More new classes of permutation trinomials over \mathbb{F}_{2^n}
aa r X i v : . [ c s . I T ] J un More new classes of permutation trinomials over F n Yanping WangISN Laboratory, Xidian University, Xi’an 710071, Chinae-mail: [email protected] ZhangISN Laboratory, Xidian University, Xi’an 710071, Chinae-mail: [email protected] ZhaSchool of Mathematical Sciences, Luoyang Normal University, Luoyang 471934, Chinae-mail: [email protected]
Abstract
Permutation polynomials over finite fields have wide applications in many areasof science and engineering. In this paper, we present six new classes of permutationtrinomials over F n which have explicit forms by determining the solutions of someequations. Keywords:
Finite field, permutation polynomial, trinomial, multivariate.
Let q be a power of prime p and F q denote the finite field with q elements. Define F ∗ q be the multiplication group of F q . A polynomial f ( x ) ∈ F q [ x ] is called a permutationpolynomial over F q if the associated polynomial function f : c → f ( c ) from F q into F q is apermutation of F q [29]. Permutation polynomials over finite fields have wide applicationsin combinational designs [29], cryptography [2, 3], and coding theory [1]. Finding newpermutation polynomials is of great interest in both theoretical and applied aspects. Thereare numerous interesting results about the study of permutation polynomials (see [4–13]).The reader may refer to [29, Chapter 7] and [30, Chapter 8] for detailed knowledge onpermutation polynomials.Permutation polynomials with few terms over finite fields are in particularly interestingfor their simple algebraic forms and important applications in the areas of mathematicsand engineering. Recently, the studying of permutation trinomials is a hot topic. Someexcellent works were done and numerous beautiful permutation trinomials were discovered.For instance, Dobbertin proposed one permutation trinomial x k +1 +1 + x + x over F k +1 , which is applied to show that the Welch power function x k +3 is almost per-fect nonlinear(APN) [14]. Afterwards, Dobbertin developed the multivariate method [15]1o confirm the permutation property of certain types of polynomials over F n . Hou [16]determined the permutation behaviors of trinomials with the form ax + bx q + x q − over F q for some explicit conditions on a and b . Moreover, Hou did a survey of permutationbinomials and trinomials [17]. Lee et al. [18] presented some permutation trinomials of theform x r f ( x s ) over finite fields. Ding et al. [19] gave more classes of permutation trinomialswith nonzero coefficients 1 over finite fields with even characteristic. Motivated by [19], Liet al. [21,22] also proposed several classes of permutation trinomials over F n . Bhattacharyaet al. [20] characterized permutation trinomials of the form x s +1 + x s − +1 + αx over F n for some α ∈ F n . After that, three classes of permutation trinomials were proposedby Ma et al. [23]. Gupta et al. [24] built four new classes of permutation trinomials of theform x r h ( x m − ) over F m . Followed the work of [24], Zha et al. [25] further presented sixclasses of permutation trinomials of the form x r h ( x m − ) over F m . Li et al. [26] constructedfour new classes of permutation trinomials over F m from Niho exponents of the form x s (2 m − + x t (2 m − + x. M.E. Zieve [27] constructed some classes of permutation trinomials over F q by exhibitingclasses of low-degree rational functions over F q which induce bijections on the set of ( q + 1)-th roots of unity. Based on the paper of M.E. Zieve [27], Wu et al. [28] gave the explicit formof all permutation trinomials over F m , and presented some classes permutation trinomialsof the form x r f ( x s ) over F m .In this paper, six new classes of permutation trinomials are proposed by determiningthe solutions of some equations. The paper is organized as follows. In Section 2, we presentfive new classes of permutation trinomials over F n by applying the multivariate method.In Section 3, we obtain a new class of permutation trinomials over F n for the case of n ≡ Inspired by the idea of [15], we present five new classes of permutation trinomials in thissection by using the multivariate method. n ≡ In this subsection, we propose two classes of permutation trinomials over F n when n ≡ Definition 1.
Let n and k be positive integers with k | n . The trace function Tr nk ( x ) from F p n to F p k is defined by Tr nk ( x ) = x + x p k + x p k + · · · + x p ( nk − k . If k = 1 , then Tr n ( x ) is called the absolute trace function. heorem 1. Let k and n = 3 k . Then f ( x ) = x k +2 k − + x k + x is a permutation polynomial on F n .Proof. We shall show that for each a ∈ F n , the equation f ( x ) = x k +2 k − + x k + x = a (2.1)has one unique solution in F n .When a = 0, we need to prove that f ( x ) = 0 has the unique solution x = 0. Assume x = 0 is a solution of the equation x k +2 k − + x k + x = 0 . (2.2)Raising both sides of (2.2) to the 2 k -th power, we have x k − k +1 + x k + x = 0 , which means that x k − k + x k − + 1 = 0 . (2.3)Setting y = x k − , equation (2.3) can be written as y k + y + 1 = 0 . (2.4)Applying the mapping Tr nk ( · ) on both sides of equation (2.4), we get Tr nk (1) = 0, whichcontradicts with Tr nk (1) = 1. Hence, equation (2.3) has no solution.When a = 0, we will show that equation (2.1) has one non-zero solution. Let y = x k , z = y k , b = a k and c = b k . Then we can obtain the system of equations x + z + yzx = a,y + x + xzy = b,z + y + xyz = c. (2.5)Let u = yzx , v = xzy , and w = xyz . Then x = vw , y = uw , and z = uv . We have u + vw + uv = a ,v + vw + uw = b ,w + uw + uv = c . (2.6)By (2.6), we deduce that u + v + w = a + b + c , which implies u + v + w = a + b + c .Setting ǫ = a + b + c , we have ǫ ∈ F k and w = ǫ + u + v .If ǫ = 0, then pluging u + v + w = 0 into (2.6), we have u = b , v = c , w = a . So theequation has a unique solution x = ( vw ) = ( ac ) .3f ǫ = 0, then from (2.6), we can get (cid:26) u + v + ǫv = a ,u + ǫu + ǫv = b . (2.7)It can be verified that u = v k and v k +1 + v + ǫv = a . (2.8)Eliminating the indeterminate u in (2.7), we obtain v + ǫ v + ǫ v + a + b + a ǫ = 0 . (2.9)Since ǫ ∈ F k and ǫ = 0, substituting v with ǫv in (2.8) and (2.9), we obtain (cid:26) v k +1 + v + v = a /ǫ ,v + v + v + ǫ ( a + b + a ǫ ) = 0 . (2.10)Suppose v and v are the solutions of (2.10) and let λ = v + v . Then we get λ = 0 or (cid:26) λ k +1 − + λ + 1 = 0 ,λ + λ + 1 = 0 , (2.11)which means that λ k +1 − = 1. Since k k +1 − , k −
1) = 2 gcd(2 k − , k ) − λ = 1. Note that λ = 1 is not the solution of (2.11). Therefore, λ = 0 and there is aunique solution v of (2.10). Hence, (2.5) has a unique solution when k Theorem 2.
Let k and n = 3 k . Then f ( x ) = x k +2 k − + x k − k +1 + x is a permutation polynomial on F n .Proof. For any fixed a ∈ F n , we need to prove that the equation f ( x ) = x k +2 k − + x k − k +1 + x = a (2.12)has one unique solution in F n . Obviously, f (0) = 0. For a = 0, we have x = 0 or x k +2 k − + x k − k + 1 = 0 . (2.13)Setting y = x k − , equation (2.13) can be written as y k +2 + y k + 1 = 0 . (2.14)Note that y = 0 and y k +2 k +1 = 1. Equation (2.14) leads to y k +2 + y k + y k +2 k +1 = 0 , y + 1 + y k +1 = 0 . Raising both sides of the above equation to the 2 k − -th power gives y k + 1 + y k − (2 k +1) = 0 . (2.15)From equations (2.14) and (2.15), we obtain y k +2 = y k − (2 k +1) , which implies that y k +3 = 1. Since k k + 2 k + 1 , k + 3) = gcd(2 k + 3 ,
7) = 1 . Then we get y = 1. But y = 1 is not the solution of (2.14). Thus, f ( x ) = 0 if and only if x = 0.When a ∈ F ∗ n . Let y = x k , z = y k , b = a k and c = b k . Then we obtain the followingsystem of equations yzx + xzy + x = a, (2.16) xzy + xyz + y = b, (2.17) xyz + yzx + z = c. (2.18)Adding the three equations above, we obtain x + y + z = a + b + c . Equations (2.16) and(2.17) lead to (cid:26) y z + x z + x y = axy, (2.19) xz + xy + y z = byz. (2.20)Adding (2.19) with (2.20) gives x ( y + z ) ǫ + y ( ax + bz ) = 0 , (2.21)where ǫ = a + b + c and ǫ ∈ F k . Plugging x = ǫ + y + z into (2.21), we get( ǫ + a ) y + ǫz + ( a + b ) yz + ǫ ( ǫ + a ) y + ǫ z = 0 . (2.22)Setting y = λz with λ = 0, and plugging it into (2.22), we obtain( ǫ + a ) λ z + ǫz + ( a + b ) λz + ǫ ( ǫ + a ) λz + ǫ z = 0 , (2.23)which can be rewritten as[( ǫ + a ) λ + ( a + b ) λ + ǫ ] z + [ ǫ ( ǫ + a ) λ + ǫ ] z = 0 . (2.24)Let ξ = ( ǫ + a ) λ + ( a + b ) λ + ǫ and ξ = ǫ ( ǫ + a ) λ + ǫ . Then we have ξ z + ξ = 0 . (2.25)5ultiplying equation (2.17) by yz and then plugging x = ǫ + y + z , it leads to y + z + yz + ǫ ( y + z ) + byz = 0 . (2.26)Plugging y = λz into (2.26) gives( λ + λ + 1) z + ( ǫλ + bλ + ǫ ) = 0 . (2.27)Let η = λ + λ + 1 and η = ǫλ + bλ + ǫ , equation (2.27) becomes η z + η = 0 . (2.28)By equations (2.25) and (2.28), we can deduce that ξ η + ξ η = 0 . Then [( ǫ + a ) λ + ( a + b ) λ + ǫ ]( ǫλ + bλ + ǫ ) + [ ǫ ( ǫ + a ) λ + ǫ ]( λ + λ + 1) = 0 . (2.29)We have ( ǫ + ab + aǫ ) λ + ( ǫ + ab + b ) λ = 0 , (2.30)which implies ζ λ + ζ = 0 , (2.31)where ζ = ǫ + ab + aǫ = ac + b + c and ζ = ǫ + ab + b = ab + a + c .Next, we verify ζ = 0. Suppose that ζ = 0. Then a k + a k +1 + a k +1 = 0 . Since a = 0, the above equation leads to a k +1 (2 k − + a (2 k − k − + 1 = 0 . Setting µ = a k − , we obtain µ k +2 k +1 = 1 and µ k +1 + µ k − + 1 = 0 . (2.32)Note that ζ + ζ k + ζ k = ab + ac + bc = 0 , which implies that a k +2 k + a k +1 + a k +1 = 0 . (2.33)It leads to a k − + a k − k + 1 = 0 , (2.34)which can be written as µ k +1 + µ k + 1 = 0 . (2.35)6ultiplying equation (2.35) by µ k − , and then adding equation (2.32), we get µ k +1 − + 1 = 0 . (2.36)It can be verified thatgcd(2 k + 2 k + 1 , k +1 −
1) = gcd(2 k +1 − , k + 2 k − + 1)= gcd(3 · k − + 1 , k + 3) = gcd(3 · k + 2 , k + 3) = gcd(7 , k + 3) = 1 . Then equation (2.36) has a unique solution µ = 1. However, µ = 1 is not the solution of(2.32). Therefore, we get ζ = 0 and λ = ζ ζ = ζ k − = ( ac + b + c ) k − , which implies that λ k +2 k +1 = 1.When k ≡ , k + 2 k + 1) = 1. If λ + λ + 1 = 0,then λ = 1, which means that λ = 1. It leads to λ + λ + 1 = 1, which is a contradiction.Therefore, λ + λ + 1 = 0. By (2.27), we get a unique solution x = z k = (cid:0) ǫ (1+ λ )+ bλλ + λ +1 (cid:1) k of f ( x ) = a in (2.12).When k ≡ k ≡ λ + λ + 1 = 0, then λ = 1, z = λ − y = λ y and x = y k = λ k z k = λ y. Plugging them into equation (2.17), we obtain( λ + λ − + 1) y = b, which is equivalent to ( λ + λ + 1) y = b. (2.37)Supposing that λ + λ + 1 = 0, we get λ + λ = 0, which implies that λ = 1. However, λ = 1 is not the solution of λ + λ + 1 = 0. So λ + λ + 1 = 0. Hence, y = bλ + λ +1 , thereby x = bλ λ + λ +1 . If λ + λ +1 = 0, we can similarly obtain one solution x = z k = (cid:0) ǫ (1+ λ )+ bλλ + λ +1 (cid:1) k of (2.12) from (2.27). This completes the proof. n ≡ In this subsection, we give one class of permutation trinomials over F n when n ≡ Theorem 3.
Let k be a positive integer and n = 3 k + 1 . Then f ( x ) = x k +1 +2 k +1 +1 + x k +1 +1 + x is a permutation polynomial on F n . roof. We prove that for any fixed a ∈ F n , the equation x k +1 +2 k +1 +1 + x k +1 +1 + x = a (2.38)has one unique solution in F n . It is obvious that f (0) = 0. For a = 0, we have x = 0 or x k +1 +2 k +1 + x k +1 + 1 = 0 . (2.39)Let y = x k , z = y k , then x = z k +1 . Equation (2.39) can be written as y z + y + 1 = 0 . (2.40)Raising both sides of equation (2.40) to 2 k -th power, we get xz + z + 1 = 0 . (2.41)Further, raising both sides of equation (2.41) to the 2 k -th power gives xy + x + 1 = 0 . (2.42)Combining equations (2.39), (2.40) and (2.41) eliminates the indeterminate y and z , weobtain x = 1. However f (1) = 1. So equation (2.39) has no solution. Hence f ( x ) = 0 onlyhas a solution x = 0.When a = 0, we prove that f ( x ) = a has one unique solution. Let y = x k , z = y k , b = a k and c = b k , then x = z k +1 and a = c k +1 . Therefore, we obtain xy z + xy + x = a, (2.43) xyz + yz + y = b, (2.44) xyz + xz + z = c. (2.45)Combining equations (2.43) and (2.44) eliminates the indeterminate z , we have x y + bxy + x + ax + x + a = 0 . (2.46)Adding equations (2.43) and (2.44) multiply by y together results in xy + y z + y + x + by + a = 0 . (2.47)Computing (2 .
43) + (2 .
44) + (2 . ∗ yz , there is xy + cyz + x + y + a + b = 0 . (2.48)By (2.45) ∗ cy +(2.48) ∗ ( xy + x + 1), this gives x y + x y + x y + ( a + b + 1) xy + x + ( a + b + 1) x + ( c + 1) y + ( a + b ) = 0 . (2.49)Multiplying equation (2.46) by y and then adding (2.49), it leads to x y + bxy + bxy + x + ( a + b + 1) x + ( a + c + 1) y + ( a + b ) = 0 . (2.50)8dding (2.46) and (2.50) derives bxy + ( a + c + 1) y + ( bx + b ) = 0 . (2.51)Further, multiplying equation (2.48) by b and then adding (2.51), we deduce bcyz + ( a + b + c + 1) y + ( ab + b + b ) = 0 . (2.52)Raising both sides of equation (2.52) to the 2 k +1 -th power results in abxy + ( a + b + c + 1) x + ( ac + a + a ) = 0 . (2.53)Multiply equation (2.46) by a b , and then adding the square of (2 . a b xy + ( a + b + c + 1) x + a b ( x + ax + x + a ) + ( a c + a + a ) = 0 . (2.54)Furthermore, multiplying equation (2.53) by ab and then adding (2.54), there is( a + a b + b + c + 1) x + ab ( a + b + c + 1) x + a ( a + b + b c + c + 1) = 0 . (2.55)Equation (2.55) can rewrite as( a + a b + b + c + 1)( x + a ) + ab ( a + b + c + 1)( x + a ) = 0 . (2.56)Supposing a + a b + b + c + 1 = 0 , (2.57)raising both sides of equation (2.57) to the 2 k -th power, this leads to a + b + b c + c + 1 = 0 . (2.58)Adding equations (2.57) and (2.58), we get a + b + c + 1 = 0 . (2.59)Again raising both sides of equation (2.59) to the 2 k -th power, and then adding equation(2.59). There is a + a = 0 , which means a = 0 or a = 1. Since a = 0, we obtain b = c = 1 from a = 1. Thereby a + a b + b + c + 1 = 0, this is a contradiction. Hence, there are two solutions ofequation (2.56), which is x = a or x = a + a + b + c + 1 a + a b + b + c + 1 . If x = a , then y = b and we deduce ab ( a + 1) = 0 from equation (2.46). Because a = 0,this leads to b = 0, and hence a = 1. Therefore equation has one unique solution x = 1.If x = a , then there is one unique solution x = a + a + b + c +1 a + a b + b + c +1 . The proof finishes.9 .3 The case of n ≡ In this subsection, two classes of permutation trinomials are proposed over F n when n ≡ Theorem 4.
Let k be a positive integer and n = 3 k − . Then f ( x ) = x k − − k +2 k + x k − + x is a permutation polynomial on F n .Proof. We shall prove that for each fixed a ∈ F n , the equation f ( x ) = x k − − k +2 k + x k − + x = a (2.60)has one unique solution in F n .When a = 0, we claim that x = 0. Otherwise x = 0 is a solution of the equation x k − − k +2 k + x k − + x = 0 . (2.61)This implies that equation x k − − k +2 k − + x k − + 1 = 0 (2.62)has nonzero solution. Let y = x k , z = y k , then x = z k . Hence there is yz + yx + 1 = 0 , (2.63) zx + zy + 1 = 0 . (2.64)Combining equations (2.63) and (2.64) to eliminate z , we obtain x y ( y + 1) = 0 . (2.65)Since x = 0, this means y = 0. Thereby we get y = 1. It leads to x = 1 forgcd(2 k − , n −
1) = 2 ( k, k − − . However f (1) = 1, this is a contradiction. Hence f ( x ) = 0 if and only if x = 0.When a = 0, we will verify that f ( x ) = a has one unique nonzero solution. Let y = x k , z = y k , b = a k and c = b k , then x = z k and a = c k . Therefore, we obtain xyz + yx + x = a, yzx + zy + y = b, x zy + x z + z = c. (2.66)Eliminating the indeterminate z in (2.66), we get (cid:26) ( x + ax + b ) y + ( x + bx + abx ) = 0 , (2.67)( x + c + 1) y + ( cx + acx ) y + a x = 0 . (2.68)10ombining equations (2.67) and (2.68) to eliminate the indeterminate y , we obtain αx + βx + γx + θ = 0 , (2.69)where α = a + b + bc + c + 1, β = abc , γ = a + a bc + a b + a c + b + bc + c + 1, θ = a bc + abc .If α = 0, then we get γ = 0. From (2.69) we can deduce that x = a + 1, which implies x = a + 1. Therefore equation (2.69) has one solution.Next, substituting x with u + α − β in (2.69), we get u + ( α − β + α − γ ) u + ( α − βγ + α − θ ) = 0 . (2.70)Since αθ = βγ , this implies α − βγ + α − θ = 0. Hence we obtain u = 0 or u = ( α − β + α − γ ) / . Thereby there is x = α − β or x = ( α − β + α − γ ) / + α − β . Since γ = ( a +1) α ,then α − γ = a + 1. It can be easily checked that x = ( α − β + α − γ ) / + α − β = a + 1.Assume that x = a + 1, then y = b + 1. Plugging y = b + 1 into equation (2.67), by somesimplifying computation we obtain( a + b )( a + b + 1) = 0 . If a = b , we have x = y = z = a + 1. Further we get x = 0 from equation (2.66), whichis a contradiction.If a = b + 1, we deduce that x = a + 1, y = a , z = a + 1. Plugging them into equation(2.66), there is a + a + 1 = 0 . (2.71)Suppose n is odd, it can be easily verified that equation (2.71) has no solution. Suppose n is even, then k is odd and 2 k ≡ a = ω and ω . If a = ω , it derives that x = ω , y = ω, z = ω . If a = ω , it leads to x = ω, y = ω , z = ω. Without loss of generality, we consider the case of a = ω . We can deduce that α = a + b + bc + c + 1 = ω and β = abc = 1. Thereby, we get another solution x = α − β = ω .But, it equals to the solution x = a + 1 = ω .Therefore, equation (2.69) has a unique solution. We complete the proof. Theorem 5.
Let k be a positive integer and n = 3 k − . Then f ( x ) = x k +2 k +1 + x k +1 + x is a permutation polynomial on F n .Proof. For each fixed a ∈ F n , it suffices to prove that equation f ( x ) = x k +2 k +1 + x k +1 + x = a (2.72)11as one unique solution in F n .We first verify that f ( x ) = 0 if and only if x = 0. Assume that x = 0 is a solution ofthe equation f ( x ) = x k +2 k +1 + x k +1 + x = 0 , (2.73)which implies that equation x k +2 k + x k + 1 = 0 (2.74)has nonzero solution. Let y = x k , z = y k , then x = z k . The above equation leads to yz + z + 1 = 0 , (2.75) x z + x + 1 = 0 , (2.76) x y + y + 1 = 0 . (2.77)Combining (2.75) and (2.76) to eliminate the indeterminate z , we obtain x y + y + 1 = 0 . (2.78)Multiplying equation (2.78) by y and then adding (2.77) gives y = 1 , this means x = 1. However f (1) = 1, it is a contradiction. So f ( x ) = 0 if and only if x = 0.When a = 0, we will show that f ( x ) = a has one unique nonzero solution. Let y = x k , z = y k , b = a k , c = b k , then x = z k and a = c k . Hence we obtain xyz + xz + x = a,x yz + x y + y = b,x y z + y z + z = c. (2.79)Eliminating the indeterminate z in (2.79), we deduce (cid:26) ( x + 1) y + ( ax + b + 1) y + b = 0 , (2.80)( x + 1) y + ( bx + b ) y + ( cx + x + 1) y + b = 0 . (2.81)Combining equations (2.80) and (2.81) results in( ax + x + ax + 1) y + ( bx + cx + x + b + 1) y + b = 0 . (2.82)By equations (2.80) and (2.82), there is( x + 1) y = a − ( bx + cx + x + a ) . (2.83)Obviously, f (1) = 1. We claim that x = 1 if and only if a = 1. Suppose that a = 1 and x = 1, then we get that b = c = 1 and y = x +1 from (2.83). Plugging y = x +1 into equation(2.80), we obtain xx +1 = 0, which implies x = 0. This contradicts the first assumption x = 0. 12ext, we consider the case of a = 0 ,
1. Plugging y = ( b + c +1) x + aa ( x +1) into equation (2.80)leads to ( a c + a + b + c + 1) x + ( a + ab + abc + ac + a ) = 0 . (2.84)If a c + a + b + c + 1 = 0 , (2.85)raising both sides of (2.85) to the 2 k -th power gives a b + a + b + c + 1 = 0 . (2.86)Adding (2.85) and (2.86), and then dividing a , there is a + b + c + 1 = 0 . (2.87)Since a = 0, we deduce that a + ab + abc + ac + a = a ( a + b + bc + c + 1) = abc = 0, whichimplies that equation (2.84) has no solution. If a c + a + b + c + 1 = 0, we have a solutionof (2.84). Therefore, equation (2.84) has at most one solution. The proof is completed. F n with n ≡ In the section, one class of permutation trinomials is presented over F n when n ≡ Theorem 6.
Let k be an odd integer and let m , n , d be positive integers satisfying n = 4 m , ≤ k ≤ n − , gcd( m, k ) = 1 , and d = P mi =0 ik . Then f ( x ) = x d + x m + x is a permutation polynomial on F n .Proof. For any fixed a ∈ F n , we need to prove that equation f ( x ) = x d + x m + x = a (3.1)has at most one solution in F n . Obviously, there is f ( x ) = 0 when x = 0. On the contrary,assume that x = 0 is a solution of the equation x d + x m + x = 0 . (3.2)Raising 2 m -th power in both sides of (3.2), we get x m · d + x m + x = 0 . (3.3)Adding (3.2) and (3.3) gives x m · d + x d = x d ( x (2 m − · d + 1) = 0 . (3.4)It leads to x = 0 or x (2 m − · d = 1. Since gcd( d, n −
1) = 1, this yields x m = x . Plugging x m = x into equation (3.2), we have x d = 0, thereby x = 0, which is impossible. So f ( x ) = 0 only has a solution x = 0. 13ince k is an odd integer with gcd( m, k ) = 1, it can be easily check that d = m X i =0 ik = 2 (2 m +1) k − k − , gcd( d, n −
1) = 1, and gcd(2 k − , m −
1) = 1.In the sequel, assume that x, a ∈ F ∗ n . We need to show that equation x d + x m + x = a (3.5)has at most one solution. The above equation can be written as x m +1) k − k − = x m + x + a. (3.6)Raising (2 k − x (2 m +1) k − = ( x m + x + a ) k − . (3.7)Let x m = θx , θ = 0, then θ m +1 = 1 and x (2 m +1) k − = x m + k − = ( θx ) k x − = θ k x k − . From equation (3.7), we obtain( θ k k − x ) k − = ((1 + θ ) x + a ) k − . (3.8)Since gcd(2 k − , m −
1) = 1, we deduce that θ k k − x = (1 + θ ) x + a , which can be writtenas (1 + θ + θ k k − ) x = a . Because of a = 0, we have x = a θ + θ k k − . (3.9)Plugging (3.9) into x m = θx , we get a m θ − + θ − k k − = θ a θ + θ k k − , which can be simplified as a m (1 + θ + θ k k − ) = a (1 + θ + θ − k − ) . Let θ k − = β , we have β m +1 = 1. The above equation can be reduced as a m (1 + β k − + β k ) = a (1 + β k − + β − ) . It leads to a m ( β + β k + β k +1 ) = a ( β + β k + 1) , a m β k +1 + ( a + a m )( β + β k ) + a = 0 . (3.10)Set β = t + w , where w = 1 and w = 1. Since k is odd, this implies 2 k ≡ a m ( t + w ) k +1 + ( a + a m )( t k + t + w k + w ) + a = 0 , which implies that a m t k +1 + ( w a m + a ) t k + ( wa m + a ) t = 0 . (3.11)If t = 0, then β = w , and β m +1 = w = 1, which is a contradiction. Dividing equation(3.11) by t k +1 , we obtain( wa m + a ) t − k + ( w a m + a ) t − + a m = 0 . (3.12)Let z = t − , then z = 0, equation (3.12) becomes( wa m + a ) z k + ( w a m + a ) z + a m = 0 . (3.13)If wa m + a = 0 and w a m + a = 0, then a = 0. This is a contradiction.If wa m + a = 0 or w a m + a = 0, then equation (3.13) has one solution z . Thereforeequation (3.9) only has one solution.If wa m + a = 0 and w a m + a = 0, assume that z , z are two solutions of equation(3.13), then ( wa m + a )( z − z ) k + ( w a m + a )( z − z ) = 0 . (3.14)Since gcd(2 k − , m −
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