010: Sum / Sums of Three Cubes, Part 1


Chapter 6:  Third Powers


I. Sum / Sums of cubes (Part 1, in blue)


  1. x3+y3 = z3
  2. x3+y3+z3+t3 = 0
  3. x3+y3+z3 = 1
  4. x3+y3+z3 = 2
  5. x3+y3+z3 = (z+m)3
  6. p(p2+bq2) = r(r2+bs2)
  7. (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k
  8. x3+y3+z3 = at3
  9. x3+y3 = 2(z3+t3)
  10. w3+x3+y3+z3 = nt3
  11. x3+y3+z3 = t2
  12. ak+bk+ck = {x2, y2, z3}, k = 1,2,3
  13. xk+yk+zk = tk+uk+vk,  k = 1,3
  14. xk+yk+zk = tk+uk+vk,  k = 2,3
  15. x3+y3+z3 = 3t3-t
  16. x3+y3+z3 = m(x+y+z)
  17. x1k+x2k+x3k+x4k = y1k+y2k,  k = 1,2,3
  18. x1k+x2k+x3k+x4k = y1k+y2k+y3k+y4k,  k = 1,2,3
  19. ax3+by3+cz3 = N
  20. ax3+by3+cz3+dxyz = 0
  21. x3+y3+z3-3xyz = tk


1. Sum of two cubes:  x3+y3 = z3




(x3+z3)3y3 + (x3-y3)3z3 = (y3+z3)3x3,   if x3+y3 = z3


Or course, by Fermat’s Last Theorem, x3+y3 = z3 has no non-trivial integer solns.  However, in general,

a(-bxy3-cxz3)3 + b(ax3y+cyz3)3 = c(-ax3z+by3z)3,   if ax3+by3 = cz3


Thus, an initial soln can give rise to a second, which in turn leads to a third ad infinitum



2. Form:  x3+y3+z3+t3 = 0


F. Vieta


(a4-2ab3)3 + (a3b+b4)3 + (2a3b-b4)3 =  (a4+ab3)3


P. Sondat


(ap-q)3 + (bp-q)3 + (cp+q)3 + (dp+q)3 = (a3+b3+c3+d3)(a+b+c+d)3


{p,q} = {a+b+c+d, a2+b2-c2-d2}


Thus, given an initial soln, subsequent ones can be found.  A similar identity exists for a3+b3+2(c3+d3) = 0.

Choudhry (complete in the integers, 1998 paper)

x13 x2x3x43 = 0 

dx1 = (a2+ab+b2)+ (2a+b)c3

dx2 = -(a2+ab+b2)+ (a-b)c3 

dx3 = c(-a3+b3+c3)

dx4 = -(a2+ab+b2)(2a+b)c - c4


where a,b,c are arbitrary integers and d  0 is an integer so chosen such that (x1x2x3x4) = 1.

Euler (complete in the rationals)


(p+q)3 + (p-q)3 = (r+s)3 + (r-s)3


p = 3(bc-ad)(c2+3d2)

q = (a2+3b2)2 - (ac+3bd)(c2+3d2)

r = 3(bc-ad)(a2+3b2),

s = -(c2+3d2)2 + (ac+3bd)(a2+3b2)


For details on how Euler derived this, see “Euler’s Extended Conjecture and ak+bk+ck = dk for k>4” by this author.


J. Binet (complete in the rationals)


a3+b3 + n(c3+d3) = 0


a = (1 - mn(p-3q))r

b = (-1 + mn(p+3q))r

c = (m2n - (p+3q))r

d = (-m2n + (p-3q))r


where m = p2+3q2 and r is just a scaling factor.


Proof (Piezas):  For any rational soln a,b,c,d, one can always find rational p,q,r, to get those particular values using the formulas,


{p,q,r} = {(bc+ad-2(bd+ac))/(2v),  (bc-ad)/(2v),  -v(c+d)/(3(bc-ad))}


where v = a2-ab+b2.  For example, for n=1, to get the soln {3,5,4,-6}, the formulas give {p,q,r} = {1,1,1/3}.  In fact, by permuting the {a,b,c,d}, one finds there can be six distinct {p,q,r} up to sign.  Binet’s original soln only discussed the case n=1 but with a little tweaking by this author, one can generalize it and a derivation will be given later.  It should be pointed out that the algebraic form a2+ab+b2, or equivalently x2+3y2, appears a lot when dealing with third and fourth powers.  It may be of interest that while x2+y2 = 1 defines a circle, x2+xy+y2 = 1 yields an ellipse.


J. Steggall (complete)


(s4-rs(p+3q))3 + (-s4+rs(p-3q))3 = (r2-s3(p+3q))3 + (-r2+s3(p-3q))3 


if r = p2+3q2.  For s=1 reduces to Binet’s (with n=1).  Note that Euler, Binet, and Stegall’s solns involve quartic polynomials.  Noam Elkies' complete soln uses only cubic polynomials,


N. Elkies (complete)




av = s3-m-2r3+(r2-s2)t+(s-2r)t2

bv = -s3+m-r3+nt-(s+r)t2

cv =   t3+m+r3+nt-(s+r)t2

dv = -t3+2rs2-r2s+2r3-nt+(s+r)t2


where {m,n} = {rs2-2r2s, s2+2r2} and v is just a scaling factor.


A.Werebrusow (complete)


If 3w2xy = a2+ab+b2, then,


(-ax+wy2)3 + (ay-wx2)3 = (-bx+wy2)3 + (by-wx2)3 = ((a+b)x+wy2)3 + (-(a+b)y-wx2)3,




(ax-wy2)3 + (-bx+wy2)3 = (ay-wx2)3 + (-by+wx2)3 


For the second formula, set {a,b,w,x,y} = {(p-3q)r,  (p+3q)r,  r,  1, p2+3q2} to get Binet’s soln.  Variations of this were found by K. Schwering and Ramanujan.




(-a2d+bc2)3 + (ad2-b2c)3 =  c3(ab-cd)3 + d3(ab-cd)3,   if a3+b3 = c3+d3


Thus, this is another identity such that one soln to a3+b3+c3+d3 = 0 leads to a second. Related to the identity below,


A. Desboves


(b3-d3)(a2d-bc2)3 + (a3-c3)(ad2-b2c)3 = (ab-cd)3(-b3c3+a3d3)


C. Hermite


((a+2b)c-1)3 + (-a-2b+c2)3 = ((a-b)c-1)3 + (-a+b+c2)3,   if c = a2+ab+b2


S. Baba


(pq)3 + (p+6q3)3 = (p-6q3)3 + ((p+12)q)3,  if p = q6-4


J. Young


(a2+6ab3-3c)3 – (a2-6ab3-3c)3 = b3(a2+6a+3c)3  b3(a2-6a+3c)3,  if c = b6-1


The above generalizes Baba’s.  Also useful for finding Martin triples.  There are also quadratic parametrizations though these are no longer complete.


J. Young


(p2+16pq-21q2)3 + (-p2+16pq+21q2)3 + (2p2-4pq+42q2)3 = (2p2+4pq+42q2)3




(3x2+5xy-5y2)3 + (4x2-4xy+6y2)3 + (5x2-5xy-3y2)3 = (6x2-4xy+4y2)3


Other binary quadratic form solns have been found by other authors such as Gerardin, Womack, etc.  A generalization has been found by this author as,




(ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-v2xy+cwy2)3 = (a3+b3+c3+d3) (x2+wy2)3


where {v1, v2, w} = {c2-d2,  a2-b2,  (a+b)(c+d)} 


thus for any soln to a3+b3+c3+d3 = N where N is zero or any number of cubes, then one can always find a quadratic parametrization.  It is also the case that,


(ax-v1y)k + (bx+v1y)k + (cx+v2y)k + (dx-v2y)k = (ax+v1y)k + (bx-v1y)k + (cx-v2y)k + (dx+v2y)k,  for k =1,3




If a+b = c+d.  Let {u1, u2} = {c-d,  a-b}, then,


(ax2+u1xy+by2)3 + (bx2-u1xy+ay2)3 + (cx2-u2xy+dy2)3 + (dx2+u2xy+cy2)3 = (a3+b3+c3+d3) (x2+y2)3 


There are many particular cubic equations with this property, one of which is 93+133+193+233 = 283, (9+23 = 13+19) as well as those in a nice arithmetic progression like,


113+123+133+143 = 203

313+333+353+373+393+413 = 663


the latter found by D. Rusin in the context of a certain elliptic curve.  There is also a quadratic identity, similar to the general one,


J. Nicholson


(ap-qx)3 + (bp+qx)3 + (cp-qy)3 + (dp+qy)3(a3+b3+c3+d3((a+b)x2+(c+d)y2)3 


where {p, q} = {(a+b)x2+(c+d)y2,  (a2-b2)x+(c2-d2)y}

Y. Perelman (submitted by Balarka Sen)

If a3+b3+c3+d3 = p3+q3+r3+s3 = 0, then,


(a-pu)3 + (b-qu)3 + (c-ru)3 + (d-su)3 = 0 


where u = (pa2+qb2+rc2+sd2) / (ap2+bq2+cr2+ds2)



2a. Form: x3+ny3+(n+1)z3 = 0



(Update, Sept 17, 2013)  James Buddenhagen solved,


x3 + ny3 + (n+1)z3 = 0




x = 4n5+10n4+28n3+32n2+8n-1

y = n5-8n4-32n3-28n2-10n-4

z = -n5-13n4-10n3+10n2+13n+1


Note that if n is a cube, and the z term is distributed, then this also solves x13 + x23 + x33 + x43 = 0.



3. Form: x3+y3+z3 = 1


It turns out solving this in the integers may involve an elliptic curve or a Pell equation.  Using Werebrusow’s identity with one term set equal to 1,


(1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1,  where,

a2+ab+b2 = 3c(ac-1)2      (eq.0)


then given any non-trivial soln {x,y,z}, one can always find rational {a,b,c} as {a,b,c} = {(y-c2)/(1-c3),  (-c2x-z)/(1-c3),  (1-x)/(y+z)}.


However, the conditional equation (eq.0) above is a quadratic in b and to make its discriminant a square, one has to solve,


y2 = 3(4c3-1)a2-24c2a+12c


which is an elliptic curve in c, or equivalently, a quadratic curve in a.  This is easier made a square if its constant term is also a square. Let c = 3v2,


y2 = 3(108v6-1)a2-216v4a+36v2


This is of the form y2 = ax2+bx+c2 where an infinite number of integral solns can be found by solving the Pell equation p2-aq2 = ±1.  (See the relevant section in Part 2, Quadratic Polynomial as a kth power.)  Sparing the reader some algebra, this gives,




(1-ac+bc)3 + (a+c2-ac3)3 + (ac3-b-c2)3 = 1    (eq.1)


1. {a,b,c} = {12qrt,  3(q-r)(3q+r)t,  3s2t2}, and,

2. {a,b,c} = {12qrt,  -3(q+r)(3q-r)t,  3s2t2}


where, for both, r = p-18qs3t3 and the condition p2-3(108s6t6-1)q2 = s.  By setting s = ±1, this is of course a Pell equation.  Note that the second addend of (eq.1), since it does not contain the b variable, is unchanged, implying a pair of solns which share a term thus solving the system,


x13+x23 = x33+x43 = x53+1


Example, for the simplest case s = t = 1, this gives p2-321q2 = 1, with fundamental {p,q} = {215, 12} yielding the pair,


{x,y,z} = {4528, 3753, -5262}

{x,y,z} = {-3230, 3753, -2676}


It turns out that for s = 1, the Pell equation,


p2-3(108t6-1)q2 = 1


has its fundamental soln given by the parametrization {p1, q1} = {216t6-1, ±12t3}.  From this one can then generate an infinite family.  Using trivial {p0, q0} = {1, 0} on the expressions for {x,y,z} yields the well-known 4-deg identity,


(1-9t3)3 + (9t4)3 + (3t-9t4)3 = 1


and {p1, ±q1} with one sign yields a 10-deg,


(1+9t3+648t6-3888t9)3 + (-135t4+3888t10)3 + (-3t-81t4+1296t7-3888t10)3 = 1


while the other sign gives a 16-deg.  The next, {p2, ±q2} gives a 22 and 28-deg, {p3, ±q3} a 34 and 40-deg, and so on for an infinite sequence with degree k = 6n+4.  One of the terms is always a polynomial with only even powers, unchanged for ±t, so a term is shared by two distinct solns to x3+y3+z3 = 1.  (This in fact is the same infinite family that can be derived by a recursion found by D.H.Lehmer.) 


Note:  Aside from s = 1, I do not know of any other non-trivial integral s such that p2-3(108s6t6-1)q2 = s is solvable in the integers {p,q} for a given t.


One can also use the binary quadratic forms together with Pell equations to find an infinite number of integral solns if an initial one is known.  To recall,


(ax2-v1xy+bwy2)3 + (bx2+v1xy+awy2)3 + (cx2+v2xy+dwy2)3 + (dx2-v2xy+cwy2)3 = (a3+b3+c3+d3)(x2+wy2)3


{v1, v2, w} = {c2-d2,  a2-b2,  (a+b)(c+d)} 


so it suffices to find one soln to a3+b3+c3+1 = 0.  It is easy to set {x,y} = {p+v2q, 2q} to transform the fourth addend x2-v2xy+cwy2 to the form p2-nq2(The discriminant n is a function of a,b,c with 3! = 6 possible values and it is the experience of this author that at least one has n > 0).  Assume d = ±1 and one can then solve the Pell equation p2-nq2 = ±1.  For example, using the famous taxicab number 1728 = 93+103 = 123+13, and after a little modification,


(9+44pq-404q2)3 + (10+45pq-417q2)3 = (12+56pq-518q2)3 + 1,  if  p2-85q2 = -4


(9+44pq+404q2)3 + (10+45pq+417q2)3 = (12+56pq+518q2)3 + 1,  if  p2-85q2 = 4


Using another initial soln, 63 + 83 = 93 – 1, we get,


(6+222pq+4014q2)3 + (8+270pq+4806q2)3 = (9+312pq+5616q2)3 - 1,  if  p2-321q2 = 1


and so on. 


A. Gerardin


(p4+9pq3)3 + (3q2)6 = (3p3q+9q4)3 + p12


For p=1 this gives a parametrization to a3+b3+c3 = 1.



4. Form: x3+y3+z3 = 2


(1+ax3)3 + (1-ax3)3 = 6a2x6 + 2


It suffices to make a = 6 to satisfy the equation, with x arbitrary so there are parametrizations for x13+x23+x33 = n for n = 1,2. 


(Update, 12/14/09):  Alain Verghote gave a simpler derivation of the above.  Given the basic identity,


(1+n)3 + (1-n)3 = 6n2 + 2


one then simply lets n = 6x3.  (End update.) 


There is a fifth power version,


x15+x25+x35+x45+x55+x65+x75 =  n,


also for n =1,2, with n=1 to be discussed in a later chapter.  For n=2 by Seiji Tomita, surprisingly this is dependent on Pythagorean triples and is analogous to the cubic case,


(1+ax5)5 + (1-ax5)5 + (1+bx5)5 + (1-bx5)5 + (-1+cx5)5 + (-1-cx5)5 + (dx4)5 = 2


where {a,b,c,d} should satisfy the two conditions a2+b2 = c2 and 20a2b2 = d5, one soln of which is {a,b,c,d} = {270, 360, 450, 180}.



5. Form: x3+y3+z3 = (z+m)3


J. Jandasek (m =1)


n3 + (3n2+2n+1)3 + (3n3+3n2+2n)3 = (3n3+3n2+2n+1)3


This implies that any integer n appears in a cubic quadruple at least once.  A similar identity exists for second powers as was already discussed.


G. Ampon (m =1)


(3n2)3 + (6n2+3n+1)3 + (9n3+6n2+3n)3 = (9n3+6n2+3n+1)3


Q: Any soln for some other constant m?



6. Form: p(p2+bq2) = r(r2+bs2)    


Derivation: To derive Binet’s version, for convenience, set,


(p+q)3 + (p-q)3 = (r+s)3 + (r-s)3


Expanding, we get  p(p2+3q2) = r(r2+3s2).  This can generalized to,


p(p2+bq2) = r(r2+bs2)               (eq.1)


Assume that,


r2+bs2 = (p2+bq2)(u2+bv2)        (eq.2)


Factor over √-b and equate factors,


r+s√-b = (p+q√-b)(u+v√-b)

r-s√-b = (p-q√-b)(u-v√-b)


Solve for {r,s},


r = pu-bqv                    (eq.3)

s = pv+qu                     (eq.4)


Substitute eq.2 into eq.1 (by eliminating p2+bq2),


p = r(u2+bv2)                (eq.5)


Substitute r from eq.3 into eq.5,


p = (pu-bqv)(u2+bv2)


Collect {p,q},


p(-1+u(u2+bv2)) = qbv(u2+bv2)


And the equation is true if,


p = bv(u2+bv2),   q = -1+u(u2+bv2)


Substitute these into eq.3 and eq.4 to get,


r = bv,   s = -u+(u2+bv2)2


and we have all unknowns {p,q,r,s} for any b.  But this can be generalized even further.



7. Form: (x+c1y)(x2+c2xy+c3y2)k = (z+c1t)(z2+c2zt+c3t2)k


Since by using the transformation {x,y} = {p-c1q, q} on the left hand side (and a similar one for the right) will transform this into the form p(p2+apq+bq2)k, one can simply assume c1 = 0 without loss of generality,


p(p2+apq+bq2)k = r(r2+ars+bs2)k




{p,q,r,s} = {bvwk,  -1+uwk,  bv,  -(u+av)+wk+1},  if w = u2+auv+bv2


This is easily be proven for k=1,2.  For k=1 and a=0, this reduces to the formula for third powers given previously.  For k=2, this is relevant to equal sums of fifth powers to be discussed later.  Using computer algebra one can see it is also true for other k>2, but I have no proof it is the case for all positive integer k.



8. Form: x3+y3+z3 = mt3


A. Werebrusow


(p2r3+qs3)3 + (p2r3-qs3)3 + (-6rs2)3 = 2(p2r3)3,  if pq = ±6


Other poly solns for m ≥ 2?  Using Ryley’s Identity below (also given in Part 1), it can be shown there are integral {x,y,z,t} for any constant m,


S. Ryley


(p3+qr)3 + (-p3+pr)3 + (-qr)3 = m(6mnp2)3, 


{p,q,r}= {m2+3n3,  m2-3n3,  36m2n3}


with arbitrary n.  Note that the first term is a 6-deg poly in m. Can smaller deg poly be found that works for any m?


(Update, 4/19/10):  A 3rd-deg polynomial in m is possible. William Ellison cited one example from the book Cubic Forms by Yuri Manin as,


(m3-36n9)3 + (-m3+35mn6+36n9)3 + (33m2n3+35mn6)3 = m(32m2n2 +34mn5+36n8)3


for arbitrary n.  Note how this has simple coefficients that are only powers of 3.  Robert Israel pointed out that a small tweak can reduce the powers as,


(27m3-n9)3 + (-27m3+9mn6+n9)3 + (27m2n3+9mn6)3 = m(27m2n2 +9mn5+3n8)3


See also Ryley's Theorem.  (End update.)



9. Form: x3+y3 = 2(z3+t3)


Given one soln to the above equation, a second one can be found as,


A. Gerardin


x3+y3+2(z3+t3) = (p3+q3+2(r3+s3)) (p3-2r3)3


{x,y,z,t} = {p(v+3r3),  q(v-3r3),  -2rv,  s(v-3r3)},  if v = p3+r3


As we’ll later see, the more general result by Desboves is that given an initial soln to the sum of cubes ax13+a2x23+...+ anxn3 = 0, then one can find a second.  While the complete soln to a3+b3 = n(c3+d3) is given by the modified Binet formula, some nice partial solns for n=2 are,


A. Gerardin


(a3+3b3)3 + (a3-3b3)3 = 2a9 + 2(3ab2)3


(a2+4ab-b2)3 + (-a2+4ab+b2)3 = 2(a+b)6 - 2(a-b)6



9b. Form: x3+y3 = 2(z3+Poly)


(Update, 12/14/09):  Alain Verghote


4(ac+bd+e)3+4(ad+bc-e)3 = x3+3xy2


where {x,y} = {(a+b)(c+d),  (a-b)(c-d)+2e}.


Note that the polynomial x2+3y2 appears often when dealing with third and fourth powers.



10. Form: w3+x3+y3+z3 = ntk



(Update, 12/14/09)Gerhard Paseman


(a-1)3 - (a+1)3 + (c+1)3 - (c-1)3 = (6b)2,   where a2 + 6b2 = c2


(Update, 12/15/09):  Alain Verghote


((s+1)2)3 + (2-(s+1)2)3 + (-(s-1)2)3 + (-2+(s-1)2)3  = 64x6,   where s = 3x2


for arbitrary x. One can choose x = y2 to make the sum a 4th power.  (End update)


A. Martin (also the case k = 3 of Boutin's Identity)


(-a+b+c)3 + (a-b+c)3 + (a+b-c)3 + d3 = (a+b+c)3,   if 24abc = d3




This is a special case of a more general identity.  Using the nice initial soln 113+123+133+143 = 203, then,


(11x2+xy+14y2)3 + (12x2-3xy+13y2)3 + (13x2+3xy+12y2)3 + (14x2-xy+11y2)3 = 203(x2+y2)3


Binary quadratic form solns to x13+x23+…+xn3 = t3 can be found for any n > 2 given an initial soln and the identity found by this author discussed previously.




(a2+ab)k + (a2-ab)k + (b2+ab)k + (b2-ab)k  = 2(a2+b2)k,  for k = 1,2,3.


By finding expressions {a,b} such that a2+b2 = cm (which is easily done), the above equation provides a template for solving,


x13 + x23 + x33 + x43 = 2t3m


for any m. A similar identity for fourth powers was found by Ramanujan,


x14 + x24 + x34 = 2t2m


to be discussed later and a fifth power version was found by this author as well.



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