A New Method in the Problem of Three Cubes

Transcription

1 Universal Journal of Computational Mathematics 5(3): 45-56, 07 DOI: 0.389/ujcmj A New Method in the Problem of Three Cubes Armen Avagyan, Gurgen Dallakyan Armenian State Pedagogical University after Khachatur Abovyan, Armenia Copyright c 07 by authors, all rights reserved. Authors agree that this article remains permanently open access under the terms of the Creative Commons Attribution License 4.0 International License Abstract In the current paper we are seeking P (y), P (y), P 3 (y) with the highest possible degree polynomials with integer coefficients, and Q(y) via the lowest possible degree polynomial, such that P 3 (y)+p 3 (y)+p 3 3 (y) = Q(y). Actually, the solution of this problem has close relation with the problem of the sum of three cubes a 3 + b 3 + c 3 = d, since deg Q(y) = 0 case coincides with above mentioned problem. It has been considered estimation of possibility of minimization of deg Q(y). As a conclusion, for specific values of d we survey a new algorithm for finding integer solutions of a 3 + b 3 + c 3 = d. Keywords Diophantine Equation, Sum of Three Cubes, Parametic Solutions 00 MSC Y50, D5, D45 Some facts on the history of the equation a 3 + b 3 + c 3 = d The question as to which integers are expressible as a sum of three integer cubes is over 60 years old. The first known reference to this problem was made by Fermat, who has offered to find the three nonzero integers, so that the sum of their nth powers is equal to zero. This framework of the problem makes the beginning of survey of equation a 3 + b 3 + c 3 = d for many mathematicians. We present some significant result schedule. 85 year S. Ryley in [] gave a parametrization of rational solutions for d Z: x = (9k6 30d k 3 + d 4 )(3k 3 + d ) + 7d 4 k 3 6kd(3k 3 + d ) y = 30d k 3 9k 6 d 4 6kd(3k 3 + d ) z = 8dk5 6d 3 k (3k 3 + d ) 908 year A.S. Werebrusov [] found the following parametric family for d = : (6t 3 + ) 3 (6t 3 ) 3 (6t ) 3 = 936 year Later in [3] Mahler discovered a first parametric solution for d = : (9t 4 ) 3 + (3t 9t 4 ) 3 + ( 9t 3 ) 3 = 94 year Mordell proved in [] that for any other d a parametric solution with rational coefficients must have degree at least year Miller and Woollett [4] discovered explicit representations for 69 values of d between and 00. Their search exhausted the region a, b, c year 963 Gardiner, Lazarus, and Stein [5] looked at the equation x 3 + y 3 = z 3 d in the range 0 x y 6, where 0 z x 6 and 0 d 999. Their search left only 70 values of d between and 000 without a known representation including eight values less than year the first solution for d = 39 was found. Heath-Brown, Lioen, and te Riele [8] determined that 39 = ( 59380) 3 with the rather deep algorithm of Heath-Brown [6]. This algorithm involved searching for

2 46 A New Method in the Problem of Three Cubes k < T , 65, 38, 366, 390, 40, 534, 564, 579, 609, 67, 633, 73, 758, 786, 789, 795, 834, , 903, 906, 9, 948, 975 solutions for a specific value of d using the class number of Q( 3 d) to eliminate values of a, b, c which would not yield a solution. 994 year Koyama [7] used modern computers to expand the search region to a, b, c and successfully found first solutions for 6 integers between 00 and 000 [9]. Also in 994, Conn and Vaserstein [8] chose specific values of d to target, and then used relations implied by each chosen value to limit the number of triples (a, b, c) searched. So doing, they found first representations for 84 and 960. Their paper also lists a solution for each d < 00 for which a representation was known. 995 year Bremner [9] devised an algorithm which uses elliptic curve arguments to narrow the search space. He discovered a solution for 75 (and thus a solution for 600), leaving only five values less than 00 for which no solution was known. Lukes then extended this search method to also find the first representations for each of the values 0, 435, and 478 [0]. 997 year Koyama, Tsuruoka, and Sekigawa [] used a new algorithm to find first solutions for five more values between 00 and 000 as well as independently finding the same solution for 75 that Bremner found. Also in the same paper, the authors discuss the complexity of the above algorithms. 999 year Bernstein [] had implemented the method of Elkies [3] and found solutions for new values of d. Summarizing the above, it can be noted that up to st century only 7 values were left unresolved. These, together with the range of their search, are presented in the table below. Only recently, in 007, Elsenhans and Yahnel [4] found the solutions for the values d = 56, 38, 366, 40, 564, 758, 789, 894, = = = = = = = = = = = = Thus, until 000 there are only the numbers 33, 4, 4, 65, 390, 579, 67, 633, 73, 795, 906, 9, 975 lefts, which have not yet been solved, all the other presentations are posted on the web, in particular, it was made by Sander Huisman [5]. Some known notes about the equation a 3 + b 3 + c 3 = d for specific values of d Consider first the equation This has infinitely many solutions because of the identity a 3 + b 3 + c 3 =. () ( + 9m 3 ) 3 + (9m 4 ) 3 + ( 9m 4 +3m) 3 = () but there are other solutions as well. Are there any other identities that give a different -parameter family of solutions? Is every solution of () a member of a family like this? In general it s known that there is no finite method for determining whether a given Diophantine equation has solutions. However, its an open problem whether if there is a general method for determining if a given Diophantine equation has algebraic solutions, i.e., an algebraic identity like the one above that gives an infinite family of solutions. More specifically, is there a proposition, that only equations of genus < can have an algebraic solution.

3 Universal Journal of Computational Mathematics 5(3): 45-56, n m a b c by It may be worth mentioning that the complete rational-solution of the equation a 3 + b 3 + c 3 = d 3 is known, and is given a = q[ (x 3y)(x + 3y )] b = q[ (x + 3y)(x + 3y )] c = q[(x + 3y ) (x + 3y)] d = q[(x + 3y ) (x 3y)] where q, x, y are any rational numbers. So if we set q equal to the inverse of [(x + 3y ) (x 3y)] we have rational solutions of (). However, the problem of finding the integer-solutions is more difficult. If d is allowed to be any integer (not just ) then Ramanujan gave the integer solutions a = 3n + 5nm 5m, b = 4n 4nm + 6m, c = 5n 5nm 3m, d = 6n 4nm + 4m. This occasionally gives a solution of equation () (with appropriate changes in sign), as in the following cases However, this doesn t cover all of the solutions given by (). By the way, the equation a 3 + b 3 + c 3 = has algebraic solutions [6], other than (). There are known to be infinitely many algebraic solutions, for example ( 9t t t 9 ) 3 + ( 35t t 0 ) 3 +(3t 8t 4 96t t 0 ) 3 = However, it s not known whether every solution of the equation lies in some family of solutions with an algebraic parameterization. Interestingly, note that if you replace by, then again there s a parametric solution: (6t 3 + ) 3 (6t 3 ) 3 (6t ) 3 = (3) but again this doesn t cover all known integer solutions. Note, that precisely one solution is known that is not given by (3) (see [6]): = It s evidently not known up todays if there are any other algebraic solutions besides the one noted above. In general it seems to be a difficult problem to characterize all the solutions of a 3 + b 3 + c 3 = d (4) for some arbitrary integer d >. In particular, the question of whether all integer solutions are given by an algebraic identity seems both difficult and interesting. Note that for d ±4(mod9) there are no solutions since, for any integer a, a 3 0,, (mod9). It is a long standing problem as to whether every rational integer d 4, 5(mod9) can be written as a sum of three integral cubes. According to the web page [] of Daniel Bernstein, the first attacks by computer were carried out as early as 955. Nevertheless, for example, for d = 3, there is still no solution known apart from the obvious ones: (,, ), (4, 4, 5), (4, 5, 4), and ( 5, 4, 4). For d = 30, the first solution was found by N. Elkies and his coworkers in 000 [7]. It is

4 48 A New Method in the Problem of Three Cubes interesting to note that, in 99, D.R. Heath-Brown [6] had made a prediction on the density of the solutions for d = 30 without knowing any solution explicitly. Over the years, a number of algorithms have been developed in order to attack the general problem. An excellent overview concerning the various approaches invented up to around 000 was given in [8], published in 007. The historically first algorithm which has a complexity of O(B +ε ) for a search bound of B is the method of R. Heath-Brown [6]. For d > Kenji Koyama [7] has generated a large table of integer solutions of a 3 + b 3 + c 3 = d for noncubes in the range d 000 and a b c consists of two tables: Table (55 pages) contains the integer solutions, sorted by d, and Table ( pages) lists the number of primitive solutions found for each d in the search range. Consider now some specific: d = m 3, d = m and d = m 9 type values of d. Multiply both sides of (3) by m 9, and apply the change of variable t t/m to obtain the more general solution (6t 3 + m 3 ) 3 (6t 3 m 3 ) 3 (6t ) 3 = m 9 (5) which is primitive for GCD(6t, m) =. If GCD(6t, m) >, then dividing (4) by (GCD(6t 3, m 3 )) 3 gives a primitive solution. For example, for l, k the solutions ( 3t 3 + 3l m 3 ) 3 ( 3t 3 3l m 3 )3 ( l 3mt ) 3 = 9l m 9 (6) ( t k m 3 ) 3 ( t 3 3 3k m 3 ) 3 ( 3 k mt ) 3 = 3 9k 3 m 9 (7) ( t 3 + 3l 3 3 k m 3 ) 3 ( t 3 3l 3 3k m 3 ) 3 ( l 3 k kmt ) 3 = = 9l 3 9k 3 m 9 (8) are primitive for GCD(3t, m) =, GCD(t, 3m) = and GCD(t, 6m) = respectively. Equations (5)-(7) give polynomial families for n =, 8, 458, 65536, 933, , ,... An analogous procedure may be applied to (4) to obtain families of solutions for numbers of the form m. Multiplying both sides by m and applying the transformation t t/m gives (9mt 3 + m 4 ) 3 (9t 4 + 3mt) 3 + (9t 4 ) 3 = m (9) which is primitive for GCD(3t, m) =. In particular, for 3 m and k, (3 k mt k m 4 ) 3 (t k m 3 t) 3 + (t 4 ) 3 = 3 k 6 m (0) is primitive for GCD(t, 3m) =. Equations (8) and (9) give families of solutions for n =, 79, 4096, , 67776, , ,... 3 New method and results In this section a new method and results are surveyed. Here we consider more general framework of the problem of sums of three cubes. We are seeking P (y), P (y), P 3 (y) with the highest possible degree polynomials with integer coefficients and Q(y) with the lowest possible degree polynomial, so that P 3 (y) + P 3 (y) + P 3 3 (y) = Q(y). Actually the solution of this problem has close relation with the above trivial problem, since the case of deg Q(y) = 0 coincides with our problem. Nevertheless the estimation of possibility of minimization of deg Q(y) itself is also an interesting problem. RESULT. The first result of this paper is devoted to the case of degrees (8, 8, 6). We search the desired polynomials within the class of polynomials of the form (ax 8 + bx 5 + cx ) 3 (ax 8 + b x 5 + c x ) 3 (Ax 6 + Bx 3 + C) 3.

5 Universal Journal of Computational Mathematics 5(3): 45-56, First we expand it C 3 3BC x 3 + (c 3 3B C 3AC c 3 )x ( B 3 + 3bc 6ABC 3b c )x ( 3AB + 3b c + 3ac 3A C 3b c 3ac )x + + (b 3 3A B b 3 + 6abc 6ab c )x ( A 3 + 3ab 3ab + 3a c 3a c )x 8 + (3a b 3a b )x. Then we take b = b, c = A3 +3a c 3a, B = Ab 3a, C = A4 aab +6a Ac 9a and obtain the form 3 A 79a 9 + A9 b 43a 8 + A6 b 4 43a 7 + A3 b 6 79a 6 A9 c 8a 7 4A6 b c 8a 6 A3 b 4 c 8a A6 c 7a 5 + 4A3 b c 7a 4 8A3 c 3 7a 3 + ( + A9 b 8a 7 4A6 b 3 8a 6 A3 b 5 8a 5 + 8A6 bc 7a 5 + 8A3 b 3 c 7a 4 8A3 bc 9a 3 ( A 6 b + 7a 5 + A3 b 4 9a 4 + A6 c 9a 4 4A3 b c 9a 3 A3 c ) 3a x ( A 6 b 9a 4 + 4A3 b 3 7a 3 A3 bc 3a ) x 9 Further considerations are devoted to the finding of cases interesting for us. CASE : b = 0. The result has the form A 79a 9 A9 c 8a 7 + 4A6 c 7a 5 8A3 c 3 7a 3 + ( A 6 c 9a 4 A3 c ) 3a x 6 ) x 3 + SUBCASE.: c = 0. The result obtains the form A 79a 9, which is a cube of an integer, so it is primitive (not interesting). SUBCASE.: c = A3 3a, the result is A 79a 9, again primitive. CASE : c = 3A3 +4ab 8a. The result: Factor the last term, then the result obtains the form A3 b a 6 A3 b 5 x 3 ( A 9 79a a 6 A3 b 4 ) 43a 4 x 6 A3 ( 3A 3 + ab )(3A 3 + ab ) 97a 6. We do the substitution a = 3A3 b. Then the result will get the form: 64b A 6 64b5 x A 3 Finally taking x = yb A we obtain the result interesting for us: 648y3. Practical considerations. Now its the time to investigate this result for applications in the solving process of the equation (). Since for max[abs[a, b, c]] 0 4 there are well known tables in [], so we seek solutions of () satisfying the condition max[abs[a, b, c]] 0 5 with possible small values of abs[d] (desirably less than 000). First rewrite the result: (54y ( + 36y y 6 )) 3 (8y ( + 08y y 6 )) 3 ( + 6y y 6 ) 3 = 648y 3

6 50 A New Method in the Problem of Three Cubes Of course, calculations expected to be significantly hard, so we ll use Mathematica.0 code: G8[y ] := /GCD [ (54y ( + 36y y 6 )), (8y ( + 08y y 6 )), ( + 6y y 6 ) ] F8[y ] := G8[y] { (54y ( + 36y y 6 )), (8y ( + 08y y 6 )), ( + 6y y 6 ) } V8[y ] :=G8[y] 3 ( 648y 3 ) For {i = 50, i 50, i + +, If[Abs[V8[i]] < , If[Max[Abs[F8[i]]] > , Print[{i, F8[i], V8[i]}]]]] The result is: {, { , , }, } { 0, { , , }, } { 9, { , , }, 47 39} {9, { , , }, } {0, { , , }, } {, { , , }, } This means that, for example = 4739 RESULT. More enhanced result is obtained for the case (9, 9, 7): Expand[(3( + 0y y y 9 )) 3 (( + 6y y y 9 )) 3 (4y(5 + 34y y 6 )) 3] y 3 G9[y ] := /GCD[(3( + 0y y y 9 )), (( + 6y y y 9 )), (4y(5 + 34y y 6 ))] F9[y ] := G9[y] {(3( + 0y y y 9 )), (( + 6y y y 9 )), (4y(5 + 34y y 6 ))} V9[y ] := G9[y] 3 (8 + 07y 3 ) For {i = 50, i 50, i + +, If[Abs[V9[i]] < , If[Max[Abs[F9[i]]] > , Print[{i, F9[i], V9[i]}]]]] The result is: {, { , , }, } { 0, { , , }, } { 9, { , , }, } { 8, { , , }, } { 7, { , , }, } { 6, { , , }, } { 5, { , , }, } { 4, { , , }, } {4, { , , }, } {5, { , , }, } {6, { , , }, } {7, { , , }, }

7 Universal Journal of Computational Mathematics 5(3): 45-56, 07 5 {8, { , , }, } {9, { , , , 34}, } {0, { , , }, } {, { , , }, } Here the most interesting triple is: = RESULT 3. Consider now the case (5, 5, 8). Through the same way we get: Expand [( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) ) 3 ( 8 y(63 36y3 + 80y y + 768y 8 + 5y 4 ) ) 3 ( 3 (3 + 0y6 + 3y + 3y 8 ) ) 3 ] 8 3y 6 Then we use the code: G5[y ] := /GCD [( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) ) (, 8 y(63 36y3 + 80y y + 768y 8 + 5y 4 ) ) (, 3 (3 + 0y6 + 3y + 3y 8 ) )] F5[y ] := G5[y] {( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) ) ( 8 y(63 36y3 + 80y y + 768y 8 + 5y 4 ) ) (, 3 (3 + 0y6 + 3y + 3y 8 ) )} V5[y ] := G5[y] 3 ( 8 3y 6 ) For {i = 50, i 0, i + +, If[Abs[V5[i]] < , If[Max[Abs[F5[i]]] > , Print[{i, F5[i], V5[i]}]]]] The result is: { 8, { , , }, } { 4, { , , }, } { 0, { , , }, } { 8, { , , }, } { 6, { , , }, } { 4, { , , }, } { 3, { , , }, } {, { , , }, } {, { , , }, } { 0, { , , }, }

8 5 A New Method in the Problem of Three Cubes { 9, { , , }, } { 8, { , , }, } { 7, { , , }, } { 6, { , , }, } RESULT 4. Case (7, 7, 0). Expand[ ( y y y y 7 ) 3 ( y y y y 7 ) 3 (34y ( y y y 8 )) 3] y 6 G7[y ] := /GCD[( y y y y 7 ), ( y y y y 7 ), (34y ( y y y 8 ))] F7[y ] := G7[y] {( y y y y 7 ), ( y y y y 7 ), (34y ( y y y 8 ))} V7[y ] := G7[y] 3 ( y 6 ) For {i =, i 0, i + +, For {j =, j 00, j + +, If[GCD[i, j] ==, If[Abs[V7[i/j]] < , If[Max[Abs[F7[i/j]]] > , Print[{i/j, F7[i/j], V7[i/j]}]]]]]] The result: {, { , , { }, }, { , ,

9 Universal Journal of Computational Mathematics 5(3): 45-56, { 3 { 6 { 3 { 6 { 3 { 39 { 3 3 { }, }, { , , }, }, { , , }, }, { , , }, }, { , , }, }, { , , }, }, { , , }, }, { , , }, }, { , , }, } 4 Summary Denote m = [ max{deg(p ), ] deg(p ), deg(p 3 )}/deg(q), where P 3 (y) + P 3 (y) + P3 3 (y) = Q(y) and n = IntegerPart Max[Abs[a,b,c]], where a 3 + b 3 + c 3 = d as a describers of solution quality. Note that if sup n(d) = +, d then the equation a 3 + b 3 + c 3 = d has infinite set of solutions. We consider the values of m and n for each result. It is clear, that as much greater are n and m as cool is result. Result. We have (54y + 944y y 8 ) 3 (8y + 944y y 8 ) 3 ( + 6y y 6 ) 3 = 648y 3 where m = 8 3. Consider the code: G8[y ] := /GCD [ (54y ( + 36y y 6 )), (8y ( + 08y y 6 )), ( + 6y y 6 ) ] F8[y ] := G8[y] { (54y ( + 36y y 6 )), (8y ( + 08y y 6 )), ( + 6y y 6 ) } V8[y ] :=G8[y] 3 ( 648y 3 ) For {i = 00, i 00, i + +, If[Abs[V8[i]] < 0 000, If[Max[Abs[F8[i]]] > , Print[{i, F8[i], V8[i]}]]]] The result is: {, { , , 47 05}, 583} {, { , , 50 56}, 585} So n = IntegerPart [ ] = 40. Result. We have ( y y y 9 ) 3 ( + 6y y y 9 ) 3 (0y + 96y y 7 ) 3 = y 3 where m = 3. Now consider the code: G9[y ] := /GCD[(3( + 0y y y 9 )), (( + 6y y y 9 )), (4y(5 + 34y y 6 ))]

10 54 A New Method in the Problem of Three Cubes F9[y ] := G9[y] {(3( + 0y y y 9 )), (( + 6y y y 9 )), (4y(5 + 34y y 6 ))} V9[y ] := G9[y] 3 (8 + 07y 3 ) For {i = 50, i 50, i + +, If[Abs[V9[i]] < 0 000, If[Max[Abs[F9[i]]] > , Print[{i, F9[i], V9[i]}]]]] The result is: {, { , , }, 8548} {, { , , 0 43}, 8604} ] = 55. So n = IntegerPart [ noindent Result 3. We have ( ) 3 ( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) 8 y(63 36y3 + 80y 6 + ( ) y + 768y 8 + 5y )) 4 3 (3 + 0y6 + 3y + 3y 8 ) = 8 3y 6 where m = 5 6. Note that here the leading coefficient of Q is significantly small. Consider the code: G5[y ] := /GCD [( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) ) (, 8 y(63 36y3 + 80y y + 768y 8 + 5y 4 ) ), ( 3 (3 + 0y6 + 3y + 3y 8 ) )] F5[y ] := G5[y] {( 8 y( y3 + 80y y + 768y 8 + 5y 4 ) ) ( 8 y(63 36y3 + 80y y + 768y 8 + 5y 4 ) ), ( 3 (3 + 0y6 + 3y + 3y 8 ) )} V5[y ] := G5[y] 3 ( 8 3y 6 ) For {i =, i 50, i + +, If[Abs[V5[i]] < , If[Max[Abs[F5[i]]] > , Print[{i, F5[i], V5[i]}]]]] The result is: {4.{ , , }, 6657} {6, { , , }, } {8, { , , }, } So n = IntegerPart [ ] = Result 4. We have ( y y y y y y 7 ) 3 ( y y y y y y 7 ) 3 (574470y y y y 0 ) 3 = y 6 where m = 9. Consider the code: G7[y ] := /GCD[( y y y y 7 ),

11 Universal Journal of Computational Mathematics 5(3): 45-56, ( y y y y 7 ), (34y ( y y y 8 ))] F7[y ] := G7[y] {( y y y y 7 ), ( y y y y 7 ), (34y ( y y y 8 ))} V7[y ] := G7[y] 3 ( y 6 ) For {i =, i 0, i + +, For {j =, j 00, j + +, If[GCD[i, j] ==, If[Abs[V7[i/j]] < , If[Max[Abs[F7[i/j]]] > , Print[{i/j, F7[i/j], V7[i/j]}]]]]]] The result: {, { , , { }, }, { , , { }, } 3, { , , { }, } 6, { , , { }, } 3, { , { , }, } 6, { , { , }, } 3, { , , { }, } 39, { , { , }, } 3 3, { , , { }, } 3 6, { , , }, } Note, that nevertheless is rather greater than the considered range, however three cubes are essentially big numbers: =

12 56 A New Method in the Problem of Three Cubes So [ ] n = IntegerPart = = Taking in account results obtained we pose the following hypotheses: Hypothesis. If deg(q) = 0, then Q(y) = d 3 or Q(y) = d 3 (d is a constant). Hypothesis. If deg(q) 0, then sup m = +. REFERENCES [] S. Ryley, The Ladies Diary (85), 35. [] L.J. Mordell, On Sums of Three Cubes, Journal of the London Mathematical Society 7 (94), MR (4:89d). [3] Kurt Mahler, Note On Hypothesis K of Hardy and Littlewood, Journal of the London Mathematical Society (936), [4] J.C.P. Miller and M.F.C. Woollett, Solution of the Diophantine Equation x 3 + y 3 + z 3 = k, Journal of the London Mathematical Society 30 (955), 0-0. MR (6:797e). [5] V.L. Gardiner, R.B. Lazarus, and P.R. Stein, Solutions of the Diophantine Equation x 3 + y 3 = z 3 d, Mathematics of Computation 8 (964), MR (3:9). [6] D.R. Heath-Brown, W.M. Lioen, and H.J.J. te Riele, On Solving the Diophantine Equation x 3 + y 3 + z 3 = k on a Vector Computer, Mathematics of Computation 6 (993), MR060 (94f:3). [7] Kenji Koyama, Tables of solutions of the Diophantine equation x 3 + y 3 + z 3 = n, Mathematics of Computation 6 (994), [8] W. Conn and L.N. Vaserstein, On Sums of Three Integral Cubes, Contemporary Mathematics 66 (994), MR84068 (95g:8). [9] Andrew Bremner, On sums of three cubes, Canadian Mathematical Society Conference Proceedings 5 (995), MR35393 (96g:04). [0] Richard F. Lukes, A Very Fast Electronic Number Sieve, University of Manitoba doctoral thesis, 995. [] Kenji Koyama, Yukio Tsuruoka and Hiroshi Sekigawa, On Searching for Solutions of the Diophantine Equation x 3 + y 3 + z 3 = n, Mathematics of Computation 66 (997), MR4094 (97m:04). [] Bernstein, D., Three cubes, available at: [3] Noam Elkies, < elkies@abel.math.harvard.edu > x 3 + y 3 + z 3 = d, 9 July 996, < nmbrthry@listserv.nodak.edu > via [4] [5] Sander G. Huisman, Newer sums of three cubes, archive.org. [6] Payne G., Vaserstein L.N., Sums of three cubes. Pages in The Arithmetic of Function Fields, de Gruyter, 99. [7] Elkies, N. D., Rational points near curves and small nonzero x3 y via lattice reduction, in: Algorithmic number theory (Leiden 000), Lecture Notes in Computer Science 838, Springer, Berlin 000, MR (00g:035). [8] Beck M., Pine E., Tarrant W. and Yarbrough Jensen K.: New integer representations as the sum of three cubes, Math. Comp. 76 (007), MR99795 (007m:70).