### 011: Sum / Sums of Three Cubes, Part 2

 Back to Index     I. Sum / Sums of cubes (Part 2, in blue)   x3+y3 = z3 x3+y3+z3+t3 = 0 x3+y3+z3 = 1 x3+y3+z3 = 2 x3+y3+z3 = (z+m)3 p(p2+bq2) = r(r2+bs2) (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k x3+y3+z3 = at3 x3+y3 = 2(z3+t3) w3+x3+y3+z3 = nt3 x3+y3+z3 = t2 ak+bk+ck = {x2, y2, z3}, k =1,2,3 (Martin triples) xk+yk+zk = tk+uk+vk,  k = 1,3 xk+yk+zk = tk+uk+vk,  k = 2,3 x3+y3+z3 = 3t3-t x3+y3+z3 = m(x+y+z) x1k+x2k+x3k+x4k = y1k+y2k,  k = 1,2,3 x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 1,2,3 ax3+by3+cz3 = N ax3+by3+cz3+dxyz = 0 x3+y3+z3-3xyz = tk   11. Form: x3+y3+z3 = t2   V. Bouniakowsky:   (n3+1)3 + (-n3+2)3 + (3n)3 = (3(n3+1))2   E. Catalan   (a4+2ab3)3 + (b4+2a3b)3 + (3a2b2)3 = (a6+7a3b3+b6)2   A. Gerardin   (9a4+8ab3)3 + (4ab3)3 + (4b4)3 = (27a6+36a3b3+8b6)2   Any other solns with addends as polynomials of small degree?  The ff identity proves one soln to x3+y3+z3 = t2 leads to another,   A. Werebrusow   If a3+b3+c3 = d2, then,   (a-px)3 + (b+px)3 + (c-x)3 = (d-qx)2,    where x = 3(a+b)p2+3c-q2,  q = (3p(a2-b2)+3c2)/(2d),  for arbitrary p.      12. Form: ak+bk+ck = {x2, y2, z3} for k =1,2,3 (aka Martin triples)   A Martin triple is a triple of integers {a,b,c} such that,    a + b + c = x2            (eq.1) a2 + b2 + c2 = y2          (eq.2) a3 + b3 + c3 = z3          (eq.3)   The smallest in positive integers is M1 = {1498, 2461, 7490}, so,   1498 + 2461 + 7490 = 1072 14982 + 24612 + 74902 = 80252 14983 + 24613 + 74903 = 75973   There are also an infinite number of distinct Martin triples unscaled by a square factor, and this can be proved via a polynomial identity or an elliptic curve.  Originally, A. Martin solved only eq.2 and eq.3, and gave a polynomial identity of high degree (see below), but any soln can be made valid for eq.1 as well.  Each term is simply multiplied with a scaling factor m which is the square-free part of the sum a+b+c = mn2.  For ex, the above was derived from,   142 + 232 + 702 = 752 143 + 233 + 703 = 713   and since a+b+c = 14+23+70 = 107, this yields M1 after multiplying all terms by m = 107.  These "pre-triples" are quite rare.  Given a3+b3+c3 = z3 (eq.3), of around 267000 positive and primitive solns with z < 100,000 in J. Wroblewski’s database here (which is up to z = 106), only ten can be transformed into Martin triples, with the next three being,  {3, 34, 114} {18, 349, 426} {145, 198, 714}    However, if positive and negative solns are allowed there is a parametrization to this,   A. Martin (using Young’s soln)   (a2+6ab3-n)k + (-a2+6ab3+n)k + (b(a2-6a+n))k   where n = 3(b6-1), for k = 3 already sums up to a cube q3.  Expanding for k = 2, if it is to be a square p2, as a polynomial in a this has the form,   (b2+2)a4+c3a3+c2a2+c1a+c02(b2+2) = p2   where the ci are polynomials in b.  This quartic is easily made a square if b2+2 = y2.  If so, two small solns are,   a = (b8+8b6+12b4-b2+4)/(2b2+4)   a = 6(b8+2b6-b2-2)/(b8+8b6+12b4-b2+4)   which can then be used to compute further rational points on this curve.  It still remains to make b2+2 = y2 which is easily done as b = (u2-2v2)/(2uv).  Considering this results in high degree polynomials it is not surprising the smallest positive soln found by Martin had 17 digits.  In general, it can be proven that finding these triples may involve an elliptic curve.  Given an initial soln to a3+b3+c3 = d3, one can always generate quadratic form parametrizations using an identity found by this author.  For ex, using the smallest positive pre-triple {14, 23, 70}, define the function M(x) as,   M(x):= (-8x2+36x+14)k + (9x2-25x+23)k + (-6x2-8x+70)k   For k = 3 this is already the perfect cube (x2-9x+71)3.  For k = 2, this yields the quartic,   M(x):= 181x4-930x3+1335x2+1262x+752 = y2   which is to be made a square. One soln, not surprisingly, is x = 0 with another as x = 68/75, and so on.  It turns out an analogous situation can be extended to fourth powers as,   ak+bk+ck+dk = {p2, q4} for k = 2,4   with the only known soln,   159352+270222+579102+592602 = 885972 159354+270224+579104+592604 = 701214   found by this author using Wroblewski’s database for fourth powers. Any others?     13. Form: ak+bk+ck = dk+ek+fk,  k =1,3   Just like for k = 1,2, the eqn   ak+bk+ck = dk+ek+fk,                (eq.1)   also has a complete soln for k = 1,3.  We can solve this in three ways.     First, is by Choudhry (Some diophantine problems concerning equal sums of integers and their cubes, 2010).  Theorem:  The complete solution to eq.1 for k = 1,3 entails solving the system,   u+v+w = x+y+z uvw = xyz   Proof:   Given the invertible linear transformation,   {a, b, c} = {-u+v+w, u-v+w, u+v-w} {d, e, f}  = {-x+y+z,  x-y+z,  x+y-z}   then eq.1 becomes,   u+v+w = x+y+z (u+v+w)3 - 24uvw = (x+y+z)3 - 24xyz   Second, using the form by Lander,  (u+x)k + zk + (v+y)k = (v+x)k + yk + (u+z)k   call this L0, a complete soln for k = 1 which he used for k = 1,5.  Expanding L0 for k = 3,   u2x-v2x+ux2-vx2+v2y+vy2-u2z-uz2 = 0   As a quadratic in z, this has discriminant D,   D:= -4(x-y)uv2-4(x2-y2)uv+u2(u+2x)2   which must be made a square.  Since this discriminant as a polynomial in v is a quadratic with a square constant term, it is easily made a square using Fermat’s method.  A special case is when one of the terms of L0 is zero, say letting y = 0, since this gives the complete symmetric ideal soln of degree four,   (t+a)k+(t+b)k+(t+c)k+(t+d)k+(t+e)k = (t-a)k+(t-b)k+(t-c)k+(t-d)k+(t-e)k   for k = 1,2,3,4 which is true for any t iff ak+bk+ck+dk+ek = 0 for k = 1,3.    Third, another approach which turns out to be simpler is to use the form,   (a+bp+q)k + (b-bp+q)k + (c+ap+q)k = (a+cp+q)k + (b+ap+q)k + (c-cp+q)k   call this L1 since this author found it while studying Lander’s work on L0 for k = 1,5.  This is a subset of L0, yet still complete for multi-grades with one k >1 (the easy proof will be given in Fifth Powers).  For k = 1,3, expanding L1 at k = 3 gives just a linear condition in q,   -a2+b2+c2+ab+ac+bc+2(b+c)q = 0   Solving for q then gives the complete soln of eq.1 in four variables {a,b,c,p}.  (And just like in the previous approach, any of the terms of L1 can be set equal to zero, say a+bp+q = 0, solve for q, and the linear condition will now be in the variable p.)   One can see the approach can easily be generalized for k = 1,4 and higher.  So L1 for k = 1,4, as a polynomial in p after removing the trivial factor p(p-1)(b-c), is,   (Poly10)p2+(Poly20)p+(Poly30) = 0   with discriminant D that is a quadratic polynomial in q, while for k = 1,5,   (Poly11)p2+(Poly21)p+(Poly31) = 0   has a D that is a quartic in q, and for k = 1,2,6,   (Poly12)p2+(Poly22)p+(Poly32) = 0   with D that is again a quartic in q, so the problem can be treated as finding rational points on the curve D = y2.  Thus, L1 is a very versatile form because when expanded for higher powers, the variable p remains of relatively low degree and for either of the systems k = 1,5 or k = 1,2,6, the complete soln involves solving only a quadratic eqn and an elliptic curve.  These will be discussed in more detail later.   Another special case of eq.1 also has abc = def.  One way to solve this is to use the form,   (ap)k + (bq)k + (cr)k =  (bp)k + (cq)k + (ar)k   which already satisfies the condition.  Expanding for k = 1,3 and solving the two eqns yields,   {p,q,r} = {ab-c2,  -a2+bc,  -b2+ac}   There is a second distinct form though,   (ap)k + (bq)k + (cr)k =  (-bp)k + (-cq)k + (ar)k   with {p,q,r} = {ab+c2,  -a2-bc,  b2-ac}.  In fact, it can easily be proven that if eq.1 has abc = def, then a+b+c = d+e+f = 0.   Proof:  The system ak+bk+ck = dk+ek+fk for k = 1,3 where abc = def, by eliminating {c,f} from this system depends on the final eqn,   ab(a+b) = de(d+e)   call this E1. However, letting {c,f} = {-a-b, -d-e} and substituting into the system, this also satisfies k = 1,3 given E1, proving the assertion. (End proof)  An earlier proof was given by Choudhry,   (Update, 3/23/10):  Choudhry (2001):  "If ak+bk+ck = dk+ek+fk,  for k = 1,3, (eq.1 & 2) and abc = def (eq.3), then a+b+c = d+e+f = 0."   Proof:  One can use the identity,   (a+b+c)3 + 2(a3+b3+c3)-6abc = 3(a+b+c)(a2+b2+c2)   and a similar one in the variables {d,e,f}.  If eqs.1,2,3 hold, then   (a+b+c)(a2+b2+c2) = (d+e+f)(d2+e2+f2)     (eq.4)   For a+b+c = d+e+f ≠ 0, this implies a2+b2+c2 = d2+e2+f2.  But Bastien’s Theorem excludes a non-trivial soln to,   ak+bk+ck = dk+ek+fk,  for k = 1,2,3,   Thus, for eq.4 be non-trivially true, it must be the case that a+b+c = d+e+f = 0.  Eqs 1,2,3 are then satisfied if,   ak+bk+(-a-b)k = dk+ek+(-d-e)k,  for k = 1,3   where ab(a+b) = de(d+e), parametric solns of which are easily found.  (End proof.)  Source: Triads of cubes with equal sums and equal products, The Mathematics Student, Vol. 70, 2001.  (End update)   Theorem (Piezas):  If ak+bk+ck = dk+ek+fk for k = 1,3 where abc = def (or equivalently, a+b+c = d+e+f = 0), then,   9abc(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9)               (eq.1)Proof:  Simply substitute {c,f} = {-a-b, -d-e} into eq.1 and this also has E1:= ab(a+b)-de(d+e) as a factor.    (Update, 6/22/10):  For the weaker system,   If a3+b3+c3 = d3+e3+f3 where abc = def, then,  3(a3+b3+c3)(a6+b6+c6-d6-e6-f6) = 2(a9+b9+c9-d9-e9-f9)    (eq.2)  This has several easy solns: 1. S. VandermergelIf a3+b3 = c3+d3, then (ac)3+(bc)3+(d2)3 = (ad)3+(bd)3+(c2)3.   2. Gerardin   (a2p)3 + (b2q)3 + (abr)3 = (a2q)3 + (-b2r)3 + (-abp)3   where {p,q,r} = {a3-2b3,  a3+b3,  2a3-b3}.   3. Choudhry gave a 3-parameter version,   (ap)3 + (bq)3 + (cr)3 = (aq)3 + (-br)3 + (-cp)3   where {p,q,r} = {a3-2b3-c3,  a3+b3+2c3,  2a3-b3+c3}.   4. Piezas   (ap)3 + (bq)3 + (cr)3 = (ar)3 + (bp)3 + (cq)3   where {p,q,r} = {ab-c2,  -a2+bc,  -b2+ac}.   The last identity, discussed in the previous paragraph, is also valid for x1+x2+x3 = y1+y2+y3 = 0.For fourth powers, if ak+bk+ck+dk = ek+fk+gk+hk where k = 2,4 and abcd = efgh, then, 4(a2+b2+c2+d2)(a6+b6+c6+d6-e6-f6-g6-h6) = 3(a8+b8+c8+d8-e8-f8-g8-h8)A related identity also exists for fifth powers.     14. Form: xk+yk+zk = tk+uk+vk, k = 2,3   Identities that are good for squares can be extended for particular values to cubes as well, just as this one,   C. Goldbach   p2 + q2 + (p+q+3r)2 = (p+2r)2 + (q+2r)2 + (p+q+r)2   Alternatively, this can be more symmetrically expressed as,   (p-r)k + (q-r)k + (p+q+r)k = (p+r)k + (q+r)k + (p+q-r)k   which is already for k = 2, but is also valid for k = 3 if 6pq = r2.     15. Form: x3+y3+z3 = 3t3-t   M. Noble   Given a3+b3+c3 = 3d3-d, then,   {a,b,c,d} = {(4p-4r)/(5q),  (p+4r)/(5q),  (3p)/(5q),  (4p)/(5q)}   where {p,q,r} = {15u2+4v2, 15u2+9uv-4v2, 10uv+3v2}   This soln has the special property such that by defining the expressions,   {x,y,z} = {a3-d3, b3-d3, c3-d3}   then {x-(x+y+z)3,  y-(x+y+z)3,  z-(x+y+z)3} are perfect cubes.  (This is a particular case of a more general identity by Noble.)     16. Form: x3+y3+z3 = m(x+y+z)   L.Aubry   x3+y3+z3 ± (x+y+z) = 0   {x,y,z} = {u+v,  -u+v,  -2v(3n+1)},  if  u2-(1+12n+36n2+36n3)v2 = ±n   with n some constant.  For example, for n=1 this entails solving either u2-85v2 = ±1 depending on which sign is involved.  In general, if x,y,z need not be co-prime, then,   {x,y,z} = {n(u+v),  n(-u+v),  -2nv(3n2+1)},  if  u2-(1+12n+36n2+36n3)v2 = ±1.     17. Form: x1k+x2k+x3k+x4k = y1k+y2k,  k = 1,2,3   By Bastien’s Theorem, the system of eqns,   x1k+x2k+…+xnk = y1k+y2k+…+ynk,  k = 1,2,3   has a non-trivial soln only for n>3.  However, one can always set one term x1 = 0.  A simple soln is,   J. Nicholson   ak + bk + (2a+4b)k + (3a+3b)k = (a+3b)k + (2a+b)k + (3a+4b)k,  k = 1,2,3   Piezas   But it is also possible two terms on one side are zero.  The complete soln for that case is,   ak + bk + ck + dk = ek + fk,   {d,e,f} = {(-abc)/x,  ((a+b)(a+c)(b+c)+y)/(2x),  ((a+b)(a+c)(b+c)-y)/(2x)}   where x = ab+ac+bc and a,b,c satisfy (a2-x)(b2-x)(c2-x) = y2.  Two easy solns can be found.  First, let b = c, and,   {a,b,c,y} = {p+q, q, q, 2(p+q)qr},  where p2 = 2q2 + r2.   Second, let a+b = c,   {a,b,c,y} = {-2p+q,  2p+q,  2q,  4(2p-q)(2p+q)r},  where p2 = q2 + r2.  Any more?     18a. Form: x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 1,3   (Update, 8/6/09):  Tony Rizzo gave the soln,   (p+1)k + (q+1)k + (r-1)k + (s-1)k = (p-1)k + (q-1)k + (r+1)k + (s+1)k,  for k = 1,3,   where p2+q2 = r2+s2 and solved the condition as,   (a-d)2+(b+c)2 = (a+d)2+(b-c)2    where a/b = c/d.  One can make this valid for k = 1,2,3 by re-arranging terms as,   (p+t)k + (-p+t)k + (q+t)k + (-q+t)k = (r+t)k + (-r+t)k + (s+t)k + (-s+t)k,  for k = 1,2,3,   for any t and hence is a special case of Theorem 5 discussed in the section on Equal Sums of Like Powers.  (End update.)    18b. Form: x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 1,2,3   The complete soln to this system is already known.  For multigrades valid for the consecutive powers k = 1,2,…m by Frolov’s Theorem it is always possible to add a constant to each term such that,   x1+x2+… +xn = y1+y2+…+yn = 0         (eq.1)   so we can make use of this property.  One approach to completely solving k = 1,2,3 is then to use the Chernick-Lander form,   (a+b+c)k + (a-b-c)k + (-a-b+c)k + (-a+b-c)k = (d+e+f)k + (d-e-f)k + (-d-e+f)k + (-d+e-f)k   which satisfies eq.1.  The solution to this, which is in binary quadratic forms, also solves k = 5 (by Gloden’s Theorem) so will be discussed in the section on Fifth Powers.  Necessarily some of the terms will be negative so this form does not parametricize the even system,   x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k   for k = 2,4,6.  To include this, we have to solve k = 1,2,3 without appealing to eq.1.  Use the general form F2,   (a+bh)k + (c+dh)k + (e+fh)k + (g+h)k = (a-bh)k + (c-dh)k + (e-fh)k + (g-h)k   To satisfy k = 1,2, set {g,f} = {-(ab+cd+ef),  -(1+b+d)}.  Expanding for k = 3, we get a quadratic in h,   (1+b)(1+d)(b+d)h2-(Poly1) = 0   where Poly1 is a quadratic in the variables a,b,c,d,e.  To find rational h, one must make its discriminant a square,   (1+b)(1+d)(b+d)(Poly1) = y2               (eq.2)   which is easily done since the expression is just a quadratic in {a,c,e}.  It is well-known that given an initial soln to F(x) = y2 where F(x) is a quadratic polynomial, then one can find its complete soln. (One way is to use Fermat’s method.)  As a polynomial in e, an initial soln to eq.2 is,   e = (a+ab-c+cd)/(b+d)   from which one can then find the complete soln.  Using this, one can prove that,   Theorem:  Given ak+bk+ck+dk = ek+fk+gk+hk for k = 1,2,3.  Define n as a2+b2+c2+d2 = n(a+b+c+d)2, then,   25(a4+b4+c4+d4-e4-f4-g4-h4)(a6+b6+c6+d6-e6-f6-g6-h6) = 12(n+1)(a5+b5+c5+d5-e5-f5-g5-h5)2   This 4-6-5 Identity, by squaring all variables, can become a 8-12-10 Identity good for the even system k = 2,4,6 and analogous to Ramanujan’s 6-10-8 which depends on a k = 2,4.  This will be discussed more in Fourth Powers.     18c.  Form: x13+x23+...+xn3 = Poly   (Update, 12/15/09):  Tony Rizzo gave the ff cubic and quintic identities,   (ax2+y)3 + (ax2-y)3 + (a-2)(ax2)3 = (a2x3)2 + (axy)2 + (6-a)ax2y2         (eq.1)   (ax2+y)5 + (ax2-y)5 + (a-2)(ax2)5 = (a3x5+10xy2)2 + 10(a-10)x2y4       (eq.2)   for arbitrary {a,x,y}.  For eq.1, when a = 6, the RHS is a sum of two squares.  This itself is a square if {36x3, 6xy} = {2m, m2-1}, thus,   (6x2+y)3 + (6x2-y)3 + 4(6x2)3 = (6xy+2)2,    if 6xy = 324x6-1   For eq.2, when a = 10, the RHS becomes Bouniakowsky's quintic-square eqn,   (10x2+y)5 + (10x2-y)5 + 8(10x2)5 = 100x2(100x4+y2)2   Note:  The relationship between the two can be appreciated more by considering the basic identities,   (a+y)3 + (a-y)3 + 4a3 = 6a(a2 +y2)        (Id.1)   (a+y)5 + (a-y)5 + 8a5 = 10a(a2 +y2)2    (Id.2)   It is easy to make the sum of Id.1 a square, while that of Id.2 a fourth power.   Piezas   (a+y)3 + (a-y)3 + 4a3 = (6m)2(2n2-y)2     where {a,y} = {6n, n2-9}, and n = m2, for arbitrary m.   (a+y)5 + (a-y)5 + 8a5 = (10m)4(2n2-y)4     where {a,y} = {10n, n2-25}, and n = 100m4, for arbitrary m.   (End update.)   19. Form: ax3+by3+cz3 = N   Theorem 1: If ax3+by3+cz3 = N has a solution, then one can generally find a second. (A. Desboves).   Proof:  ap3+bq3+cr3 = (ax3+by3+cz3)(ax3-by3)3   {p,q,r} = {(ax4+2bxy3),  -(2ax3y+by4),  z(ax3-by3)}   Special cases a=b=1 were found by J.Prestet and A. Legendre while a different version was given by Cauchy.  (Desboves also found a rather similar identity for fourth powers.)   Special cases of the general form are given by,   S. Realis (quadratic)   (2x2-4xy+9yz-9z2)3 + (2y2-xy+9xz-18z2)3 = 9(2x2-4xz-yz+y2)3,  if x3+y3 = 9z3   Piezas   Let q = p3+1, then   (px2-p2xy+qyz-qz2)3 + (py2-xy+qxz-pqz2)3 = q(px2-p2xz-yz+y2)3,  if x3+y3 = qz3   (Realis’s was the case p=2 of this identity.)   E.Lucas (nonic)   p3+q3+cr3 = 27 (x3+y3+cz3) (x7y+x4y4+xy7)3   {p,q,r} = {x9+6x6y3+3x3y6-y9,  -x9+3x6y3+6x3y6+y9,   3xyz(x6+x3y3+y6)}   More generally,   Theorem 2: If ax13+a2x23+...+ amxm3 = N has a solution, then this generally leads to a second.   Proof:  Desboves’ identity in Theorem 1 is essentially,   a(ax4+2bxy3)3  + b(-2ax3y-by4)3 = (ax3+by3)(ax3-by3)3   If ax3+by3 is the sum of cubes c1z13+c2z23+…+ cmzm3 = N, then,   a(ax4+2bxy3)3 + b(-2ax3y-by4)3 = (c1z13+c2z23+…+ cmzm3)(ax3-by3)3 = 0   so, by distribution, one gets a new identity of the form ax13+a2x23+...+ amxm3 = 0 if there are initial solns x,y, and zi. (End proof)   Form p3+q3 = nr3   Expressing a given number n as the sum of two rational cubes is simply the case a = b = 1, and c = 0, of Desboves' identity above.  The fact that an initial soln leads to more can be seen in a better light by using the birational transformation, {p,q,r} = {36n-y,  36n+y,  6x}, to reduce the eqn to the elliptic curve,   y2 = x3-432n2   and can be reversed as,   {x,y} = {12n/(p+q),  -36n(p-q)/(p+q)}   More generally though,   Theorem 3: If au3+bv3+cw3 = 0 has a solution, then so does y2 = x3-432a2b2c2.   Proof:  y2-x3+432a2b2c2 = (au3+bv3+cw3)(au3-bv3-cw3)(432b2c2/u6)   {x,y} = {4(p2+q)/r2,  4(2p3+3pq)/r3},  where {p,q,r} = {bv3-cw3,  3bcv3w3,  uvw}   Theorem 4: If au3+bv3+cw3 = 0 has a solution, then so does x3+y3+abcz3 = 0.   Proof:  x3+y3+abcz3 = 27bcv3w3(au3+bv3+cw3)(p2+q)3   {x,y,z} = {p3+3bqv3,  -p3+3cqw3,  3(p2+q)r},  where {p,q,r} as in theorem 3.   A generalization was found by J.Sylvester.  (Who found theorems 3 and 4?).  }      20. Form: ax3+by3+cz3 = dxyz   S. Realis (complete, other than the case x = y = z)   x3+y3+z3 = 3xyz   {x,y,z} = {(a-b)3+(a-c)3,  (b-a)3+(b-c)3,  (c-a)3+(c-b)3}   All positive integers N other than those div by 3 but not by 9 are representable as N = x3+y3+z3-3xyz with integral x,y,z => 0.  The primes (other than 3) are representable in this manner in one and only one way (Carmichael).   Theorem 1: If ax3+by3+cz3+dxyz = 0 has a solution, then one can generally find a second. (A. Cauchy)   Proof:  ap3+bq3+cr3+dpqr = (ax3+by3+cz3+dxyz)(pqr)/(xyz)   {p,q,r} = {x(by3-cz3),  y(-ax3+cz3),  z(ax3-by3)}   This generalizes Desboves’ result in a certain way.  Cauchy original statement was that given an initial soln {x,y,z}, a second soln {p,q,r} is,   p/(v1x) = q/(v2y) = r/(v3z)   {v1, v2, v3} =  {by3-cz3,  -ax3+cz3,  ax3-by3}   After a little algebraic manipulation, one can get the above identity.   E. Lucas   Similarly, given an initial soln {x,y,z}, a second {p,q,r} satisfies the elegant relations p/x+q/y+r/z = 0 and apx2+bqy2+ crz2 = 0.  One can then solve for {p,q} which gives essentially the same result as Cauchy’s.   Theorem 2: If ax3+by3+cz3+dxyz = 0 has a soln, then so does p3+q3+abcr3+dpqr = 0. (J. Sylvester)   Proof:  {p,q,r} = {f2g+g2h+h2f-3fgh,  fg2+gh2+hf2-3fgh,  xyz(f2+g2+h2-fg-fh-gh)},  where {f,g,h} = {ax3,  by3,  cz3}.   Note that this generalizes theorem 4 of the previous section which is just the case d=0.  For a=b=1, d=0, and after some simplification this also gives the nonic parametrization of Lucas.   C.Souillart, E. Mathieu   x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)(u3+nv3+n2w3-3nuvw)   {x,y,z} = {pu+n(qw+rv),  pv+qu+nrw,  pw+qv+ru}   which proves that, like the quadratic form x2+y2, the product of two forms x3+ny3+n2z3-3nxyz is of like form.  In general, the result extends to cyclic determinants of order m.   21. Form: x3+y3+z3-3xyz = tk   Krafft, Lagrange   To recall, the algebraic form we can designate as f2:= x2-dy2 factors over the square root extension d1/2 (which may involve the complex unit).  For its cubic analogue, use the general form f3:= x3+ny3+n2z3-3nxyz which factors linearly over a complex cube root of unity w as,   f3:= x3+ny3+n2z3-3nxyz = (x+my+m2z)(x+mwy+m2w2z)(x+mw2y+m2wz)   where w3-1 = 0 and m = n1/3 for convenience.  Or alternatively, if one has Mathematica,   f3:= Resultant[x+wy+w2z, w3-n, w]   Because it has three linear factors and variables, one can then easily solve the equation,   x3+ny3+n2z3-3nxyz = (p3+nq3+n2r3-3npqr)k   for any positive integer k (just like for f2) by factoring both sides.  In fact, for the special case f3 = ±1, this is sometimes known as the cubic Pell equation, an analogue to the Pell equation f2 = ±1 since starting with one initial soln {p,q,r}, one can then find an infinite number of solutions {x,y,z}.  Specifically, one solves the system,   x+my+m2z = (p+mq+m2r)k x+mwy+m2w2z = (p+mwq+m2w2r)k x+mw2y+m2wz = (p+mw2q+m2wr)k   for x,y,z and where m = n1/3.  For k=2, this gives,   {x,y,z} = {p2+2nqr,  nr2+2pq,  q2+2pr}   and so on for other k.       ◄
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