomials, available online: http://arxiv.org/pdf/1310.0337v1.pdf.

[19] Z. Tu, X. Zeng and L. Hu, Several classes of complete permutation polynomials, Finite Fields and Appl., 25 (2014), pp. 182-193.

k [20] X. Zeng, X. Zhu, and Lei Hu, Two new permutation polynomials with the form (x2 + x + d)s + x

over F2n . Appl. Algebra Eng. Commun. Comput. 21(2)(2010), pp. 145-150.

[21] X. Zeng, S. Tian, and Z. Tu, Permutation polynomials from trace functions over finite fields, Finite Fields Appl. 35(2015), pp. 36-51.

[22] X. Zhu, X. Zeng, and Y. Chen, Some binomial and trinomial differentially 4-uniform permutation polynomials, Int. J. Found. Comput. Sci. 26(4)(2015), pp, 487-498.

9 [23] M. Zieve, Permutation polynomials on Fq induced form R´edei function on subgroups of F∗ q , available online: http://arxiv.org/pdf/1310.0776v2.pdf.

r (q−1)/d [24] M. Zieve, On some permutation polynomials over Fq of the form x h(x ), Proc. Amer. Math. Soc. 137 (2009), pp. 2209-2216.

10