I. Sum / Sums of cubes (Part 2, in blue)
11. Form: x^{3}+y^{3}+z^{3} = t^{2}
V. Bouniakowsky:
(n^{3}+1)^{3} + (n^{3}+2)^{3} + (3n)^{3} = (3(n^{3}+1))^{2}
E. Catalan
(a^{4}+2ab^{3})^{3} + (b^{4}+2a^{3}b)^{3} + (3a^{2}b^{2})^{3} = (a^{6}+7a^{3}b^{3}+b^{6})^{2}
A. Gerardin
(9a^{4}+8ab^{3})^{3} + (4ab^{3})^{3} + (4b^{4})^{3} = (27a^{6}+36a^{3}b^{3}+8b^{6})^{2}
Any other solns with addends as polynomials of small degree? The ff identity proves one soln to x^{3}+y^{3}+z^{3} = t^{2} leads to another,
A. Werebrusow
If a^{3}+b^{3}+c^{3} = d^{2}, then,
(apx)^{3} + (b+px)^{3} + (cx)^{3} = (dqx)^{2},
where x = 3(a+b)p^{2}+3cq^{2}, q = (3p(a^{2}b^{2})+3c^{2})/(2d), for arbitrary p.
12. Form: a^{k}+b^{k}+c^{k} = {x^{2}, y^{2}, z^{3}} for k =1,2,3 (aka Martin triples)
A Martin triple is a triple of integers {a,b,c} such that,
a^{ }+ b^{ }+ c = x^{2} (eq.1) a^{2 }+ b^{2 }+ c^{2} = y^{2 }(eq.2) a^{3 }+ b^{3 }+ c^{3} = z^{3 }(eq.3)
The smallest in positive integers is M_{1} = {1498, 2461, 7490}, so,
1498 + 2461 + 7490 = 107^{2} 1498^{2} + 2461^{2} + 7490^{2} = 8025^{2} 1498^{3} + 2461^{3} + 7490^{3} = 7597^{3}
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 squarefree part of the sum a+b+c = mn^{2}. For ex, the above was derived from,
14^{2} + 23^{2} + 70^{2} = 75^{2} 14^{3} + 23^{3} + 70^{3} = 71^{3}
and since a+b+c = 14+23+70 = 107, this yields M_{1} after multiplying all terms by m = 107. These "pretriples" are quite rare. Given a^{3}+b^{3}+c^{3} = z^{3} (eq.3), of around 267000 positive and primitive solns with z < 100,000 in J. Wroblewski’s database here (which is up to z = 10^{6}), 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)
(a^{2}+6ab^{3}n)^{k} + (a^{2}+6ab^{3}+n)^{k} + (b(a^{2}6a+n))^{k}
where n = 3(b^{6}1), for k = 3 already sums up to a cube q^{3}. Expanding for k = 2, if it is to be a square p^{2}, as a polynomial in a this has the form,
(b^{2}+2)a^{4}+c_{3}a^{3}+c_{2}a^{2}+c_{1}a+c_{0}^{2}(b^{2}+2) = p^{2}
where the c_{i} are polynomials in b. This quartic is easily made a square if b^{2}+2 = y^{2}. If so, two small solns are,
a = (b^{8}+8b^{6}+12b^{4}b^{2}+4)/(2b^{2}+4)
a = 6(b^{8}+2b^{6}b^{2}2)/(b^{8}+8b^{6}+12b^{4}b^{2}+4)
which can then be used to compute further rational points on this curve. It still remains to make b^{2}+2 = y^{2} which is easily done as b = (u^{2}2v^{2})/(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 a^{3}+b^{3}+c^{3} = d^{3}, one can always generate quadratic form parametrizations using an identity found by this author. For ex, using the smallest positive pretriple {14, 23, 70}, define the function M(x) as,
M(x):= (8x^{2}+36x+14)^{k} + (9x^{2}25x+23)^{k} + (6x^{2}8x+70)^{k}
For k = 3 this is already the perfect cube (x^{2}9x+71)^{3}. For k = 2, this yields the quartic,
M(x):= 181x^{4}930x^{3}+1335x^{2}+1262x+75^{2} = y^{2}
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,
a^{k}+b^{k}+c^{k}+d^{k} = {p^{2}, q^{4}} for k = 2,4
with the only known soln,
15935^{2}+27022^{2}+57910^{2}+59260^{2} = 88597^{2} 15935^{4}+27022^{4}+57910^{4}+59260^{4} = 70121^{4} found by this author using Wroblewski’s database for fourth powers. Any others? 13. Form: a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k =1,3
Just like for k = 1,2, the eqn
a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, (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, uv+w, u+vw} {d, e, f} = {x+y+z, xy+z, x+yz}
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} + z^{k} + (v+y)^{k} = (v+x)^{k} + y^{k} + (u+z)^{k}
call this L_{0}, a complete soln for k = 1 which he used for k = 1,5. Expanding L_{0} for k = 3,
u^{2}xv^{2}x+ux^{2}vx^{2}+v^{2}y+vy^{2}u^{2}zuz^{2} = 0
As a quadratic in z, this has discriminant D,
D:= 4(xy)uv^{2}4(x^{2}y^{2})uv+u^{2}(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 L_{0} 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} = (ta)^{k}+(tb)^{k}+(tc)^{k}+(td)^{k}+(te)^{k}
for k = 1,2,3,4 which is true for any t iff a^{k}+b^{k}+c^{k}+d^{k}+e^{k} = 0 for k = 1,3.
Third, another approach which turns out to be simpler is to use the form,
(a+bp+q)^{k} + (bbp+q)^{k} + (c+ap+q)^{k} = (a+cp+q)^{k} + (b+ap+q)^{k} + (ccp+q)^{k}
call this L_{1} since this author found it while studying Lander’s work on L_{0} for k = 1,5. This is a subset of L_{0}, yet still complete for multigrades with one k >1 (the easy proof will be given in Fifth Powers). For k = 1,3, expanding L_{1} at k = 3 gives just a linear condition in q,
a^{2}+b^{2}+c^{2}+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 L_{1} 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 L_{1} for k = 1,4, as a polynomial in p after removing the trivial factor p(p1)(bc), is,
(Poly10)p^{2}+(Poly20)p+(Poly30) = 0
with discriminant D that is a quadratic polynomial in q, while for k = 1,5,
(Poly11)p^{2}+(Poly21)p+(Poly31) = 0
has a D that is a quartic in q, and for k = 1,2,6,
(Poly12)p^{2}+(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 = y^{2}. Thus, L_{1} 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} = {abc^{2}, a^{2}+bc, b^{2}+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+c^{2}, a^{2}bc, b^{2}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 a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} 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 E_{1}. However, letting {c,f} = {ab, de} and substituting into the system, this also satisfies k = 1,3 given E_{1}, proving the assertion. (End proof) An earlier proof was given by Choudhry,
(Update, 3/23/10): Choudhry (2001): "If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, 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(a^{3}+b^{3}+c^{3})6abc = 3(a+b+c)(a^{2}+b^{2}+c^{2})
and a similar one in the variables {d,e,f}. If eqs.1,2,3 hold, then
(a+b+c)(a^{2}+b^{2}+c^{2}) = (d+e+f)(d^{2}+e^{2}+f^{2}) (eq.4)
For a+b+c = d+e+f ≠ 0, this implies a^{2}+b^{2}+c^{2} = d^{2}+e^{2}+f^{2}. But Bastien’s Theorem excludes a nontrivial soln to,
a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 1,2,3,
Thus, for eq.4 be nontrivially true, it must be the case that a+b+c = d+e+f = 0. Eqs 1,2,3 are then satisfied if,
a^{k}+b^{k}+(ab)^{k} = d^{k}+e^{k}+(de)^{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 a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 1,3 where a+b+c = d+e+f = 0 (or equivalently, abc = def), then,
9abc(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6}) = 2(a^{9}+b^{9}+c^{9}d^{9}e^{9}f^{9}) (eq.1)
Proof: Simply substitute {c,f} = {ab, de} into eq.1 and this also has E_{1}:= ab(a+b)de(d+e) as a factor.
(Update, 6/22/10): The system,
x_{1}x_{2}x_{3} = y_{1}y_{2}y_{3} x_{1}^{3}+x_{2}^{3}+x_{3}^{3} = y_{1}^{3}+y_{2}^{3}+y_{3}^{3}
has several solns:
1. Gerardin
(a^{2}p)^{3} + (b^{2}q)^{3} + (abr)^{3} = (a^{2}q)^{3} + (b^{2}r)^{3} + (abp)^{3}
where {p,q,r} = {a^{3}2b^{3}, a^{3}+b^{3}, 2a^{3}b^{3}}.
2. Choudhry gave a 3parameter version,
(ap)^{3} + (bq)^{3} + (cr)^{3} = (aq)^{3} + (br)^{3} + (cp)^{3}
where {p,q,r} = {a^{3}2b^{3}c^{3}, a^{3}+b^{3}+2c^{3}, 2a^{3}b^{3}+c^{3}}.
3. Piezas
(ap)^{3} + (bq)^{3} + (cr)^{3} = (ar)^{3} + (bp)^{3} + (cq)^{3}
where {p,q,r} = {abc^{2}, a^{2}+bc, b^{2}+ac}.
The last identity, discussed in the previous paragraph, is also valid for x_{1}+x_{2}+x_{3} = y_{1}+y_{2}+y_{3} = 0, as well as,
9x_{1}x_{2}x_{3} (x_{1}^{6}+x_{2}^{6}+x_{3}^{6}y_{1}^{6}y_{2}^{6}y_{3}^{6}) = 2(x_{1}^{9}+x_{2}^{9}+x_{3}^{9}y_{1}^{9}y_{2}^{9}y_{3}^{9})
14. Form: x^{k}+y^{k}+z^{k} = t^{k}+u^{k}+v^{k}, 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
p^{2} + q^{2} + (p+q+3r)^{2} = (p+2r)^{2} + (q+2r)^{2} + (p+q+r)^{2}
Alternatively, this can be more symmetrically expressed as,
(pr)^{k} + (qr)^{k} + (p+q+r)^{k} = (p+r)^{k} + (q+r)^{k} + (p+qr)^{k} which is already for k = 2, but is also valid for k = 3 if 6pq = r^{2}.
15. Form: x^{3}+y^{3}+z^{3} = 3t^{3}t
M. Noble
Given a^{3}+b^{3}+c^{3} = 3d^{3}d, then,
{a,b,c,d} = {(4p4r)/(5q), (p+4r)/(5q), (3p)/(5q), (4p)/(5q)}
where {p,q,r} = {15u^{2}+4v^{2}, 15u^{2}+9uv4v^{2}, 10uv+3v^{2}}
This soln has the special property such that by defining the expressions,
{x,y,z} = {a^{3}d^{3}, b^{3}d^{3}, c^{3}d^{3}}
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: x^{3}+y^{3}+z^{3} = m(x+y+z)
L.Aubry
x^{3}+y^{3}+z^{3} ± (x+y+z) = 0
{x,y,z} = {u+v, u+v, 2v(3n+1)}, if u^{2}(1+12n+36n^{2}+36n^{3})v^{2} = ±n
with n some constant. For example, for n=1 this entails solving either u^{2}85v^{2} = ±1 depending on which sign is involved. In general, if x,y,z need not be coprime, then,
{x,y,z} = {n(u+v), n(u+v), 2nv(3n^{2}+1)}, if u^{2}(1+12n+36n^{2}+36n^{3})v^{2} = ±1.
17. Form: x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = y_{1}^{k}+y_{2}^{k}, k = 1,2,3
By Bastien’s Theorem, the system of eqns,
x_{1}^{k}+x_{2}^{k}+…+x_{n}^{k} = y_{1}^{k}+y_{2}^{k}+…+y_{n}^{k}, k = 1,2,3
has a nontrivial soln only for n>3. However, one can always set one term x_{1} = 0. A simple soln is,
J. Nicholson
a^{k} + b^{k} + (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,
a^{k} + b^{k} + c^{k} + d^{k} = e^{k} + f^{k},
{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 (a^{2}x)(b^{2}x)(c^{2}x) = y^{2}. Two easy solns can be found. First, let b = c, and,
{a,b,c,y} = {p+q, q, q, 2(p+q)qr}, where p^{2} = 2q^{2} + r^{2}.
Second, let a+b = c,
{a,b,c,y} = {2p+q, 2p+q, 2q, 4(2pq)(2p+q)r}, where p^{2} = q^{2} + r^{2}. Any more?
18a. Form: x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}, k = 1,3
(Update, 8/6/09): Tony Rizzo gave the soln,
(p+1)^{k }+ (q+1)^{k }+ (r1)^{k }+ (s1)^{k} = (p1)^{k }+ (q1)^{k }+ (r+1)^{k }+ (s+1)^{k}, for k = 1,3,
where p^{2}+q^{2} = r^{2}+s^{2} and solved the condition as,
(ad)^{2}+(b+c)^{2} = (a+d)^{2}+(bc)^{2}
where a/b = c/d. One can make this valid for k = 1,2,3 by rearranging 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: x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}, 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,
x_{1}+x_{2}+… +x_{n} = y_{1}+y_{2}+…+y_{n} = 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 ChernickLander form,
(a+b+c)^{k} + (abc)^{k} + (ab+c)^{k} + (a+bc)^{k} = (d+e+f)^{k} + (def)^{k} + (de+f)^{k} + (d+ef)^{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,
x_{1}^{k}+x_{2}^{k}+x_{3}^{k}+x_{4}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}+y_{4}^{k}
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 F_{2},
(a+bh)^{k} + (c+dh)^{k} + (e+fh)^{k} + (g+h)^{k} = (abh)^{k} + (cdh)^{k} + (efh)^{k} + (gh)^{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)h^{2}(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) = y^{2} (eq.2)
which is easily done since the expression is just a quadratic in {a,c,e}. It is wellknown that given an initial soln to F(x) = y^{2} 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+abc+cd)/(b+d)
from which one can then find the complete soln. Using this, one can prove that,
Theorem: Given a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k} for k = 1,2,3. Define n as a^{2}+b^{2}+c^{2}+d^{2} = n(a+b+c+d)^{2}, then,
25(a^{4}+b^{4}+c^{4}+d^{4}e^{4}f^{4}g^{4}h^{4})(a^{6}+b^{6}+c^{6}+d^{6}e^{6}f^{6}g^{6}h^{6}) = 12(n+1)(a^{5}+b^{5}+c^{5}+d^{5}e^{5}f^{5}g^{5}h^{5})^{2}
This 465 Identity, by squaring all variables, can become a 81210 Identity good for the even system k = 2,4,6 and analogous to Ramanujan’s 6108 which depends on a k = 2,4. This will be discussed more in Fourth Powers.
18c. Form: x_{1}^{3}+x_{2}^{3}+...+x_{n}^{3} = Poly (Update, 12/15/09): Tony Rizzo gave the ff cubic and quintic identities,
(ax^{2}+y)^{3} + (ax^{2}y)^{3} + (a2)(ax^{2})^{3} = (a^{2}x^{3})^{2} + (axy)^{2} + (6a)ax^{2}y^{2 } (eq.1)
(ax^{2}+y)^{5} + (ax^{2}y)^{5} + (a2)(ax^{2})^{5} = (a^{3}x^{5}+10xy^{2})^{2} + 10(a10)x^{2}y^{4 } (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 {36x^{3}, 6xy} = {2m, m^{2}1}, thus,
(6x^{2}+y)^{3} + (6x^{2}y)^{3} + 4(6x^{2})^{3} = (6xy+2)^{2}, if 6xy = 324x^{6}1
For eq.2, when a = 10, the RHS becomes Bouniakowsky's quinticsquare eqn,
(10x^{2}+y)^{5} + (10x^{2}y)^{5} + 8(10x^{2})^{5} = 100x^{2}(100x^{4}+y^{2})^{2}
Note: The relationship between the two can be appreciated more by considering the basic identities,
(a+y)^{3} + (ay)^{3} + 4a^{3} = 6a(a^{2} +y^{2}) (Id.1)
(a+y)^{5} + (ay)^{5} + 8a^{5} = 10a(a^{2} +y^{2})^{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} + (ay)^{3} + 4a^{3} = (6m)^{2}(2n^{2}y)^{2}
where {a,y} = {6n, n^{2}9}, and n = m^{2}, for arbitrary m.
(a+y)^{5} + (ay)^{5} + 8a^{5} = (10m)^{4}(2n^{2}y)^{4}
where {a,y} = {10n, n^{2}25}, and n = 100m^{4}, for arbitrary m.
(End update.)
19. Form: ax^{3}+by^{3}+cz^{3} = N
Theorem 1: If ax^{3}+by^{3}+cz^{3} = N has a solution, then one can generally find a second. (A. Desboves).
Proof: ap^{3}+bq^{3}+cr^{3} = (ax^{3}+by^{3}+cz^{3})(ax^{3}by^{3})^{3}
{p,q,r} = {(ax^{4}+2bxy^{3}), (2ax^{3}y+by^{4}), z(ax^{3}by^{3})}
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)
(2x^{2}4xy+9yz9z^{2})^{3} + (2y^{2}xy+9xz18z^{2})^{3} = 9(2x^{2}4xzyz+y^{2})^{3}, if x^{3}+y^{3} = 9z^{3}
Piezas
Let q = p^{3}+1, then
(px^{2}p^{2}xy+qyzqz^{2})^{3} + (py^{2}xy+qxzpqz^{2})^{3} = q(px^{2}p^{2}xzyz+y^{2})^{3}, if x^{3}+y^{3} = qz^{3}
(Realis’s was the case p=2 of this identity.)
E.Lucas (nonic)
p^{3}+q^{3}+cr^{3} = 27 (x^{3}+y^{3}+cz^{3}) (x^{7}y+x^{4}y^{4}+xy^{7})^{3}
{p,q,r} = {x^{9}+6x^{6}y^{3}+3x^{3}y^{6}y^{9}, x^{9}+3x^{6}y^{3}+6x^{3}y^{6}+y^{9}, 3xyz(x^{6}+x^{3}y^{3}+y^{6})}
More generally,
Theorem 2: If ax_{1}^{3}+a_{2}x_{2}^{3}+...+ a_{m}x_{m}^{3} = N has a solution, then this generally leads to a second.
Proof: Desboves’ identity in Theorem 1 is essentially,
a(ax^{4}+2bxy^{3})^{3 } + b(2ax^{3}yby^{4})^{3} = (ax^{3}+by^{3})(ax^{3}by^{3})^{3}
If ax^{3}+by^{3} is the sum of cubes c_{1}z_{1}^{3}+c_{2}z_{2}^{3}+…+ c_{m}z_{m}^{3} = N, then,
a(ax^{4}+2bxy^{3})^{3 }+ b(2ax^{3}yby^{4})^{3} = (c_{1}z_{1}^{3}+c_{2}z_{2}^{3}+…+ c_{m}z_{m}^{3})(ax^{3}by^{3})^{3} = 0
so, by distribution, one gets a new identity of the form ax_{1}^{3}+a_{2}x_{2}^{3}+...+ a_{m}x_{m}^{3} = 0 if there are initial solns x,y, and z_{i}. (End proof)
Form p^{3}+q^{3} = nr^{3}
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} = {36ny, 36n+y, 6x}, to reduce the eqn to the elliptic curve,
y^{2} = x^{3}432n^{2}
and can be reversed as,
{x,y} = {12n/(p+q), 36n(pq)/(p+q)}
More generally though,
Theorem 3: If au^{3}+bv^{3}+cw^{3} = 0 has a solution, then so does y^{2} = x^{3}432a^{2}b^{2}c^{2}.
Proof: y^{2}x^{3}+432a^{2}b^{2}c^{2} = (au^{3}+bv^{3}+cw^{3})(au^{3}bv^{3}cw^{3})(432b^{2}c^{2}/u^{6})
{x,y} = {4(p^{2}+q)/r^{2}, 4(2p^{3}+3pq)/r^{3}}, where {p,q,r} = {bv^{3}cw^{3}, 3bcv^{3}w^{3}, uvw}
Theorem 4: If au^{3}+bv^{3}+cw^{3} = 0 has a solution, then so does x^{3}+y^{3}+abcz^{3} = 0.
Proof: x^{3}+y^{3}+abcz^{3} = 27bcv^{3}w^{3}(au^{3}+bv^{3}+cw^{3})(p^{2}+q)^{3}
{x,y,z} = {p^{3}+3bqv^{3}, p^{3}+3cqw^{3}, 3(p^{2}+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: ax^{3}+by^{3}+cz^{3} = dxyz
S. Realis (complete, other than the case x = y = z)
x^{3}+y^{3}+z^{3} = 3xyz
{x,y,z} = {(ab)^{3}+(ac)^{3}, (ba)^{3}+(bc)^{3}, (ca)^{3}+(cb)^{3}}
All positive integers N other than those div by 3 but not by 9 are representable as N = x^{3}+y^{3}+z^{3}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 ax^{3}+by^{3}+cz^{3}+dxyz = 0 has a solution, then one can generally find a second. (A. Cauchy)
Proof: ap^{3}+bq^{3}+cr^{3}+dpqr = (ax^{3}+by^{3}+cz^{3}+dxyz)(pqr)/(xyz)
{p,q,r} = {x(by^{3}cz^{3}), y(ax^{3}+cz^{3}), z(ax^{3}by^{3})}
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/(v_{1}x) = q/(v_{2}y) = r/(v_{3}z)
{v_{1}, v_{2}, v_{3}} = {by^{3}cz^{3}, ax^{3}+cz^{3}, ax^{3}by^{3}}
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 apx^{2}+bqy^{2}+ crz^{2} = 0. One can then solve for {p,q} which gives essentially the same result as Cauchy’s.
Theorem 2: If ax^{3}+by^{3}+cz^{3}+dxyz = 0 has a soln, then so does p^{3}+q^{3}+abcr^{3}+dpqr = 0. (J. Sylvester)
Proof: {p,q,r} = {f^{2}g+g^{2}h+h^{2}f3fgh, fg^{2}+gh^{2}+hf^{2}3fgh, xyz(f^{2}+g^{2}+h^{2}fgfhgh)}, where {f,g,h} = {ax^{3}, by^{3}, cz^{3}}.
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
x^{3}+ny^{3}+n^{2}z^{3}3nxyz = (p^{3}+nq^{3}+n^{2}r^{3}3npqr)(u^{3}+nv^{3}+n^{2}w^{3}3nuvw)
{x,y,z} = {pu+n(qw+rv), pv+qu+nrw, pw+qv+ru}
which proves that, like the quadratic form x^{2}+y^{2}, the product of two forms x^{3}+ny^{3}+n^{2}z^{3}3nxyz is of like form. In general, the result extends to cyclic determinants of order m.
21. Form: x^{3}+y^{3}+z^{3}3xyz = t^{k}
Krafft, Lagrange
To recall, the algebraic form we can designate as f_{2}:= x^{2}dy^{2} factors over the square root extension d^{1/2} (which may involve the complex unit). For its cubic analogue, use the general form f_{3}:= x^{3}+ny^{3}+n^{2}z^{3}3nxyz which factors linearly over a complex cube root of unity w as,
f_{3}:= x^{3}+ny^{3}+n^{2}z^{3}3nxyz = (x+my+m^{2}z)(x+mwy+m^{2}w^{2}z)(x+mw^{2}y+m^{2}wz)
where w^{3}1 = 0 and m = n^{1/3} for convenience. Or alternatively, if one has Mathematica,
f_{3}:= Resultant[x+wy+w^{2}z, w^{3}n, w]
Because it has three linear factors and variables, one can then easily solve the equation,
x^{3}+ny^{3}+n^{2}z^{3}3nxyz = (p^{3}+nq^{3}+n^{2}r^{3}3npqr)^{k}
for any positive integer k (just like for f_{2}) by factoring both sides. In fact, for the special case f_{3} = ±1, this is sometimes known as the cubic Pell equation, an analogue to the Pell equation f_{2} = ±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+m^{2}z = (p+mq+m^{2}r)^{k} x+mwy+m^{2}w^{2}z = (p+mwq+m^{2}w^{2}r)^{k} x+mw^{2}y+m^{2}wz = (p+mw^{2}q+m^{2}wr)^{k}
for x,y,z and where m = n^{1/3}. For k=2, this gives,
{x,y,z} = {p^{2}+2nqr, nr^{2}+2pq, q^{2}+2pr}
and so on for other k.
