Deducing Polynomial Division Algorithms using a Groebner Basis

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Martin Clifton Goodman
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1 Deducing Polynomial Division Algorithms using a Groebner Basis In the previous Chapter I discussed a method of finding the real root of an associated cubic polynomial using a modular function, the eta function. In this Chapter I will find the form of higher order polynomials with integer coefficients which are divisible by a given irreducible cubic polynomial of a general form Ax 3 +Bx 2 +Cx+D where A, B, C, D are integers. The problem is expressible in the following form: Find the integer coefficients a, b, c, d,.. and n for which the polynomial F(x) = ax n + bx (n-1) +c x (n-2) + dx ( n-3) +. constant is divisible by G= Ax 3 +Bx 2 +Cx+D. Consequently, x is any of the real or complex roots of G such that G(x) = F(x) = 0. Also, the coefficients a, b, c, d,.. are integers derived from the integer sequence associated with the cubic polynomial G. The polynomial G will be any of the 16 irreducible polynomials of class number 3 mentioned in the previous chapter. This definition of the problem puts several constraints on the form of the polynomial F{x) and discards general polynomials H(x) with G(x)*H(x) = F(x). The problem will be solved with the aid of a computerized algebra system available with Mathematica. Given a set of initial conditions on the polynomial G, an algorithm called the Groebner Basis will be used to derive equivalent conditions that simplify the derivation of the polynomial F(x). The process is equivalent to that of solving a system of simultaneous polynomials but relies on the systematic elimination of variables using a division algorithm to remove remainders. The Groebner Basis can be applied to solve several types of problems in algebraic geometry, analytical geometry, geometric proofs, sudoku puzzles, and robotics. I will not attempt to further describe this algorithm since it is beyond the scope of this chapter. A good introduction can be found in Cox [1]. The above problem has been demonstrated for the associated Perrin polynomial x 3 -x-1. [See R. Witula and S. Gorczyca s entry in OEIS A001608]. From their analysis it is unclear whether the higher order polynomials can be determined for other cubic equations G, having general coefficients A, B, C, and D. I will show that the method is general for any irreducible equation G. First we find the set of polynomial ideals {f 1, f 2,f 3, f n} which best describes the problem. This is done by describing the equation which is to be eventually represented as F(x), followed by setting constraints on the function G(x). Next, we need to identify the variables necessary to solve the problem. Let x, y, and z be variables in a complex field. We know that if x, y, and z are the three roots of a polynomial G(x) then for any G(x), the infinite integer sequence a(n), a(n+1), a(n+2).. a(n+i) can be determined by powers of x, y, and z, by definition: [1] a(n) = (x n + y n +z n ) 1 This is a simple definition of the problem with a(n) raised to the 1 st power. To make it more interesting we need to raise a(n) to various powers. The function F(x) will be derived for powers of 2, 3 and 4. At powers >4 the algebra becomes complicated but the general method of derivation can be used for any power. Starting with the second power the first ideal is: [2] f 1 = (x n + y n +z n ) 2 a(n) 2

2 Where x, y and z and f lie in a ring of complex space C 3 and a(n) an integer. Since there will be powers of x 2, y 2 and z 2 in the expansion of f 1, we define; [3] f 2= (x 2n + y 2n +z 2n ) a(2n) Next the form of G(x) is considered. For the monic cubic polynomial x 3 +Bx 2 +Cx+D it is known that the coefficients are expressed in terms of the roots x, y, and z. [4 a,b,c] B = x + y + z C = xy+xz +yz D = -xyz Using the cubic polynomial for the Perrin sequence we know that B = 0, C = -1 and D = -1. Define the three ideas as; [5a,b,c] f 3 = xyz - 1 f 4 = xy + xz + yz - 1 f 5 = x +y + z The last ideal defines an algebraic relation between a(n) and powers of x, y and z. It can be shown from equation [1]; [6] a(n)*x n x 2n = x n *y n + x n *z n The associated ideal is: [7] f 6 = a(n)*x n x 2n x n *y n x n *z n We have found the ideal of the polynomials defined by I = {f 1, f 2, f 3, f 4, f 5, f 6} in C[x,y,z]. We now want to calculate the solution set {a(n), a(2n), x, y, z} for which f i[a(n), a(2n), x, y, z] = 0, i = For this solution set we can use the Groebner basis of the ideal I. The calculated basis depends on the ordering of the solution set and it turns out that the most practical ordering is {x, a(n), y, z, a(2n)}. The reason for this is obvious when the calculations are completed since for other orderings the resulting basis is just a copy of the input ideal. Let g i [x,a(n), y, z, a(2n)] be the output basis from Mathematica. The results show: [8a, b, c, d, e,f] g 1 = (x 2n + y 2n +z 2n ) a(2n) = 0 g 2 = z 3 z -1 = 0 g 3 = z 2 + yz + y 2 1 = 0 g 4 = a(n)*x n x 2n x n *y n x n *z n = 0 g 5 = a(n) 2 x 2n -2x n y n y 2n 2x n z n 2y n z n -z 2n = 0 g 6 = x + y + z = 0 Although this basis does not give the final result directly, there is enough information to derive the functions F(x) and G(x). The first basis g 1 is a re-iteration of the ideal. This equation will be important to the final results. The basis g 2 shows that z is also a solution to the problem and defines on of the cubic polynomials that divides F[z]. The basis g 3 is used to calculate y from the known value of z. It also may be used to find x since multiplying g 3 by x and applying the equations [5a] and [1] results in the equation G[x] = x 3 -x-1 = 0. Using g 6 the values of all variables x, y, and z can be determined.

3 We can find F[x] by substituting g 4 into g 5 and use g 1: a(n) 2 - (x 2n + y 2n +z 2n ) 2(x n *y n + x n *z n ) -2(y n *z n ) a(n) 2 a(2n) 2x 2n 2/x n [(a(n) 2 a(2n))/2]* x n a(n) x 2n + x 3n 1 and rearranging, [9] F[x] = a(n) x 2n + [(a(2n) a(n) 2 )/2]*x n +1 - x 3n Note that the polynomial F[x] is also true if the complex solutions y or z are substituted for x. This can be proven due to the symmetric nature of the original definition for equation [2]. Equation [9] is also true for any value of n and since we know that the a(n) are integers of the Perrin sequence all coefficients are real numbers. It remains to show that a(2n) a(n) 2 is even and the coefficients are integers. If equations [2] and [3] are expanded for a(n) 2 - a(2n) the result is: [10] a(n) 2 - a(2n) = 2x n y n + 2x n z n + 2y n z n = 2(x n y n + x n z n + y n z n ) It only remains to show that the value in parenthesis is an integer. Using [5a] [11] 2(x n y n + x n z n + y n z n ) = 2(1/z n + 1/y n + 1/x n ) It was shown in Chapter 1 that the value in parenthesis is defined as the inverted Perrin sequence and is a positive or negative integer. For any n the expression in [11] is even proving that the coefficients in [9] are always integers. In equation [9] the term, x 3n, can be replaced with (1+x) n since G[x] = 0. [12] F[x] = a(n) x 2n + [(a(2n) a(n) 2 )/2]*x n +1 (1+x) n This term is the binomial expansion of the polynomial (1+x) n. For any value of n the coefficients are integers based on Pascal s triangle. Since the final term of the binomial expansion is 1, F[x] is a product of monomial terms in x with the leading coefficient in x n as (a(2n) a(n) 2-2)/2 and lower monomial coefficients following Pascal s triangle for n. It can be shown by using [1] that the sum of the quotients is an integer constant dependent on n: [13] F[x]/G[x] + F[y]/G[y] + F[z]/G[z] = integer[n] However, the sum of the function F over all roots is a constant dependent only on the sequence. For the Perrin sequence and sequences Discriminant=31,59 and 83* for any value of n: [13a,b] F[x] + F[y] + F[z] = 3 a(n) 3 /2 + a(3n)- 3*a(n)*a(2n)/2 3 = 0 For all cubic Class 3 sequences: a(n) 3 +2*a(3n)-3*a(n)*a(2n)-6*D n = 0 (See Below)

4 Example- Using Groebner Basis to solve for the Coefficients To show how the Groebner basis eliminates variables and finds a solution to this problem, equation [12] is generalized and solved for a given value of n. Let n= 6, then the system of equations to solve are; [14a,b] a x 12 + bx 6 +1 (1+x) 6 = 0 x 3 (x + 1) = 0 where a, b are the coefficients to be found. Using these equations as the ideal {f 1, f 2} and a single variable x, Mathematica gives the following basis: [15] g 1 = a 11a 2 + a 3 +62b 2a 2 b -31b 2 + 5ab 2 +b 3 g 2 = a +2a 2-39b + 7ab + b x 50ax + 5a 2 x g 3 = 9 a - 2b + 8x - 2ax + bx g 4 = -7 + a + b - 10x + 2ax - 5x 2 + ax 2 g 5 = -1 x - x 3 Notice that the Groebner basis in this case results in what appears to be a more complicated solution to the problem. However, the first basis has eliminated x from the problem. The original problem solves for the three variables {x, a, b} where only two equations are given. The basis contains five equations leading to a solution. The second and third basis are a function of x, a, and b. If we use the system {g 1=0, g 2=0, g 3=0} as the equations to solve for a, b, and x with Mathematica, the variable results a = 5 and b = 2 are obtained and x is approximated. If g 5 is used to find x then a and b are again correctly numerically obtained. Fortunately, we have previously shown that a and b are integers using [12] with a = a(6) = 5 and b = (a(12) a(6) 2 )/2 = 2 and do not require a further solution. Using the function PolynomialReduce the quotient of F[x]/G[x] is found for [14a]; [16] [5 x x 6 +1 (1+x) 6 ]/ [x 3 (x + 1)] = 6x + 9x x x 4 + 5x 5 + 5x 6 +5x 7 + 5x 9 As expected, the quotient is exact and the remainder is zero. Applying equation [13] the sum of the real and complex quotients is the integer 216. This sequence of sums for n = 3 to 8 is 9,20,95,216,567,1792. Application of equation [12] to additional irreducible polynomials G[x]. We investigate the application of the Groebner division algorithm to other polynomials G[x]. Proceeding as in the example above, let G[x] = x 3-2x 2-2x-1 which is the irreducible polynomial with discriminant = 83. Choosing n= 7 let the system of equations to be solved be as follows; [17a,b] a x 14 + bx 7 +1 (2x 2 +2x+1) 7 = 0 x 3 (2x 2 +2x+1) = 0 where the binomial term is chosen to match the terms I the cubic polynomial. Applying the Groebner basis algorithm to solve this system we get: [18] g 1 = a 1010a 2 + a b 58a 2 b b ab 2 + b 3

5 g 2 = a +7248a b ab -136b x ax a 2 x g 3 = a - 385b x + 15ax + 136bx g 4 = a + b x + 113ax x ax 2 g 5 = -1 2x - 2x 2 - x 3 Solving for x with basis g 5 and using {g 1=0, g 2=0} to solve for a and b, a = 1458 and b = 58. If the integer sequence for G[x] is generated from powers of the real and complex solutions as in {1], a[7] = 1458 and a[14] = Inserting into b = (a(14) a(7) 2 )/2 = 58, suggesting equation [12] can modified as in [17a] to be divisible by G[x]. A surprising benefit of the Groebner basis is that the solution to the problem is given in the first basis g 1. A glance at the terms a 2 b and ab 2 shows that the values of a and b are already calculated e.g. (58)a 2 b and 1458ab 2! Note: a(2n) = coefficient of a 2 b coefficient of b 2 term. Also notice that the correct answer is obtain despite the incorrect ideal given in [5b and 5c]. For this G[x] the correct ideals are: [19a,b,c] f 3 = xyz - 1 f 4 = xy + xz + yz + 2 f 5 = x +y + z - 2 However, in the rearrangement of the equations leading to [12] these terms cancel. Also, not shown is the value of A in G[x] = Ax 3 +Bx 2 +Cx+D since we assume it is always equal to 1. In the next example, it will be shown that only the value of D different than (-1) will require the following correction to [12]; [20] F[x] = a(n) x 2n + [(a(2n) a(n) 2 )/2]*x n + D n (1+x) n Let G[x] = x 3-5x 2 + x 2 (discriminant= 907). For n = 4 the first term of the Groebner basis g 1 is, [21] g 1 = a a a b + 177a 2 b b ab 2 + b 3 Reading off terms for a 2 b, ab 2, and a 3 and calculating the quotient F[x]/G[x] it can be shown that: [22] [567 x 8-177x (5x 2 - x +2) 4 ]/ [x 3 (5x 2 - x +2)] = -16x + 84x x x 4-58x 5 In the above sections we have found using a computerized algebra algorithm the integer coefficients a, b and n for which the polynomial F(x) = a x 2n + bx n + D n (Bx 2 -Cx-D) n is divisible by G(x) = x 3 +Bx 2 +Cx+D. Consequently, x is any of the real or complex roots of G such that G(x) = F(x) = 0. Also, the coefficients a, b are integers derived from the integer sequence associated with the cubic polynomial G. Solving F[x] for higher powers of a(n) In this section, we find the form of the polynomials F[x] derived from higher powers of a(n) p, where p = 3 and 4. Knowing that the constant term D influences the form of F[x] that is divisible by x 3 +Bx 2 +Cx+D the Groebner basis for higher powers of a(n) are solved using, respectively for p = 3 and 4; [23] f 13 = (x n + y n +z n ) 3 a(n) 3 [24] f 14 = (x n + y n +z n ) 4 a(n) 4 With the condition

6 [25] f 3 = D n x n y n z n The resulting basis elements g i similar to those found in equations [8 a,b,c,d,e,f] are rearranged as in the example above. For f 13, F[x] is a function of {x,a(n),a(2n),a(3n)} and for F 14, F[x] is a function of {x, a(n),a(2n),a(3n),a(4n)} where; [26 a,b] (x 3n + y 3n +z 3n ) = a(3n) (x 4n + y 4n +z 4n ) = a(4n) The resulting equations are respectively, [27] F[x] = 2x 4n a(n)x 3n a(2n)x 2n + ((a(n) 3 -a(3n)-3d n )/3)x n a(n) D n [28] F[x] = -3D 2n 2a(2n)D n x n + ((a(n) 4 a(4n) 12a(n)D n )/2)x 2n +2x 3n (D n -a(3n)) 3a(2n)x 4n -2a(n)x 5n + 7x 6n As a test to the accuracy of these equations let G(x) = x 3-5x 2 +x-2 and apply with n = 5 the following constants from the corresponding integer sequence. [29 a,b,c,d] a(5) = 2765 a(10) = a(15) = a(20) = Using PolynomialReduce the following quotients are obtained for p = 3 and p = 4 with no remainder. F[x]/G[x] = x 99540x x x x x x x x x x x x x x x x 17 F[x]/G[x] = x 3456x x x x x x x x x x x x x x x x x x x x x x x x x x 27 Generating New Recurrence Relationships Equations [20], [27] and [28] can be summed as in equation [13a] resulting in the following recurrence relations for powers of 2, 3 and 4, respectively: [30] a(n) 3 +2*a(3n)-3*a(n)*a(2n)-6*D n = 0 [31] a(n) 4 + 6*a(4n) 4*a(3n)*a(n) 3*a(2n)^2 12*a(n)*D n = 0 [32] 14*a(6n) 4*a(n)*a(5n) -6*a(2n)*a(4n) + 4*a(3n)*(D n a(3n)) +a(n) 4 *a(2n) a(4n)*a(2n) -16*a(n)*a(2n)*D n 18*D 2n = 0

7 These equations assume an integer sequence generated from any cubic equation G(x) = x 3 +Bx 2 +Cx-D where B and C are positive or negative integers and D is assumed a positive integer with the form of G(x) given above. The sign before the D n and D 2n terms are modified if D is negative. These equations are universal for any value of n. Note that for the Perrin sequence D =1, B = 0 and C = -1. The Groebner basis can be used to generate new recurrence relationships between the various sequence integer coefficients for any value of n and any equation of form G(x). Using equations [30] and [31] as ideals in the Mathematica Groebner Basis command, the following relationships are quickly generated where D = D1, a(n) = a1, a(2n) = a2, a(3n) = a3, a(4n) = a4, a(5n) = a5, a(6n) = a6: [33] a2 6 4a2 3 a3 2 4a a2a3 2 a4 3a2 2 a4 2 2a4 3 12a2 3 D1 2n + 24a3 2 D1 2n 12a2a4D1 2n + 32a3D1 3n + 12D1 4n = 0 [34] a2 5 a a2 2 a a1a3 6 a2 3 a3 3 a4 8a3 5 a4 a2 4 a3a a2a3 3 a4 2 + a2 2 a3a4 3 4a1a3 2 a a3a4 4 + a2 5 a3 2 D1 n 6a2 2 a3 4 D1 n a2 3 a3 2 a4d1 n + 12a3 4 a4d1 n 6a2a3 2 a4 2 D1 n + a2 5 a3d1 2n + 12a2 2 a3 3 D1 2n 12a1a3 4 D1 2n + a2 3 a3a4d1 2n + 24a3 3 a4d1 2n + 6a2a3a4 2 D1 2n a2 5 D1 3n 4a2 2 a3 2 D1 3n + a2 3 a4d1 3n 16a3 2 a4d1 3n + 2a2a4 2 D1 3n 14a2 2 a3d1 4n + 12a1a3 2 D1 4n 16a3a4D1 4n + 10a2 2 D1 5n + 4a4D1 5n 4a1D1 6n = 0 [35] a2 5 a a2 2 a a1a3 5 a2 3 a3 2 a4 8a3 4 a4 a2 4 a a2a3 2 a4 2 + a2 2 a4 3 4a1a3a a4 4 + a2 5 a3d1 n 6a2 2 a3 3 D1 n a2 3 a3a4d1 n + 12a3 3 a4d1 n 6a2a3a4 2 D1 n + a2 5 D1 2n + 12a2 2 a3 2 D1 2n 12a1a3 3 D1 2n + a2 3 a4d1 2n + 24a3 2 a4d1 2n + 6a2a4 2 D1 2n 6a2 2 a3d1 3n 4a1a3 2 D1 3n + 4a1a2a4D1 3n 12a3a4D1 3n 10a2 2 D1 4n + 8a1a3D1 4n 16a4D1 4n + 4a1D1 5n = 0 [36] a2 5 2a2 2 a3 2 4a1a3 3 a2 3 a4 + 4a1a2a3a4 + 4a3 2 a4 2a2a a2 2 a3d1 n 4a1a3 2 D1 n 10a2 2 D1 2n + 4a1a3D1 2n 4a4D1 2n + 4a1D1 3n = 0 [37] 2a2 3 a a1a2a3 3 4a3 4 a2 4 a4 + 4a2a3 2 a4 + a2 2 a4 2 4a1a3a a4 3 4a1a2a3 2 D1 n + 8a3 3 D1 n 4a2a3a4D1 n + 2a2 3 D1 2n 4a1a2a3D1 2n + 16a3 2 D1 2n + 8a2a4D1 2n + 4a1a2D1 3n 8a3D1 3n 12D1 4n = 0 [38] a2 3 a3 2a1a2a a3 3 + a1a2 2 a4 3a2a3a4 + a1a4 2 + a2 3 D1 n 2a3 2 D1 n + a2a4d1 n + 2a1a2D1 2n 10a3D1 2n 6D1 3n = 0 [39] a a1a2 2 a3 2a2a3 2 + a2 2 a4 2a1a3a4 + 2a4 2 2a1a2 2 D1 n + 4a2a3D1 n 2a1a4D1 n + 6a2D1 2n = 0 [40] a1a2 3 2a2 2 a3 2a1a3 2 + a1a2a4 + 2a3a4 + 2a2 2 D1 n 4a1a3D1 n + 2a4D1 n 2a1D1 2n = 0 [41] a2 3 2a1a2a3 + 2a3 2 + a1 2 a4 2a2a4 + 2a1a2D1 n 4a3D1 n 6D1 2n = 0 [42] a1a2 2 + a1 2 a3 + a2a3 a1a4 + a1 2 D1 n 3a2D1 n = 0 [43] a1 2 a2 a2 2 2a1a3 + 2a4 2a1D1 n = 0 Using equations [31] and [32] for the ideals the following are obtained:

8 [44] 3a1 3 a2 3 4a1 3 a a3 3 13a1 3 a2a4 + 28a2a3a4 12a2 2 a5 + 24a4a5 + 14a1 3 a6 56a3a6 48a2 3 D1 n + 4a1 3 a3d1 n + 32a3 2 D1 n + 180a2a4D1 n 168a6D1 n 18a1 3 D1 2n + 24a3D1 2n + 216D1 3n = 0 [45] 36a1 2 a a2 6 72a2 3 a3 2 64a1 3 a a a1 2 a2 3 a4 234a2 4 a4 208a1 2 a3 2 a a2a3 2 a4 676a1 2 a2a a2 2 a a1 3 a2 2 a5 72a1a2 3 a5 192a2 2 a3a5 + 96a1a3 2 a5 208a1 3 a4a a1a2a4a a3a4a5 + 48a1 2 a a1 2 a2 2 a a2 3 a a1 3 a3a6 1232a3 2 a a1 2 a4a6 1092a2a4a6 336a1a5a a6 2 36a1a2 4 D1 n 732a2 3 a3d1 n + 128a1 3 a3 2 D1 n + 208a3 3 D1 n + 156a1a2 2 a4d1 n + 208a1 2 a3a4d1 n a2a3a4D1 n + 48a1 2 a2a5d1 n + 192a2 2 a5d1 n 48a1a3a5D1 n 384a4a5D1 n 224a1 3 a6d1 n 168a1a2a6D1 n 1624a3a6D1 n + 216a1 2 a2 2 D1 2n + 444a2 3 D1 2n 352a1 3 a3d1 2n + 304a3 2 D1 2n 936a1 2 a4d1 2n 1476a2a4D1 2n + 432a1a5D1 2n a6D1 2n + 288a1 3 D1 3n + 216a1a2D1 3n a3D1 3n 2484D1 4n = 0 The remaining recurrence relations had over 200 terms! These equations [30] to [45] were tested with the cubic equation G(x) = x 3-4x 2 + 2x - 3, discriminant = 547 for n = 5, confirming the relations. 1. Cox, D., Little, J., and O Shea, D., Ideals, Varieties, and Algorithms. An Introduction to Computational Algebraic Geometry and Commutative Algebra, 3 rd Ed. Springer Science + Business Media, LCC (2007). Richard Turk April 15, 2017

Polynomials. Chapter 4

Polynomials. Chapter 4 Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation