gcd(n, k) = 1 Kwang Ho Kim 1 and Sihem Mesnager 2 Pyongyang, Democratic People s Republic of Korea

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1 Solvig x 2k + + x + a = 0 i F 2 gcd, k = with Kwag Ho Kim ad Sihem Mesager 2 Istitute of Mathematics, State Academy of Scieces ad PGItech Corp., Pyogyag, Democratic People s Republic of Korea khk.cryptech@gmail.com 2 LAGA, Departmet of Mathematics, Uiversities of Paris VIII ad Paris XIII, CNRS, UMR 7539 ad Telecom ParisTech, Frace smesager@uiv-paris8.fr March 6, 209 Abstract. Let N a be the umber of solutios to the euatio x 2k + + x+a = 0 i F 2 where gcdk, =. I 2004, by Bluher [2] it was kow that possible values of N a are oly 0, ad 3. I 2008, Helleseth ad Kholosha [] have got criteria for N a = ad a explicit expressio of the uiue solutio whe gcdk, =. I 204, Bracke, Ta ad Ta [5] preseted a criterio for N a = 0 whe is eve ad gcdk, =. This paper completely solves this euatio x 2k + + x + a = 0 with oly coditio gcd, k =. We explicitly calculate all possible zeros i F 2 of P ax. New criterio for which a, N a is eual to 0, or 3 is a byproduct of our result. Keywords Euatio Müller-Cohe-Matthews MCM polyomials Dickso polyomial Zeros of polyomials Irreducible polyomials. Itroductio Let be a positive iteger ad F 2 be the fiite field of order 2. The zeros of the polyomial P a x = x 2k + + x + a, a F 2 has bee studied i [2,, 2]. This polyomial has arise i several differet cotexts icludig the iverse Galois problem [], the costructio of differece sets with Siger parameters [7], to fid cross-correlatio betwee m-seueces [9, ] ad to costruct error correctig codes [4]. More geeral polyomial forms x 2k + +rx 2k +sx+t are also trasformed ito this form by a simple substitutio of variable x with r + s 2 k x + r. It is clear that P a x have o multiple roots. I 2004, Bluher [2] proved followig result. Theorem. For ay a F 2 ad a positive iteger k, the polyomial P ax has either oe, oe, three or 2 gcdk, + zeros i F 2.

2 2 Kwag Ho Kim ad Sihem Mesager I this paper, we will cosider a particular case with gcd, k =. I this case, Theorem says that P a x has oe, oe or three zeros i F 2. I 2008, Helleseth ad Kholosha [] have provided criteria for which a P a x has exactly oe zero i F 2 ad a explicit expressio of the uiue zero whe gcdk, =. I 204, Bracke, Ta ad Ta [5] preseted a criterio for which a P a x has o zero i F 2 whe is eve ad gcdk, =. I this paper, we explicitly calculate all possible zeros i F 2 of P a x whe gcd, k =. New criterio for which a, P a x has oe, oe or three zeros is a by-product of this result. We begi with showig that we ca reduce the study to the case whe k is odd. I the odd k case, oe core of our approach is to exploit a recet polyomial idetity special to characteristic 2, preseted i [3] Theorem 3. This polyomial idetity eables us to divide the problem of fidig zeros i F 2 of P a ito two idepedet problems: Problem to fid the uiue preimage of a elemet i F 2 uder a Müller-Cohe-Matthews MCM polyomial ad Problem 2 to fid preimages of a elemet i F 2 uder a Dickso polyomial subsectio 3.. There are two key stages to solve Problem. Oe is to establish a relatio of the MCM polyomial with the Dobberti polyomial. Other is to fid a explicit solutio formula for the affie euatio x 2k + x = b, b F 2. These are doe i subsectio 3.2 ad Problem is solved by Theorem 5. Problem 2 is relatively easy which is aswered by Theorem 6 ad Theorem 7 i subsectio 3.3. Fially, we collect together all these results to give explicit expressio of all possible zeros of P a i F 2 by Theorem 8, Theorem 9 ad Theorem 0. 2 Prelimiaries I this sectio, we state some results o fiite fields ad itroduce classical polyomials that we shall eed i the seuel. We begi with the followig result that will play a importat role i our study. Propositio. Let be a positive iteger. The, every elemet z of F 2 := F 2 \{0} ca be writte twice z = c+ c where c F 2 := F 2 \F 2 if T r z = 0 ad c µ 2 + := {ζ F 2 2 ζ 2 + = } \ {} if T r z =. Proof. For z F 2, z = c + c is euivalet to z = c 2 z + c 2, z ad thus this euatio has a solutio i F 2 if ad oly if T r z = 0. Hece, mappig c c + c is 2-to- from F 2 oto {z F 2 T r z = 0} with covetio 0 := 0. Also, sice c + 2 c = c c = c + c for c µ 2 +, the mappig c c+ c is 2-to- from µ 2 + with cardiality 2 oto {z F 2 T r z = } with cardiality 2. We shall also eed two classical families of polyomials, Dickso polyomials of the first kid ad Müller-Cohe-Matthews polyomials.

3 Solvig x 2k + + x + a = 0 i F 2 with gcd, k = 3 The Dickso polyomial of the first kid of degree k i idetermiate x ad with parameter a F 2 is D k x, a = k/2 i=0 k k i k i i a k x k 2i, where k/2 deotes the largest iteger less tha or eual to k/2. I this paper, we cosider oly Dickso polyomials of the first kid D k x,, that we shall deote D k x throughout the paper. A classical property of Dickso polyomial that we shall use extesively is Propositio 2. For ay positive iteger k ad ay x F 2, we have D k x + = x k + x x k. 2 Müller-Cohe-Matthews polyomials are aother classical polyomials defied as follows [6], f k,d X := T kx c d X 2k where k T k X := X 2i ad cd = 2 k +. i=0 A basic property for such polyomials that we shall eed i this paper is the followig statemet. Theorem 2. Let k ad be two positive itegers with gcdk, =.. If k is odd, the f k,2k + is a permutatio o F If k is eve, the f k,2 k + is a 2-to- o F 2. Proof. For odd k, see [6]. Whe k is eve, f k,2 k + is ot a permutatio of F 2. Ideed, Theorem 0 of [7] states that f k, is 2-to-, ad the the statemet follows from the facts that f k,2 k +x 2k + = f k, x 2k + ad gcd2 k +, 2 = whe gcdk, =. We will exploit a recet polyomial idetity ivolvig Dickso polyomials established i [3, Theorem 2.2]. Theorem 3. I the polyomial rig F 2 k [X, ], we have the idetity k X 22k + 2k 2 i X 2k + 2k = D 2k +wx. i= w F 2 k Fially we remark that the idetity by Abhyakar, Cohe, ad Zieve [, Theorem.] tatalizigly similar to this idetity treats ay characteristic, while this idetity is special to characteristic 2 this may happe because the Dickso polyomials are ramified at the prime 2. However, the Abhyakar-Cohe-Zieve idetity has ot lead us to solvig P a x = 0.

4 4 Kwag Ho Kim ad Sihem Mesager 3 Solvig P a x = 0 Throughout this sectio, k ad are coprime ad we set = 2 k. 3. Splittig the problem Oe core of our approach is to exploit Theorem 3 to the study of zeros i F 2 of P a. To this ed, we observe firstly that k X 2 + 2i X + i= = X 2 + T k 2 X +. Substitutig tx to X i the above idetity with t 2 = T 2, k we get k X 2 + 2i X + i= 2 = T k t x 2 + x + T 2. k t Now, t 2 = T k T k By all these calculatios, we get x 2 + x + fk,+ = f k,+ 2 is euivalet to t = T k 2. Therefore t = 2 T k = f k, k X 2 + 2i X +. i= 3 If k is odd, f k,+ is a permutatio polyomial of F 2 by Theorem 2. Therefore, for ay a F 2, there exists a uiue i F 2 such that a =. Hece, by Theorem 3 ad euatio 3, we have P a x = x 2 +x +a = f k,+ 2 w F where is the uiue elemet of F 2 such that a = T k 2 f k,+ 2 D + wtx 4 f k,+ 2 ad t =. Now, sice gcd, 2 =, the zeros of P a x are the images

5 Solvig x 2k + + x + a = 0 i F 2 with gcd, k = 5 of the zeros of P a x by the map x x. Therefore, whe k is odd, euatio 4 states that fidig the zeros of P a x amouts to determie preimages of uder the Dickso polyomial D +. Whe k is eve, f k,+ is o loger a permutatio ad we caot repeat agai the precedig argumet ideed, whe k is eve, f k,+ is 2-to-, see Theorem 2. Fortuately, we ca go back to the odd case by rewritig the euatio. Ideed, for x F 2, ad so P a x = x 2k + + x + a = x 2 k + + x 2 k + a 2 k 2 k {x F 2 P a x = 0} = = x + 2 k + + x + + a 2 k 2 k { } x + x 2 k + + x + a 2 k = 0, x F 2. 5 If k is eve, the is odd as gcdk, =, ad so k is odd ad we ca reduce to the odd case. We ow summarize all the above discussios i the followig theorem. Theorem 4. Let k ad be two positive itegers such that gcdk, =.. Let k be odd ad = 2 k. Let F 2 be uiuely defied by a =. The, f k,+ 2 z {x F 2 P a x = 0} = T k 2 D + z =, z F Let k be eve ad = 2 k. Let F 2 be uiuely defied by a = f k, +. The, 2 {x F 2 P a x = 0} = + z T 2 k D +z =, z F 2. Proof. Suppose that k is odd. Euatio 4 shows that the zeros of P a i F 2 are x for the elemets x F 2 such that D +wtx = where t = T 2 k. Set z = wtx. The, sice w F, x = z wt = z t = z. Item 2 follows from Item ad euality 5. T k 2 Theorem 4 shows that we ca split the problem of fidig the zeros i F 2 of P a ito two idepedet problems with odd k.

6 6 Kwag Ho Kim ad Sihem Mesager Problem. For a F 2, fid the uiue elemet i F 2 Problem 2. For F 2, fid the preimages i F 2 polyomial D +, that is, fid the elemets of the set such that a 2 = f. 6 k,+ of uder the Dickso D + = {z F 2 D +z = }. 7 I the followig two subsectios, we shall study those two problems oly whe k is odd sice, if k is eve, it suffices to replace k by k, by = 2 k, ad a by a i all the results of the odd case. 3.2 O problem Defie Q k,k x = x+ k i= xi 8 where k < 2 is the iverse of k modulo 2, that is, s.t. kk = mod 2. Note that k is odd sice gcdk, 2 =. It is kow that if gcd2, k = ad k is odd, the Q k,k is permutatio o F 2 2 see [7] or [8] where Q k,k = /Q k,k is istead cosidered. Ideed, due to [7], defiig the followig seueces of polyomials A x = x, A 2 x = x +, A i+2 x = x i+ A i+ x + x i+ i A i x, i, B x = 0, B 2 x = x, B i+2 x = x i+ B i+ x + x i+ i B i x, i, the the polyomial expressio of the iverse R k,k of the mappig iduced by Q k,k o F 22 is k R k,k x = A i x + B k x. 9 ad i= Directively from the defiitios, it follow f k,+ x + x 2 = x + x + x + x 2 Q k,k x + x = x + x + x + x k +. Sice x 2 = x k + x = x, 2kk it holds that f k,+ x + x 2 = Q k,k x + x 0

7 Solvig x 2k + + x + a = 0 i F 2 with gcd, k = 7 for ay x F 2 2. Let x be a elemet of F 2 2 such that By usig 0 we ca rewrite 6 as Therefore, we have = x + x2. a 2 = Q k,k x + x. Propositio 3. Let a F 2. Let x F 2 2 R k,k be a solutio of a 2 = x + x. The, =. fk,+ x+x 2 = + x + + x is the uiue solutio i F 2 of a 2 = Propositio 3 shows that solvig Problem amouts to fid a solutio of a affie euatio x + x = b, which is doe i the followig. Propositio 4. Let k be odd ad gcd, k =. The, for ay b F 2, b {x F 2 2 x + x = b} = S,k + F 2, ζ + where S,k x = i=0 xi ad ζ is a elemet of µ 2 +. Proof. As it was assumed that k is odd ad gcd, k =, it holds gcd2, k = ad so the liear mappig x F 2 2 x + x has kerel of dimesio, i.e. the euatio x+x = b has at most 2 solutios i F 2 2. Sice S,k x+s,k x = x + x, we have b S,k + S,k ζ + ad thus really S,k euatio x + x = b. b ζ+, S,k b ζ + b ζ+ + b = b b ζ + + ζ + = b ζ + + b ζ + + b + b = b ζ + + b /ζ + + b = 0 + F 2 2 are the F 2 2 solutios of the By Propositio 3 ad Propositio 4, we ca ow explicit the solutios of Problem.

8 8 Kwag Ho Kim ad Sihem Mesager Theorem 5. Let a F 2. Let k be odd with gcd, k = ad k be the iverse of k modulo 2. The, the uiue solutio of 6 i F 2 is = S,k Rk,k a 2 ζ+ + Rk,k S,k a 2 ζ+ 2 where ζ deotes ay elemet of F 2 such that ζ =, S,k x = i=0 xi ad R k,k is defied by 9. Furthermore, we have = T + T for T = + S,k Rk,k a 2 ζ O Problem 2 or c µ 2 +. Eua- By Propositio, oe ca write z = c + c where c F 2 tio 2 applied to z leads the to D + z = c + +. c+ Thus, we ca be reduced to solve firstly euatio T + T =, the euatio c + = T i F 2 µ 2 +, ad set z = c + c. Here, let us poit out that c+ = T is euivalet to + c = T ad that c ad c defie the same elemet z = c + c of F 2. Propositio says that the euatio T + T = has two solutios i F 2 if T r = 0 ad i µ 2 + if T r =. I fact, Propositio 4 gives a explicit solutio expressio, that is, T = S, 2 ad T = S, ζ + 2 +, 2 ζ + where S, x = i=0 ad ζ is ay elemet of µ x2i 2 +. Now, let us cosider solutios of c + = T i F 2 µ 2 +. First, ote that if T F 2, the ecessarily c F 2 ideed, if c µ 2 +, we get T 2 = T T = T 2 T = T 2 + = c = cotradictig T / F 2. Recall that if k is odd ad gcd, k =, the { gcd +, 2, if is odd = 3 3, if is eve ad gcd +, 2 + = {, if is eve 3, if is odd. 4

9 Solvig x 2k + + x + a = 0 i F 2 with gcd, k = 9 Therefore, if T F 2, the there are 0 if T is a o-cube i F 2 or 3 if T is a cube i F 2 elemets c i F 2 such that c+ = T whe is eve while there is a uiue c i.e. T + mod 2 whe is odd. Ad, if T µ 2 +, the there are 0 if T is a o-cube i µ 2 + or 3 if T is a cube i µ 2 + elemets c i µ 2 + such that c + = T whe is odd while there is a uiue c i.e. T + mod 2 + whe is eve. It remais to show i the case whe there are three solutios c, they defie three differet elemets z F 2. Deote w a primitive elemet of F 4. The these three solutios of c + = T are of form c, cw ad cw 2. Now, cw + cw = cw 2 + cw 2 implies that cw = cw 2 or cw = cw 2 because A + A = B + B is euivalet to A + BAB + = 0. The secod case is impossible because it implies that + T = c + = = because 3 divides + whe k is odd. w 2 w 22 We ca thus state the followig aswer to Problem 2. Theorem 6. Let k be odd ad be eve. Let F 2. Let T be ay elemet of F 2 2 such that T + T = this ca be give by 2.. If T is a o-cube i F 2, the D+ =. 2. If T is a cube i F 2, the { D+ = cw + } cw c+ = T, c F 2, w F If T is ot i F 2, the } D+ {T = + mod T + mod 2 + Remark. Item of Theorem 6 recovers [5, Theorem 2.] which states: whe is eve ad gcd, k = so k is odd, P a has o zeros i F 2 if ad oly 2 if a = f k,+ for some o-cube T of F T + 2. Ideed, the statemet of T Theorem 2. i [5] is ot exactly what we write but it is worth oticig that the 2 uatity that is deoted Ab i [5] satisfies Ab = f k,+. b 4 + Theorem 7. Let k be odd ad be odd. Let F 2. Let T be ay elemet of F 2 2 such that T + T = this ca be give by 2.. If T is a o-cube i µ 2 +, the D+ =. b 4

10 0 Kwag Ho Kim ad Sihem Mesager 2. If T is a cube i µ 2 +, the { D+ = cw + } cw c+ = T, c µ 2 +, w F If T is i F 2, the D + = {T + mod 2 + T + mod 2 }. 3.4 O the roots i F 2 of P a x We sum up the results of previous subsectios to give a explicit expressio of the roots i F 2 of P a x. Let k deote ay positive iteger coprime with ad a F 2. First, let us cosider the case of odd k. Let k be the iverse of k modulo 2. Defie T = + Rk,k a, 2 S,k where ζ is ay elemet of F 2 such that ζ =, S,k x = i=0 ad xi R k,k is defied by 9. Accordig to Theorem 5, Theorem 6 ad Theorem 7, we have followigs. ζ+ Theorem 8. Let be eve, gcd, k = ad a F 2.. If T is a o-cube i F 2, the P a x has o zeros i F If T is a cube i F 2, the P a x has three distict zeros cw+ cw where c + = T, w F 4 ad = T + T. 3. If T is ot i F 2, the P a x has a uiue zero c+ c c = T + mod 2 + ad = T + T. T k 2 T k 2 i F 2, i F 2, where Remark 2. Whe k =, that is, P a x = x 3 + x + a, Item of Theorem 8 is exactly Corollary 2.2 of [5] which states that, whe is eve, P a is irreducible over F 2 if ad oly if a = c + c for some o-cube c of F 2. Theorem 9. Let ad k be odds with gcd, k = ad a F 2.. If T is a o-cube i µ 2 +, the P a x has o zeros i F If T is a cube i µ 2 +, the P a x has three distict zeros cw+ cw i T k 2 F 2, where c + = T, w F 4 ad = T + T. 3. If T is i F 2, the P a x has a uiue zero c+ c i F T k 2 2, where c = T + mod 2 ad = T + T.

11 Solvig x 2k + + x + a = 0 i F 2 with gcd, k = Whe k is eve, followig Item 2 of Theorem 4, we itroduce l = k, = 2 l ad l the iverse of l modulo 2. Defie T = + S,l R l,l a 2 2 ζ+, where ζ is ay elemet of F 2 2 R l,l is defied by 9. such that ζ 2 + =, S,l x = i=0 x i ad Theorem 0. Let be odd ad k be eve with gcd, k =. Let a F 2.. If T is a o-cube µ 2 +, the P a x has o zeros i F If T is a cube i µ 2 +, the P a x has three distict zeros + dw+ dw i F 2, where d + = T, w F 4 ad = T + T. 3. If T is i F 2, the P a x has a uiue zero + c+ c c = T + mod 2 ad = T + T. T l 2 T l 2 i F 2, where Remark 3. Whe is eve, Theorem 8 shows that P a has a uiue solutio if ad oly if T is ot i F 2. Accordig to Propositio 3, this is euivalet to T r R k,k a 2 =, that is, T r R k,k a =. Whe is odd ad k is odd resp. eve, Theorem 9 ad Theorem 0 show that P a has a uiue zero i F 2 if ad oly if T resp. T is i F 2. Accordig to Propositio 3, this is euivalet to T r R k,k a = 0 or T r R l,l a = 0 for odd k or eve k, respectively. By the way, for x F 2, Q l,l x + x = x+x + = x+x + 2 k 2 x +x 2 x +x = 2 Q 2 k,k x + x 2 k. Hece if T F 2, the R l,l a = R k,k a 2 k2, ad so T r R l,l a = 0 is euivalet to T r R k,k a = 0. After all, we ca recover [, Theorem ] which states that P a has a uiue zero i F 2 if ad oly if T r R k,k a + =. 4 Coclusio I [2, 3, 5,, 2], partial results about the zeros of P a x = x 2k + + x + a i F 2 have bee obtaied. I this paper, we provided explicit expressio of all possible roots i F 2 of P a x i terms of a ad thus fiish the study iitiated i these papers whe gcd, k =. We showed that the problem of fidig zeros i F 2 of P a x i fact ca be divided ito two problems with odd k: to fid the uiue preimage of a elemet i F 2 uder a Müller-Cohe-Matthews MCM polyomial ad to fid preimages of a elemet i F 2 uder a Dickso polyomial. We completely solved these two idepedet problems. We also preseted a explicit solutio formula for the affie euatio x 2k +x = b, b F 2.

12 2 Kwag Ho Kim ad Sihem Mesager Refereces. S.S. Abhyakar, S.D.Cohe, M.E. Zieve. Bivariate factorizatios coectig Dickso polyomials ad Galois theory, Trasactios of the America Mathematical Society, 3526: , A.W. Bluher. O x + + ax + b. Fiite Fields ad Their Applicatios, 03: , A.W. Bluher. A New Idetity of Dickso polyomials. eprit arxiv: v, October C. Bracke, T. Helleseth. Triple-error-correctig bch-like codes. I IEEE Iteratioal Symposium o Iformatio Theory, ISIT 2009, Jue 28 - July 3, 2009, Seoul, Korea, Proceedigs, pages , C. Bracke, C.H. Ta,. Ta. O a class of uadratic polyomials with o zeros ad its applicatio to APN fuctios. Fiite Fields ad Their Applicatios, 25:26 36, S.D. Cohe, R.W. Matthews. A class of exceptioal polyomials. Trasactios of the America Mathematical Society, 345: , J.F. Dillo, H, Dobberti. New cyclic differece sets with siger parameters. Fiite Fields ad Their Applicatios, 03: , J.F. Dillo. Multiplicative differece sets via additive characters. Desigs, Codes ad Cryptography, 7: , H. Dobberti, P. Felke, T. Helleseth, P. Rosedhal. Niho type cross-correlatio fuctios via Dickso polyomials ad Kloosterma sums. IEEE Trasactios o Iformatio Theory, 522: , T. Helleseth, A. Kholosha, G.J. Ness. Characterizatio of m-seueces of legths 2 22k ad 2 k with three-valued crosscorrelatio. IEEE Trasactios o Iformatio Theory, 536: , T. Helleseth, A. Kholosha. O the euatio x 2l + + x + a = 0 over GF 2k. Fiite Fields ad Their Applicatios, 4:59 76, T. Helleseth, A. Kholosha. x 2l + +x+a ad related affie polyomials over GF 2k. Cryptography ad Commuicatios, 2:85 09, 200.