011: Sum / Sums of Three Cubes, Part 2

(u+x)^k + z^k + (v+y)^k = (v+x)^k + y^k + (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 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 + (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 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 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 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(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,

a^k+b^k+c^k = d^k+e^k+f^k, 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,

a^k+b^k+(-a-b)^k = d^k+e^k+(-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 a^k+b^k+c^k = d^k+e^k+f^k 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.

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, 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,

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

15935²+27022²+57910²+59260² = 88597²

15935⁴+27022⁴+57910⁴+59260⁴ = 70121⁴

{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 c_i 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,

a^k+b^k+c^k+d^k = {p², q⁴} for k = 2,4

with the only known soln,

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,

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: a^k+b^k+c^k = {x², y², z³} for k =1,2,3 (aka Martin triples)

A Martin triple is a triple of integers {a,b,c} such that,

I. Sum / Sums of cubes (Part 2, 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. a^k+b^k+c^k = {x², y², z³}, k =1,2,3 (Martin triples)

13. x^k+y^k+z^k = t^k+u^k+v^k, k = 1,3

14. x^k+y^k+z^k = t^k+u^k+v^k, k = 2,3

15. x³+y³+z³ = 3t³-t

16. x³+y³+z³ = m(x+y+z)

17. x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k, k = 1,2,3

18. x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k, k = 1,2,3

19. ax³+by³+cz³ = N

20. ax³+by³+cz³+dxyz = 0

21. x³+y³+z³-3xyz = t^k

(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+...+ a_mx_m³ = 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+…+ c_mz_m³ = N, then,

a(ax⁴+2bxy³)³ + b(-2ax³y-by⁴)³ = (c₁z₁³+c₂z₂³+…+ c_mz_m³)(ax³-by³)³ = 0

so, by distribution, one gets a new identity of the form ax₁³+a₂x₂³+...+ a_mx_m³ = 0 if there are initial solns x,y, and z_i. (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 = t^k

Krafft, Lagrange

To recall, the algebraic form we can designate as f₂:= x²-dy² factors over the square root extension d^1/2 (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 = n^1/3 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 = n^1/3. For k=2, this gives,

{x,y,z} = {p²+2nqr, nr²+2pq, q²+2pr}

and so on for other k.

18b. Form: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^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,

x₁+x₂+… +x_n = y₁+y₂+…+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 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,

x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^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 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 a^k+b^k+c^k+d^k = e^k+f^k+g^k+h^k 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+...+x_n³ = Poly

(Update, 12/15/09): Tony Rizzo gave the ff cubic and quintic identities,

(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.)

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

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: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k, k = 1,2,3

By Bastien’s Theorem, the system of eqns,

x₁^k+x₂^k+…+x_n^k = y₁^k+y₂^k+…+y_n^k, 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

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 (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: x₁^k+x₂^k+x₃^k+x₄^k = y₁^k+y₂^k+y₃^k+y₄^k, 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,