Complete solution over \GF{p^n} of the equation X^{p^k+1}+X+a=0
aa r X i v : . [ c s . I T ] J a n Complete solution over F p n of the equation X p k +1 + X + a = 0 Kwang Ho Kim , , Jong Hyok Choe , and Sihem Mesnager Institute of Mathematics, State Academy of Sciences, Pyongyang, DemocraticPeople’s Republic of Korea [email protected] PGItech Corp., Pyongyang, Democratic People’s Republic of Korea Department of Mathematics, University of Paris VIII, F-93526 Saint-Denis,University Sorbonne Paris Cit´e, LAGA, UMR 7539, CNRS, 93430 Villetaneuse andT´el´ecom Paris, 91120 Palaiseau, France. [email protected]
Abstract.
The problem of solving explicitly the equation P a ( X ) := X q +1 + X + a = 0 over the finite field F Q , where Q = p n , q = p k and p isa prime, arises in many different contexts including finite geometry, theinverse Galois problem [1], the construction of difference sets with Singerparameters [9], determining cross-correlation between m -sequences [12]and to construct error correcting codes [4], cryptographic APN functions[5,6], designs [26], as well as to speed up the index calculus method forcomputing discrete logarithms on finite fields [13,14] and on algebraiccurves [23].Subsequently, in [2,15,16,5,3,20,8,24,19], the F Q -zeros of P a ( X ) havebeen studied. In [2], it was shown that the possible values of the numberof the zeros that P a ( X ) has in F Q is 0, 1, 2 or p gcd( n,k ) + 1. Some criteriafor the number of the F Q -zeros of P a ( x ) were found in [15,16,5,20,24].However, while the ultimate goal is to explicit all the F Q -zeros, even inthe case p = 2, it was solved only under the condition gcd( n, k ) = 1 [20].In this article, we discuss this equation without any restriction on p andgcd( n, k ). In [19], for the cases of one or two F Q -zeros, explicit expres-sions for these rational zeros in terms of a were provided, but for the caseof p gcd( n,k ) +1 F Q − zeros it was remained open to explicitly compute thezeros. This paper solves the remained problem, thus now the equation X p k +1 + X + a = 0 over F p n is completely solved for any prime p , anyintegers n and k . Keywords:
Equation · Finite field · Zeros of a polynomial.
Mathematics Subject Classification.
Introduction
Let n and k be any positive integers with gcd( n, k ) = d . Let Q = p n and q = p k where p is a prime. We consider the polynomial P a ( X ) := X q +1 + X + a, a ∈ F ∗ Q . Notice the more general polynomial forms X q +1 + rX q + sX + t with s = r q and t = rs can be transformed into this form by the substitution X = ( s − r q ) q X − r .It is clear that P a ( X ) have no multiple roots.These polynomials have arisen in several different contexts including finitegeometry, the inverse Galois problem [1], the construction of difference sets withSinger parameters [9], determining cross-correlation between m -sequences [12]and to construct error correcting codes [4], APN functions [5,6], designs [26].These polynomials are also exploited to speed up (the relation generation phasein) the index calculus method for computation of discrete logarithms on finitefields [13,14] and on algebraic curves [23].Let N a denote the number of zeros in F Q of polynomial P a ( X ) and M i denotethe number of a ∈ F ∗ Q such that P a ( X ) has exactly i zeros in F Q . In 2004, Bluher[2] proved that N a takes either of 0, 1, 2 and p d + 1 where d = gcd( k, n ) andcomputed M i for every i . She also stated some criteria for the number of the F Q -zeros of P a ( X ).The ultimate goal in this direction of research is to identify all the F Q -zerosof P a ( X ). Subsequently, there were much efforts for this goal, specifically fora particular instance of the problem over binary fields i.e. p = 2. In 2008 and2010, Helleseth and Kholosha [15,16] found new criteria for the number of F n -zeros of P a ( X ). In the cases when there is a unique zero or exactly two zerosand d is odd, they provided explicit expressions of these zeros as polynomialsof a [16]. In 2014, Bracken, Tan, and Tan [5] presented a criterion for N a = 0in F n when d = 1 and n is even. In 2019, Kim and Mesnager [20] completelysolved this equation X k +1 + X + a = 0 over F n when d = 1. They showed thatthe problem of finding zeros in F n of P a ( X ), in fact, can be divided into twoproblems with odd k : to find the unique preimage of an element in F n underan M¨uller-Cohen-Matthews polynomial and to find preimages of an element in F n under a Dickson polynomial. By completely solving these two independentproblems, they explicitly calculated all possible zeros in F n of P a ( X ), with newcriteria for which N a is equal to 0, 1 or p d + 1 as a by-product.Very recently, new criteria for which P a ( X ) has 0, 1, 2 or p d + 1 roots werestated by [19,24] for any characteristic. In [19], for the cases of one or two F Q -zeros, explicit expressions for these rational zeros in terms of a are provides. Forthe case of p gcd( n,k ) + 1 rational zeros, [19] provides a parametrization of such a ’s and expresses the p gcd( n,k ) + 1 rational zeros by using that parametrization,but it was remained open to explicitly represent the zeros.Following [19], this paper discuss the equation X p k +1 + X + a = 0 , a ∈ F p n ,without any restriction on p and gcd( n, k ). After introducing some prerequisitesfrom [19] (Sec. 2), we solve the open problem remained in [19] to explicitly2epresent the F Q − zeros for the case of p gcd( n,k ) + 1 rational zeros (Sec. 3). Afterall, it is concluded that the equation X p k +1 + X + a = 0 over F p n is completelysolved for any prime p , any integers n and k . Throughout this paper, we maintain the following notations. • p is any prime. • n and k are any positive integers. • d = gcd( n, k ). • m := n/d . • q = p k . • Q = p n . • a is any element of the finite field F ∗ Q .Given positive integers L and l , define a polynomial T LlL ( X ) := X + X p L + · · · + X p L ( l − + X p L ( l − . Usually we will abbreviate T l ( · ) as T l ( · ). For x ∈ F p l , T l ( x ) is the absolute trace T r l ( x ) of x .In [19], the sequence of polynomials { A r ( X ) } in F p [ X ] is defined as follows: A ( X ) = 1 , A ( X ) = − ,A r +2 ( X ) = − A r +1 ( X ) q − X q A r ( X ) q for r ≥ . (1)The following lemma gives another identity which can be used as an alternativedefinition of { A r ( X ) } and an interesting property of this polynomial sequencewhich will be importantly applied afterwards. Lemma 1 ([19]).
For any r ≥ , the following are true.1. A r +2 ( X ) = − A r +1 ( X ) − X q r A r ( X ) . (2) A r +1 ( X ) q +1 − A r ( X ) q A r +2 ( X ) = X q ( qr − q − . (3)The zero set of A r ( X ) can be completely determined for all r : Proposition 2 ([19]).
For any r ≥ , { x ∈ F p | A r ( x ) = 0 } = ( ( u − u q ) q +1 ( u − u q ) q +1 , u ∈ F q r \ F q ) . F ( X ) := A m ( X ) ,G ( X ) := − A m +1 ( X ) − XA qm − ( X ) . It can be shown that if F ( a ) = 0 then the F Q -zeros of P a ( X ) satisfy aquadratic equation and therefore necessarily N a ≤ Lemma 3 ([19]).
Let a ∈ F ∗ Q . If P a ( x ) = 0 for x ∈ F Q , then F ( a ) x + G ( a ) x + aF q ( a ) = 0 . (4)By exploiting these definitions and facts, the following results have been got. N a ≤
2: Odd p Theorem 4 ([19]).
Let p be odd. Let a ∈ F Q and E = G ( a ) − aF ( a ) q +1 .1. N a = 0 if and only if E is not a quadratic residue in F p d (i.e. E pd − = 0 , . N a = 1 if and only if F ( a ) = 0 and E = 0 . In this case, the unique zero in F Q of P a ( X ) is − G ( a )2 F ( a ) . N a = 2 if and only if E is a non-zero quadratic residue in F p d (i.e. E pd − =1 ). In this case, the two zeros in F Q of P a ( X ) are x , = ± E − G ( a )2 F ( a ) , where E represents a quadratic root in F p d of E . N a ≤ p = 2 When p = 2, in [19] it is proved that G ( x ) ∈ F q for any x ∈ F q m and using it Theorem 5 ([19]).
Let p = 2 and a ∈ F Q . Let H = T r d (cid:16) Nr nd ( a ) G ( a ) (cid:17) and E = aF ( a ) q +1 G ( a ) .1. N a = 0 if and only if G ( a ) = 0 and H = 0 .2. N a = 1 if and only if F ( a ) = 0 and G ( a ) = 0 . In this case, ( aF ( a ) q − ) isthe unique zero in F Q of P a ( X ) .3. N a = 2 if and only if G ( a ) = 0 and H = 0 . In this case the two zeros in F Q are x = G ( a ) F ( a ) · T n (cid:16) Eζ +1 (cid:17) and x = x + G ( a ) F ( a ) , where ζ ∈ µ Q +1 \ { } . N a = p d + 1: Auxiliary resultsLemma 6 ([19]). Let a ∈ F ∗ Q . The following are equivalent.1. N a = p d + 1 i.e. P a ( X ) has exactly p d + 1 zeros in F Q . . F ( a ) = 0 , or equivalently by Proposition 2, there exists u ∈ F q m \ F q suchthat a = ( u − u q ) q ( u − u q ) q +1 .3. There exists u ∈ F Q \ F p d such that a = ( u − u q ) q ( u − u q ) q +1 . Then the p d + 1 zerosin F Q of P a ( X ) are x = − u − u q ) q − and x α = − ( u + α ) q − q u − u q ) q − for α ∈ F p d . Lemma 7 ([19]). If A m ( a ) = 0 , then for any x ∈ F Q such that x q +1 + x + a = 0 ,it holds A m +1 ( a ) = N r kmk ( x ) ∈ F p d . Furthermore, for any t ≥ A m + t ( a ) = A m +1 ( a ) · A t ( a ) . (5)In [19], it is remained as an open problem to explicitly compute the p d + 1rational zeros. N a = p d + 1 Thanks to Lemma 6, throughout this section we assume F ( a ) = 0 i.e. A m ( a ) = 0 . Let L a ( X ) := X q + X q + aX ∈ F Q [ X ] . Define the sequence of polynomials { B r ( X ) } as follows: B ( X ) = 0 , B r +1 ( X ) = − a · A r ( X ) q . (6)From Lemma 7 and the definition (1) it follows B m ( a ) = − aA m − ( a ) q = A m +1 ( a ) q ∈ F p d . (7)Using (5) and an induction on l it is easy to check: Proposition 8. B l · m ( a ) = B m ( a ) l . (8) for any integer l ≥ . The first step to solve the open problem is to induce
Lemma 9.
For any integer r ≥ , in the ring F Q [ X ] it holds X q r = r − X i =1 A r − i ( a ) q i · L a ( X ) q i − + A r ( a ) · X q + B r ( a ) · X. (9)5 roof. The equality (9) for r = 2 is X q = L a ( X ) − X q − aX which is validby the definition of L a ( X ). Suppose the equality (9) holds for r ≥
2. By raising q − th power to both sides of the equality (9), we get X q r +1 = r − X i =1 A r − i ( a ) q i +1 · L a ( X ) q i + A r ( a ) q · X q + B r ( a ) q · X q = r X i =2 A r +1 − i ( a ) q i · L a ( X ) q i − + A r ( a ) q · X q + B r ( a ) q · X q = ( r +1) − X i =2 A r +1 − i ( a ) q i · L a ( X ) q i − + A r ( a ) q · L a ( X ) − A r ( a ) q · X q − a · A r ( a ) q · x + B r ( a ) q · X q = ( r +1) − X i =1 A r +1 − i ( a ) q i · L a ( X ) q i − + A r +1 ( a ) · X q + B r +1 ( a ) · X, where the last equality follows from the definitions (6) and (1). This shows thatthe equality (9) holds also for r + 1 and so for all r ≥ ⊓⊔ For r = m , under the assumption A m ( a ) = 0, Lemma 9 gives X q m = m − X i =1 A m − i ( a ) q i · L a ( X ) q i − + B m ( a ) · X. Now, we define F ( X ) := X q m − B m ( a ) · X = m − X i =1 A m − i ( a ) q i · L a ( X ) q i − ∈ F p d [ X ] (10)and G ( X ) = m − X i =1 A m − i ( a ) q i · X q i − . (11)Then, evidently, F ( X ) = G ◦ L a ( X ) . (12)Furthermore, we can show Proposition 10. F ( X ) = L a ◦ G ( X ) . Proof.
When m = 3, A ( a ) = 0 is equivalent to a = 1. Therefore, one has F ( X ) = X q − X = ( X q − X ) q + ( X q − X ) q + ( X q − X ) = L a ◦ G ( X ).6ow, suppose m ≥
4. Then, by using Definition (6) L a ◦ G ( X ) = m − X i =1 A m − i ( a ) q i +2 · X q i +1 + m − X i =1 A m − i ( a ) q i +1 · X q i + m − X i =1 aA m − i ( a ) q i · X q i − = m X i =2 A m +1 − i ( a ) q i +1 · X q i + m − X i =1 A m − i ( a ) q i +1 · X q i + m − X i =0 aA m − − i ( a ) q i +1 · X q i = X q m − B m ( a ) · X = F ( X ) , where Equality (2) was exploited to deduce the last second equality. ⊓⊔ By (5), from A m ( a ) = 0 it follows A l · m ( a ) = 0 for any l ≥
1. Therefore, (8)and (9) for r = lm yield that for any l ≥ X q l · m − B m ( a ) l · X = l · m − X i =1 A l · m − i ( a ) q i · L a ( X ) q i − . (13) Proposition 11.
Relation (13) can be rewritten by using F ( X ) as follows: X q l · m − B m ( a ) l · X = l − X i =0 B m ( a ) l − − i · F ( X ) q m · i . (14) Proof. If l = 1, the equality is equivalent to the definition of F ( X ). Supposethat it holds for l ≥
2. By raising q m − th power to both sides of (14), we have X q ( l +1) m − B m ( a ) l · X q m = l − X i =0 B m ( a ) l − − i · F ( X ) q m · ( i +1) = ( l +1) − X i =1 B m ( a ) ( l +1) − − i · F ( X ) q m · i . Since X q ( l +1) m − B m ( a ) l · X q m = X q ( l +1) m − B m ( a ) l · F ( X ) − B m ( a ) l +1 · X, one has X q ( l +1) m − B m ( a ) l +1 · X = ( l +1) − X i =1 B m ( a ) ( l +1) − − i · F ( X ) q m · i + B m ( a ) l · F ( X )= ( l +1) − X i =0 B m ( a ) ( l +1) − − i · F ( X ) q m · i This shows that Equality (14) holds for all l ≥ ⊓⊔ N := ( p d − · m,G ( X ) = P p d − i =0 B m ( a ) p d − − i · X q m · i . Since F ( X ) and G ( X ) are p d − linearized polynomials over F p d , they are com-mutative under the symbolic multiplication “ ◦ ” (see e.g. 115 page in [22]). There-fore, regarding Equation (14) and Proposition 10, one has X q N − X = G ◦ F ( X ) = F ◦ G ( X ) = L a ◦ G ◦ G ( X ) (15)and consequently ker( F ) = G ( F q N ) , (16)ker( L a ) = G ◦ G ( F q N ) . (17)Since L a ( X ) = XP a ( X q − ), here we can state: Proposition 12.
For a ∈ F ∗ Q , { x ∈ F p | x q +1 + x + a = 0 } = { x q − | x ∈ G ◦ G ( F q N ) } \ { } . (18)Our goal is to determine S a = { x ∈ F Q | P a ( x ) = 0 } , the set of all F Q − zerosto P a ( X ) = X q +1 + X + a = 0 , a ∈ F Q . Remark 13.
In order to find the F Q − zeros of P a ( X ) it is not enough to considerthe F Q − zeros of L a ( X ). In fact, one can see that B m ( a ) = 1 in general. However,it holds: Proposition 14. L a ( X ) = 0 has a solution in F ∗ Q if and only if B m ( a ) = 1 .Proof. If L a ( x ) = 0 for x ∈ F ∗ Q , then by (12) F ( x ) = 0 i.e. x q m − B m ( a ) · x = (1 − B m ( a )) · x = 0 and consequently B m ( a ) = 1. Conversely, assume B m ( a ) = 1. Then F ( X ) = X q m − X = L a ◦ G ( X ) and ker( L a ) = G ( F q m ).Assume G ( F Q ) = { } . Then, since G is q − linearized, it holds G ( F q m ) = G ([ F q , F Q ]) = { } which contradicts to deg( G ) < q m . Thus there exists sucha x ∈ F ∗ Q that G ( x ) = 0. Then G ( x ) ∈ ker( L a ) ∩ F ∗ Q .To achieve the goal, we will further need the following lemmas. Lemma 15.
Let L ( X ) be any q − linearized polynomial over F Q . If x q − ∈ F Q ,then L ( x ) q − ∈ F Q .Proof. If x q − ∈ F Q i.e. x q − = λ for some λ ∈ F Q , then x q = λx andsubsequently x q i = Q i − j =0 λ q j x for every i ≥
1. Therefore, when L ( X ) is a q − linearized polynomial over F Q , one can write L ( x ) = λx for some λ ∈ F Q .Thus, L ( x ) q − = λ q − λ ∈ F Q . ⊓⊔ Lemma 16.
Let s = ( q m − · ( p d − Q − · ( q − . If A m ( a ) = 0 and x ∈ ker( F ) , then x s ∈ ker( F ) and ( x s ) q − ∈ F Q . roof. For x = 0, the statement is trivial. Therefore, we can assume x = 0.Then, x ∈ ker( F ) implies B m ( a ) = x q m − = ( x s ) ( q − · Q − pd − . (19)Since B m ( a ) ∈ F p d , therefore ( x s ) q − ∈ F Q .Now, we will show B m ( a ) = B m ( a ) s . Since P a ( X ) has p d + 1 rational solutions when A m ( a ) = 0, there exists such anon-zero x that L a ( x ) = 0 , x q − ∈ F Q . Then (12) gives F ( x ) = 0 i.e. x q m − = B m ( a ) , and on the other hand x q m − = ( N F Q | F pd ( x q − )) s = ( N F qm | F q ( x q − )) s = ( x q m − ) s = B m ( a ) s , where the second equality followed from the fact that N F Q | F pd ( y ) = N F qm | F q ( y )for any y ∈ F Q . Thus, B m ( a ) = B m ( a ) s .Hence, ( x s ) q m − = ( x q m − ) s = B m ( a ) s = B m ( a ) i.e. F ( x s ) = 0 . ⊓⊔ Now, take any x ∈ ker( F ). The definition (10) and Lemma 16 shows x s · F ∗ Q := { x s · α | α ∈ F ∗ Q } ⊂ ker( F ) = G ( F p N )and ( x s · F ∗ Q ) q − ⊂ F Q . Subsequently, Lemma 15 and Equality (18) prove G ( x s · F ∗ Q ) q − ⊂ S a . In order to avoid the trivial zero solution, we need G ( x s · F ∗ Q ) = { } . In fact, this is the case. Really, if we assume G ( x s · F ∗ Q ) = { } , then G ( x s · F q m ) = { } (because G is F q − linear, and F q m is generated by F q and F Q ) whichcontradicts to deg( G ) < q m .Next, in order to explicit all p d + 1 elements in S a , we need to deduce thefollowing lemma. Lemma 17.
Let A m ( a ) = 0 and x be a F Q − solution to P a ( X ) = 0 . Then, x a is a ( q − − th power in F Q . For β ∈ F Q with β q − = x a , w q − w + 1 βx = 0 (20)9 as exactly p d solutions in F Q . Let w ∈ F Q be a F Q − solution to Equation (20) .Then, the p d + 1 solutions in F Q to P a ( X ) = 0 are x , ( w + α ) q − · x where α runs over F p d .Proof. We substitute x in P a ( x ) with x − x to get( x − x ) q +1 + ( x − x ) + a = 0or x q +1 − x x q − x q x − x + x q +10 + x + a = 0which implies x q +1 − x x q − ( x q + 1) x = 0 , or equivalently, x q +1 − x x q + ax x = 0 . Since x = 0 corresponds to x being a zero of P a ( X ), we can the latterequation by x q +1 to get ax y q − x y + 1 = 0 (21)where y = x . Now, let y = tw where t q − = x a . (22)Then, Equation (21) is equivalent to w q − w + 1 tx = 0 . (23)If t is a solution to Equation (22), then the set of all q − t · F ∗ q . For every λ ∈ F ∗ q , when w is a solution to Equation (23)for t = t , λw is a solution to Equation (23) for t = t /λ . By the way, ( t , w )and ( t /λ, λw ) give the same y = t · w = t /λ · λw . Therefore, to find all F Q − solutions to Equation (21) one can consider Equation (23) for any fixedsolution t of Equation (22).Now, we will show that any solution t to Equation (22) lies in F q · F Q := { α · β | α ∈ F q , β ∈ F Q } . In fact, we know that Equation (23) has p d solu-tions w with y = wt ∈ F Q . Let’s fix a solution w with y = w t ∈ F Q ofEquation (23). Then, the set of all solutions to Equation (23) can be writtenas w + F q . Therefore, it follows that there exist p d ≥ λ ∈ F q with( w + λ ) t ∈ F Q . As w t ∈ F Q and ( w + λ ) t ∈ F Q , we have λt ∈ F Q i.e. t ∈ λ F Q ⊂ F q · F Q .Hence, we can write t = α · β , where α ∈ F q , β ∈ F Q , and it follows that theset of all solutions to Equation (22) are F ∗ q · β . This means that Equation (22) has p d − F ∗ p d · β ) in F Q , i.e., x a is a ( q − − th power in F Q . Moreover,Equation (20) has exactly p d solutions in F Q (because Equation (21) has exactly10 d solutions y = wβ in F Q ). When w ∈ F Q is such a solution, the set of all p d solutions in F Q is w + F p d . Since Equation (23) yields y = wt = − w q − ) x , wehave x − x = x − y = x − (1 − w q − ) x = w q − x . The proof is over. ⊓⊔ Finally, all discussion of this section are summed up in the following theorem.
Theorem 18.
Assume A m ( a ) = 0 . Let N = m ( p d − , s = ( q m − · ( p d − Q − · ( q − , G ( X ) = P m − i =0 A m − − i ( a ) q i +1 · X q i and G ( X ) = P p d − i =0 B m ( a ) p d − − i · X q mi . Itholds G ( G ( F ∗ p N ) s · F ∗ q · F ∗ Q ) q − = { } . Take a x ∈ G ( G ( F ∗ p N ) s · F ∗ q · F ∗ Q ) q − \{ } . x a is a ( q − − th power in F Q . For β ∈ F Q with β q − = x a , w q − w + 1 βx = 0 (24) has exactly p d solutions in F Q . Let w ∈ F Q be a F Q − solution to Equation (20) .Then, the p d + 1 solutions in F Q of P a ( X ) are x , ( w + α ) q − · x where α runsover F p d . Note that one can also explicit w by an immediate corollary of Theorem 4and Theorem 5 in [25]. In [2,15,16,5,3,20,8,24,19], partial results about the zeros of P a ( X ) = X p k +1 + X + a over F p n have been obtained. In this paper, we provided explicit expressionsfor all possible zeros in F p n of P a ( X ) in terms of a and thus finalize the studyinitiated in these papers. Acknowledgement
The authors deeply thank Professor Dok Nam Lee for his many helpful sugges-tions and careful checking.
References
1. S.S. Abhyankar, S.D. Cohen, and M.E. Zieve. Bivariate factorizations connectingDickson polynomials and Galois theory.
Transactions of the American Mathemat-ical Society , 352(6): 2871 – 2887, 2000.2. A.W. Bluher. On x q +1 + ax + b . Finite Fields and Their Applications , 10(3) pp.285 – 305, 2004.3. A.W. Bluher. A New Identity of Dickson Polynomials. arXiv:1610.05853[math.NT] , 2016.4. C. Bracken andT. Helleseth. Triple-error-correcting BCH-like codes. in:
IEEE Int.Symp. Inf. Theory , pp. 1723 – 1725, 2009. . C. Bracken, C.H. Tan and Y. Tan. On a class of quadratic polynomials with nozeros and its application to APN functions. Finite Fields and Their Applications ,25: pp. 26 – 36, 2014.6. L. Budaghyan and C. Carlet. Classes of quadratic APN trinomials and hexanomialsand related structures, In
IEEE Trans. Inform. Theory
54 (5), pp. 2354–2357, 2008.7. S. D. Cohen and R. W. Matthews. A class of exceptional polynomials.
Transactionsof the American Mathematical Society , 345(2), pp. 897 – 909, 1994.8. B. Csajb´ok, G. Marino, O. Polverino, and F. Zullo. A characterization of linearizedpolynomials with maximum kernel.
Finite Fields and Their Applications , 56, pp.109 – 130, 2019.9. J. Dillon and H. Dobbertin. New cyclic difference sets with singer parameters.
Finite Fields Appl. , 10, pp. 342 – 389, 2004.10. H. Dobbertin. Almost perfect nonlinear power functions on GF (2 n ): the Welchcase. IEEE Trans. Inform. Theory , 45, pp. 1271 – 1275, 1999.11. H. Dobbertin. Kasami power functions, permutation polynomials and cyclic differ-ence sets. in: A. Pott, P.V. Kumar, T. Helleseth, D. Jungnickel (Eds.), DifferenceSets, Sequences and their Correlation Properties, Proceedings of the NATO Ad-vanced Study Institute on Difference Sets, Sequences and their Correlation Prop-erties, Bad Windsheim, 2-14 August 1998, Kluwer, Dordrecht, pp. 133 – 158, 1999.12. H. Dobbertin, P. Felke, T. Helleseth and P. Rosendhal. Niho type cross-correlationfunctions via Dickson polynomials and Kloosterman sums.
IEEE Transactions onInformation Theory , 52(2): pp. 613 – 627, 2006.13. F. G¨olo˘glu, R. Granger, G. McGuire and J. Zumbr¨agel. On the function field sieveand the impact of higher splitting probabilities application to discrete logarithmsin F and F . R. Canetti and J.A. Garay (Eds.): CRYPTO 2013, Part II ,LNCS 8043, pp. 109 – 128, 2013.14. F. G¨olo˘glu, R. Granger, G. McGuire and J. Zumbr¨agel. Solving a 6120-bit DLPon a desktop computer.
Cryptology ePrint Archive 2013/306
15. T. Helleseth, and A. Kholosha. On the equation x l +1 + x + a over GF (2 k ). FiniteFields and Their Applications , 14(1), pp. 159-176, 2008.16. T. Helleseth, and A. Kholosha. x l +1 + x + a and related affine polynomials over GF (2 k ). Cryptogr. Commun. , 2, pp. 85 – 109, 2010.17. T. Helleseth, A. Kholosha and G.J. Ness. Characterization of m-sequences oflengths 2 k − k − IEEE Trans. Inform.Theory , 53(6), pp. 2236 – 2245, 2007.18. T. Helleseth and V. Zinoviev. Codes with the same coset weight distributions asthe Z -linear Goethals codes. IEEE Trans. Inform. Theory , 47(4), pp. 1589 – 1595,2001.19. K.H. Kim, J. Choe and S. Mesnager. Solving X q +1 + X + a = 0 over finitefields. Finite Fields and Their Applications. To appear (Cryptology ePrint Archive2019/1493, arXiv:1912.12648).20. K.H. Kim and S. Mesnager. Solving x k +1 + x + a = 0 in F n with gcd( n, k ) = 1. Finite Fields and Their Applications, 63: 1016302020 (https://doi.org/10.1016/j.ffa.2019.101630 and Cryptology ePrint Archive2019/307).21. R. Lidl, G.L. Mullen and G. Turnwald. Dickson Polynomials , Pitman Monographsin Pure and Applied Mathematics, Vol. 65, Addison-Wesley, Reading, MA 1993.22. R. Lidl and H. Niederreiter, Finite Fields, volume 20 of Encyclopedia of Mathemat-ics and its Applications, Cambridge University Press, Cambridge, second edition,1997.
3. M. Massierer. Some experiments investigating a possible L (1 /
4) algorithm forthe discrete logarithm problem in algebraic curves.
Cryptology ePrint Archive2014/996
24. G. McGuire and J. Sheekey. A characterization of the number of roots of linearizedand projective polynomials in the field of coefficients.
Finite Fields and TheirApplications , 57, pp. 68 – 91, 2019.25. S. Mesnager, K.H. Kim, J. H. Choe and D. N. Lee. Solving Some Affine Equationsover Finite. Finite Fields and Their Applications, 68: 101746 , 2020. (CryptologyePrint Archive 2020/160)26. C. Tang. Infinite families of 3-designs from APN functions. Journal of Combinato-rial Designs Vol 28, Issue 2 Pages 97–117, 2020 (arXiv preprint arXiv:1904.04071,2019)., 57, pp. 68 – 91, 2019.25. S. Mesnager, K.H. Kim, J. H. Choe and D. N. Lee. Solving Some Affine Equationsover Finite. Finite Fields and Their Applications, 68: 101746 , 2020. (CryptologyePrint Archive 2020/160)26. C. Tang. Infinite families of 3-designs from APN functions. Journal of Combinato-rial Designs Vol 28, Issue 2 Pages 97–117, 2020 (arXiv preprint arXiv:1904.04071,2019).