Reversed Dickson permutation polynomials over finite fields

Transcription

1 Reversed Dickson permutation polynomials over finite fields Department of Mathematics Northeastern University Boston Algebra, Number Theory and Combinatorics Seminar Claremont Center for the Mathematical Sciences Claremont, CA March 6, 2018

2 Introduction Let p be a prime and q a power of p. Let F q be the finite field with q elements. A polynomial f F q [x] is called a permutation polynomial (PP) of F q if the associated mapping x f (x) from F q to F q is a permutation of F q. Example 1. Every linear polynomial is a PP of F q. Example 2. The monomial x n is a PP of F q if and only if (n, q 1) = 1.

3 Background The n-th Dickson polynomial of the first kind D n (x, a) is defined by D n (x, a) = n 2 i=0 where a F q is a parameter. n n i ( n i i ) ( a) i x n 2i, When a = 0, D n (x, a) is a PP of F q if and only if (n, q 1) = 1. When a 0, D n (x, a) is a PP of F q if and only if (n, q 2 1) = 1.

4 Background (contd.) The n-th reversed Dickson polynomial of the first kind D n (a, x) is defined by D n (a, x) = n 2 i=0 where a F q is a parameter. n n i ( n i i ) ( x) i a n 2i, X. Hou, G. L. Mullen, J. A. Sellers, J. L. Yucas, Reversed Dickson polynomials over finite fields, Finite Fields Appl. 15 (2009),

5 Background (contd.) The n-th Dickson polynomial of the second kind E n (x, a) can be defined by n 2 ( ) n i E n (x, a) = ( a) i x n 2i, i i=0 where a F q is a parameter. Stephen D. Cohen (University of Glasgow, UK) Mihai Cipu (The Institute of Mathematics of the Romanian Academy) Rex Matthews (University of Queensland, Australia) Robert S. Coulter (University of Delaware, USA) Marie Henderson (New Zealand)

6 Background (contd.) The n-th reversed Dickson polynomial of the second kind E n (a, x) can be defined by n 2 ( ) n i E n (a, x) = ( x) i a n 2i, i i=0 where a F q is a parameter. S. Hong, X. Qin, W. Zhao, Necessary conditions for reversed Dickson polynomials of the second kind to be permutational, Finite Fields Appl. 37 (2016),

7 Background (contd.) For a F q, the n-th Dickson polynomial of the (k + 1)-th kind D n,k (x, a) is defined by D n,k (x, a) = and D 0,k (x, a) = 2 k. n 2 i=0 n ki n i ( n i i ) ( a) i x n 2i, Q. Wang, J. L. Yucas, Dickson polynomials over finite fields, Finite Fields Appl. 18 (2012),

8 Background (contd.) For a F q, the n-th reversed Dickson polynomial of the (k + 1)-th kind D n,k (a, x) is defined by D n,k (a, x) = and D 0,k (a, x) = 2 k. n 2 i=0 n ki n i ( n i i ) ( x) i a n 2i, Q. Wang, J. L. Yucas, Dickson polynomials over finite fields, Finite Fields Appl. 18 (2012),

9 Reversed Dickson Polynomials of the (k + 1)-th kind For a F q, the n-th reversed Dickson polynomial of the (k + 1)-th kind D n,k (a, x) is defined by D n,k (a, x) = and D 0,k (a, x) = 2 k. n 2 i=0 n ki n i ( n i i ) ( x) i a n 2i, I am primarily interested in the question: When is D n,k (a, x) a PP of F q? D n,0 (a, x) = D n (a, x) and D n,1 (a, x) = E n (a, x). Only need to consider 0 k p 1. D n,k (a, x) = ke n (a, x) (k 1)D n (a, x).

10 D n,k (a, x) in characteristic 2 D n,k (a, x) = ke n (a, x) (k 1)D n (a, x). D n,k (a, x) = { E n (a, x) D n (a, x) if k is odd, if k is even. Hereafter always assume, unless specified, in this talk that p is odd.

11 Outline of the rest of the talk 1. The case a = 0 2. Some general properties of D n,k (a, x). 3. The case n = p l, where l 0 is an integer. 4. The case n = p l + 1, where l 0 is an integer. 5. The case n = p l + 2, where l 0 is an integer. 6. An explicit expression for D n,k (1, x). 7. The sum a F q D n,k (1, a). 8. The case n = p l + 3, where l 0 is an integer. 9. A generalization to n = p l 1 + p l p l i. 10. Permutation behaviour of D p l 1 +p l 2,k. 11. More results on D n,k (1, x). 12. A Matrix Form of D n,k (1, x).

12 The case a = 0 D n,k (0, x) = { 0 if n is odd, (2 k) ( x) l if n = 2l. Theorem When a = 0, D n,k (a, x) is a PP of F q if and only if k 2 and n = 2l with (l, q 1) = 1. Hereafter, assume that a F q.

13 Why only consider a = 1? Let a F q. Then it follows from the definition that D n,k (a, x) = a n D n,k (1, x a 2 ). D n,k (a, x) is a PP on F q if and only if D n,k (1, x) is a PP on F q.

14 Functional expression [ y n+1 (a y) n+1 ] D n,k (a, x) = k (k 1){y n + (a y) n }, 2y a where y a 2. This can be simplified to [ y n (a y) y(a y) n ] D n,k (a, x) = k + D n (a, x). 2y a

15 Functional expression with a = 1 Let x = y(1 y). The functional expression can be written as [ y n (1 y) y(1 y) n ] D n,k (1, y(1 y)) = k + D n (1, y(1 y)), 2y 1 where y 1 2. When y = 1 2, D n,k ( ) 1, 1 4 = k(n 1)+2 2 n. if n 1 n 2 (mod q 2 1), then D n1,k(1, x) = D n2,k(1, x) for any x F q \ { 1 4 }.

16 Recursion Proposition Let p be an odd prime and n be a non-negative integer. Then D 0,k (1, x) = 2 k, D 1,k (1, x) = 1, and D n,k (1, x) = D n 1,k (1, x) x D n 2,k (1, x), for n 2.

17 The Case n = p l, l 0 is an integer Let x = y(1 y). When y 1 2, [ y p l (1 y) y(1 y) pl D p l,k(1, y(1 y)) = k 2y 1 ((2y 1) 2) pl 1 2 = k 2 Note that (2y 1) 2 = 1 4x. Hence, + 1 k 2 ] + D p l(1, y(1 y)) D p l,k(1, x) = k 2 (1 4x) pl k 2

18 The case n = p l, l 0 is an integer (contd.) When y = 1 2, D p l,k ( 1, 1 ) 4 = k(pl 1) + 2 = 2 k 2 pl 2 Hence for all x F q, we have = 1 k 2 = k 2 (1 4x) pl k 2. D p l,k(1, x) = k 2 (1 4x) pl k 2

19 Theorem Consider D p l,k(1, x) = k 2 (1 4x) pl k 2. Theorem Let 0 < l e. Then D 3 l,k(1, x) is a PP of F 3 e if and only if k 0 and ( 3l 1 2, 3e 1) = 1. Also, D p l,k(1, x) is not a PP of F p e when p > 3.

20 The case n = p l + 1, l 0 is an integer For all x F q, we have D p l +1,k(1, x) = ( 1 2 k 4 ) (1 4x) pl k 4 (1 4x) pl Theorem Let k = 2. Then D p l +1,k(1, x) is a PP of F q if and only if ( ) p l 1 2, q 1 = 1. Theorem Let n = p l + 1 and k 0, 2. Then D n,k (1, x) is a PP of F q if and only if l = 0. Remark Let k = 0. Then, D p l +1,k(1, x) = 1 2 (1 4x) pl +1 ( ) of F q if and only if p l +1 2, q 1 = which is a PP

21 The case n = p l + 2, l 0 is an integer For all x F q, we have D p l +2,k(1, x) = 1 2 (1 4x) pl k ( 2 x (1 4x) pl k ) 1 x Remark Let k = 0 and l = e. Then we have D p e +2,0(1, x) = 1 2 (1 4x) pe +1 2 x which is a PP of F p e if and only if p e 1 (mod 3). Let u = 1 4x. Then we have D p l +2,k(1, x) = ( 1 2 k ) u pl k ( 8 8 u pl k ) u k

22 The case n = p l + 2, l 0 is an integer (contd.) D p l +2,k(1, x) = ( 1 2 k ) u pl k ( 8 8 u pl k ) u k Theorem Let k = 2. Then D p l +2,k(1, x) = x pl x pl 1 2 is a PP of F q if and only if l = 0. Theorem Let p > 3 and k = 4. Then D p l +2,k(1, x) = x pl x is a PP of F q if and only if l = 0. Theorem Let n = p l + 2 and k 0, 2, 4. Then D n,k (1, x) is a PP of F q if and only if the trinomial (4 k) x pl k x pl (2 k)x is a PP of F q.

23 Theorem Theorem Let p be an odd prime. Then D n,k (1, x) is a PP of F q if and only if the function y k y n (1 y) y(1 y) n 2y 1 is a 2-to-1 mapping on (F q V ) \ { 1 2 } and + y n + (1 y) n k y n (1 y) y(1 y) n 2y 1 for any y (F q V ) \ { 1 2 }. + y n + (1 y) n k(n 1) n

24 An explicit expression for D n,k (1, x) For n 1, define f n,k (x) = k j 0 and ( ) n 1 (x j x j+1 ) + 2 2j + 1 j 0 f 0,k (x) = 2 k. ( ) n x j 2j Z[x], Then for n 0 ( 1 ) n D n,k (1, x) = fn,k (1 4x). 2

25 An explicit expression for D n,k (1, x) (contd.) Let p be an odd prime and n > 0 be an integer. Then in F q [x], ( 1 ) n D n,k (1, x) = fn,k (1 4x). 2 In particular, D n,k (1, x) is a PP of F q if and only if f n,k (x) is a PP of F q. Self-reciprocal property of f n,k is completely explained in N. F., Self-reciprocal polynomials and coterm polynomials, Designs, Codes and Cryptography (2018), Available Online.

26 The generating function The generating function of D n,k (1, x) is given by D n,k (1, x) z n = n=0 2 k + (k 1)z 1 z + xz 2.

27 Computation of a F q D n,k (1, a) D n,k (1, x) z n n=0 2 k + (k 1)z 1 z (mod x q x) [ q 1 (z 1) q 1 m z 2m 1 + x m] (z 1) q 1 z2(q 1) m=1 Since D n1,k(1, x) = D n2,k(1, x) for any x F q \ { 1 4 } when n 1, n 2 > 0 and n 1 n 2 (mod q 2 1), we have the following for all x F q \ { 1 4 }. q D n,k (1, x) z n = 2 k + D 1 z q2 1 n,k (1, x) z n n=1 n 0

28 Computation of a F q D n,k (1, a) (contd.) q 2 1 n=1 D n,k (1, x) z n = z (zq2 1 1) z 1 for all x F q \ { 1 4 }, + h(z) q 1 (z 1) q 1 m z 2m x m m=1 where Write h(z) = [2 k + (k 1)z] ( 1 (z zq ) q 1 ) z q z q 1. 1 h(z) = 1 z q z q 1 1 q 2 q+1 j=0 b j z j.

29 Computation of a F q D n,k (1, a) (contd.) Write j = α + βq where 0 α, β q 1. Then we have the following. ( 1) β+1 (2 k) ( ) q 1 β if α + β = q 1, ( 1) β+1 (k 1) ( ) q 1 β if α + β = q, b j = 1 k if α + β = 1, k 2 if α + β = 0, 0 otherwise.

30 Computation of a F q D n,k (1, a) (contd.) q 2 1 n=1 ( a F q ) a) z D n,k (1, n = q 2 1 n=1 k(n 1) + 2 z n z(1 z q 2 n 1 z q 1 ( h(z) (z 1) q 1 m z 2m 1 ) m, 4 m=1 2 1 ) h(z) z 2(q 1) q 2 1 (z q z q 1 1) n=1 = (1 + z q 1 z q ) [( ) D n,k (1, a) a F q q 2 1 i=1 z i ( k(n 1) + 2 )] z n 2 n ( q 1 ( z 2(q 1) + (z 1) q 1 m z 2m 1 ) m ) ( q2 q+1 b j z j) 4 m=1 j=0 q 2 1 (z q z q 1 1) n=1 [( ) D n,k (1, a) a F q ( k(n 1) + 2 )] q 2 +q 1 z n = 2 n i=1 c i z i

31 Theorem Let c j be defined as on the previous slide for 1 j q 2 + q 1. Then we have the following. k(j 1) + 2 D j,k (1, a) = c j + if 1 j q 1; 2 j a F q D q,k (1, a) = c 1 c q + 2 k ; 2 q a F q D lq+j,k = D (l 1)q+j,k D (l 1)q+j+1,k c lq+j a F q a F q a F q + (kj + 2)(1 2q + 2 q 1 ) + k(2 q 1) 2 lq+j if 1 l q 2 and 1 j q 1; D lq,k = D (l 1)q,k D (l 1)q+1,k c lq + a F q a F q a F q 2 l q 2; q 1 k(j 1) + 2 D q 2 q+j,k = c q 2 +i + 2 q2 q+j a F q i=j if 0 j q 1. (k 2)(2 q 1) + 2 q if 2 lq

32 For further details N. F., Reversed Dickson polynomials of the (k + 1)-th kind over finite fields, J. Number Theory 172 (2017),

33 Remember this Theorem? Theorem Let n = p l + 2 and k 0, 2, 4. Then D n,k (1, x) is a PP of F q if and only if the trinomial (4 k) x pl k x pl (2 k)x is a PP of F q. Xiang-dong Hou asked me when is the trinomial above a PP of F q?

34 My answer Theorem Let p > 3 be an odd prime and q = p e, where e is a positive integer. Let k be an integer such that k 0, 2, 4 and 0 k p 1. Let f (x) = (4 k)x pl kx pl (2 k)x. Then f (x) is a PP of F q if and only if l = 0 and k 3.

35 My answer (contd.) Let p = 3. Since k 0, 2, we have k = 1. Then f (x) = (4 k)x pl kx pl (2 k)x = x pl x. Theorem Let p = 3 and q = 3 e, where e is a positive integer. Let f (x) = x pl x. Then f (x) is a PP of F q if and only if (i) l = 0, or (ii) l = me + 1, where m is a non-negative even integer.

36 For further details N. F., A note on permutation binomials and trinomials over finite fields, to appear in New Zealand Journal of Mathemaics.

37 The Case n = p l + 3 D p l +3,k(1, x) is a PP of F p e if and only if f (x) is a PP of F p e, where f (x) = (2 k) x pl x pl k x pl (3 k) x.

38 The Case n = p l + 3 with p = 3 In this case, k = 0, 1, or 2. Theorem D 3 l +3,0(1, x) is a PP of F p e if and only if gcd( 3l +3 2, 3e 1) = 1. Theorem D 3 l +3,1(1, x) is not a PP of F p e. Theorem D 3 l +3,2(1, x) is a PP of F p e (i) l = 0, or if and only if (ii) l = me + 1, where m is a non-negative even integer.

39 The Case n = p l + 3 with p > 3 Theorem Let p > 5 and k = 2. Assume that ( 1 4 ) is a quadratic residue of p. Then D p l +3,k(1, x) is a PP of F p e if and only if l = 0. Remark D p l +3,k(1, x) is not a PP of F p e when p = 5 and k = 2. Theorem Let p > 5, k = 2, and l > 0. Assume that ( 1 4 ) is a quadratic non-residue of p. Then D p l +3,k(1, x) is a PP of F p e if and only if 3x pl x pl x is a PP of F p e.

40 The Case n = p l + 3 with p > 3 (contd.) Theorem Let p > 5, k = 0, and 6 be a quadratic non-residue of p. Then D p l +3,k(1, x) is a PP of F p e if and only if x pl x pl x is a PP of F p e. Remark Let p > 7, k = 7. Then D p l +3,k(1, x) is not a PP of F p e. Remark Let p > 5, k = 0, and 6 be a quadratic residue of p. Then D p l +3,k(1, x) is not a PP of F p e.

41 The Case n = p l + 3 with p > 3 (contd.) Theorem Assume that 1. p = 5 and k 2, 2. p > 5 and k 0, 2, or 3. p > 7 and k 7. Then D p l +3,k(1, x) is a PP of F p e if and only if is a PP of F p e. (2 k) x pl x pl k x pl (3 k) x

42 More Cases 1. n = p l 1 + p l 2 + p l 3 2. n = p l 1 + p l 2 + p l 3 + p l 4 How about a generalization to n = p l 1 + p l p l i?

43 A Generalization Case 1. Let i be odd and n = p l 1 + p l p l i. For all x F q, we have D n,k (1, x) = k (1 4x) pl 1 +p l 2 + +p l i 1 (2 k) 2 i 2 + (1 4x) pj 1 +p j j 2 + +p i 1 2 i 2 j 1,j 2,...,j i 1 {l 1,l 2,...,l i } + k 2 i + + j 1,j 2,...,j i 2 {l 1,l 2,...,l i } (1 4x) pj 1 +p j j 2 + +p i (2 k) (1 4x) pj 1 +p j j 2 + +p i 3 2 i j 1,j 2,...,j i 3 {l 1,l 2,...,l i } (2 k) 2 i j 1,j 2 {l 1,l 2,...,l i } (1 4x) pj 1 +p j k 2 i j 1 {l 1,l 2,...,l i } (1 4x) pj 1 1 (2 k) i

44 A Generalization (contd.) Case 2. Let i be even and n = p l 1 + p l p l i. For all x F q, we have D n,k (1, x) = + (2 k) (1 4x) p l 1 +p l 2 + +p l i 2 i 2 + k 2 i j 1,j 2,...,j i 1 {l 1,l 2,...,l i } (2 k) (1 4x) pj 1 +p j j 2 + +p i 2 2 i 2 j 1,j 2,...,j i 2 {l 1,l 2,...,l i } + k 2 i + j 1,j 2,...,j i 3 {l 1,l 2,...,l i } (2 k) 2 i j 1,j 2 {l 1,l 2,...,l i } (1 4x) pj 1 +p j j 2 + +p i (1 4x) pj 1 +p j k 2 i j 1 {l 1,l 2,...,l i } (1 4x) pj 1 +p j j 2 + +p i (1 4x) pj 1 1 (2 k) i

45 Permutation behaviour of D p l 1 +p l 2,k D p l 1 +p l 2,k (1, x) = (2 k) 4 (1 4x) pl 1 +p l k 4 (1 4x) pl k 4 (1 4x) pl Corollary Let k = 0. Then D p l 1 +p l 2,k (1, x) is a PP of F p e gcd( pl 1 +p l 2 2, p e 1) = 1. if and only if Corollary Let p = 3 and k = 2. Assume that both l 1 and l 2 are odd. Then D p l 1 +p l 2,k (1, x) is a PP of F p e if and only if the binomial x pl x pl is a PP of F q. Theorem Let p > 3 and k = 2. Then D p l 1 +p l 2,k (1, x) is not a PP of F p e.

46 Permutation behaviour of D p l 1 +p l 2,k (contd.) Corollary Let k 0, 2 and p > 3. Assume that of p. Then D p l 1 +p l 2,k (1, x) is a PP of F p e l 1 = l 2 = 0. 2k (k 2) is a quadratic residue if and only if Corollary 2k Let k 0, 2 and p > 3. Assume that (k 2) is a quadratic non-residue of p. Then D p l 1 +p l 2,k (1, x) is a PP of F p e if and only if the trinomial (2 k) x pl 1 +p l k x pl k x pl is a PP of F p e.

47 More results on D n,k (1, x) Lemma Let l be a positive odd integer and let n = 3l Then in F 3[x], D n,k (1, 1 x 2 ) = ( k 2 1 ) D n (x, 1) + k 2 D n 1 (x, 1). x Remark This result generalizes Lemma 5.5 in X. Hou, G. L. Mullen, J. A. Sellers, J. L. Yucas, Reversed Dickson polynomials over finite fields, Finite Fields Appl. 15 (2009),

48 More results on D n,k (1, x) (contd.) For all x F q we have and D n,k (1, x) = k x D n 2,1 (1, x) + D n (1, x), n 2, D n,k (1, x) = k x D n 1,2 (1, x) + D n (1, x), n 1.

49 Recursion Proposition Let p be an odd prime and n be a non-negative integer. Then D 0,k (1, x) = 2 k, D 1,k (1, x) = 1, and D n,k (1, x) = D n 1,k (1, x) x D n 2,k (1, x), for n 2.

50 A Matrix Form of D n,k (1, x) D n,k (1, x) = ( 2 k, 1 ) ( 0 x 1 1 ) n ( ) 1 0

51 For further details N. F., Reversed Dickson polynomials of the (k + 1)-th kind over finite fields, II, submitted for publication. arxiv:

52 Acknowledgement 1. Ariane Masuda at CUNY. 2. Anthony Iarrobino at Northeastern University.

53 Thank you!