The Equation q₁³ + q₂³ + q₃³ + q₄³ = (a³ + b³ + c³ + d³)q₅³ in Quadratic forms q_i

By Titus Piezas III

Contents: diophantine equations, equal sums of like powers, cubic quadruples.

This is a very short paper that is essentially an epilogue to “Ramanujan and the Cubic Equation 3³ + 4³ + 5³ = 6³” [1]. In that paper, we found an algebraic identity such that given a solution to the cubic quadruple a³ + b³ + c³ + d³ = 0 we can automatically find one in terms of quadratic forms. However, we mentioned it seems there was an even more general identity that contains the one we gave. In this continuation, we finally find this general form.

Our method consists of solving the equation,

q₁³ + q₂³ + q₃³ + q₄³ = 0 (eq.1)

where the q_i are quadratic forms. Explicitly,

(ax² + v₁xy + v₂y²)³ + (bx² + v₃xy + v₄y²)³ + (cx² + v₅xy + v₆y²)³ + (dx² + v₇xy + v₈y²)³ = 0

where the v_i are unknowns. By expanding and collecting terms in powers of x and y, we get the system of equations, call it S₁,

p₀ = (a³ + b³ + c³ + d³) x⁶

p₁ = (a²v₁ + b²v₃ + c²v₅ + d²v₇) x⁵y

p₂ = (av₁² + a²v₂ + bv₃² + b²v₄ + cv₅² + c²v₆ + dv₇² + d²v₈) x⁴y²

p₃ = (v₁³ + 6av₁v₂ + v₃³ + 6bv₃v₄ + v₅³ + 6cv₅v₆ + v₇³ + 6dv₇v₈) x³y³

p₄ = (v₁²v₂ + av₂² + v₃²v₄ + bv₄² + v₅²v₆ + cv₆² + v₇²v₈ + dv₈²) x²y⁴

p₅ = (v₁v₂² + v₃v₄² + v₅v₆² + v₇v₈²) xy⁵

p₆ = (v₂³ + v₄³ + v₆³ + v₈³) y⁶

The details can be found in [1] but by equating all p_i = 0 and solving for the v_i, we get what we called Id.2:

Identity 2 (Id.2)

If a³ + b³ + c³ + d³ = 0, then,

(ax² + v₁xy + v₂y²)³ + (bx² + v₃xy + v₄y²)³ + (cx² + v₅xy + v₆y²)³ + (dx² + v₇xy + v₈y²)³ = 0

with the solution, using the simplifying substitution a = p+s, b = p-s, c = q-r, d = q+r,

v₁, v₃ (free variables)

v₅ (6pqr) = (p³+q³+3q²r+3p²s)v₁ + (p³+q³+3q²r-3p²s)v₃

v₇ (6pqr) = –(p³+q³-3q²r+3p²s)v₁ – (p³+q³-3q²r-3p²s)v₃

v₁²-4av₂ = f₁, v₃²-4bv₄ = f₁, v₅²-4cv₆ = f₂, v₇²-4dv₈ = f₂

and discriminants f_i,

f₁ = –(4p³+q³)(pv₁-sv₁-pv₃-sv₃)²/(12p²qr²)

f₂ = –(p³+4q³)(pv₁-sv₁-pv₃-sv₃)²/(12pq²r²)

One can easily solve for the v_i with even subscripts as they are only linear (and since the v_i with odd subscripts are already given). The new variables (p,q,r,s) can be expressed in terms of the originals as,

p = (a+b)/2, s = (a-b)/2, q = (c+d)/2, r = (-c+d)/2

If we substitute the formulas for the v_i into eq.1, what we get is the factorization,

q₁³ + q₂³ + q₃³ + q₄³ = (a³ + b³ + c³ + d³) P(x,y)

We pointed out that P(x,y) is a complicated polynomial, not a perfect cube, that is irrelevant if a³ + b³ + c³ + d³ = 0 since the right hand side of the equation vanishes anyway. It was not a cube using these particular versions for the expressions for the v_i but with a small substitution, it turns out we can make this into a cube!

It was simple, really. Using the simplifying substitutions, then a³ + b³ + c³ + d³ = 0 becomes,

p³ + q³ + 3qr² + 3ps² = 0

or p³ + q³ = -(3qr² + 3ps²). Since v₅ is given by,

v₅ (6pqr) = (p³+q³+3q²r+3p²s)v₁ + (p³+q³+3q²r-3p²s)v₃

then this is equivalent to,

v₅ (2pqr) = (-qr²-ps²+q²r+p²s)v₁ + (-qr² -ps²+q²r-p²s)v₃

Same for v₇. (I actually did this once before. But I neglected to do the same for the discriminants f_i!) Since f₁ is,

f₁ = –(4p³+q³)(pv₁-sv₁-pv₃-sv₃)²/(12p²qr²)

then f₁ = –(p³-qr²-ps²)(pv₁-sv₁-pv₃-sv₃)²/(4p²qr²) and similarly for f₂. Substituting these new versions for the v_i, we get the fifth identity which covers all four given in [1]:

Identity 5 (Id.5)

q₁³ + q₂³ + q₃³ + q₄³ = (a³ + b³ + c³ + d³)q₅³ (eq.2)

or, explicitly,

(ax² + v₁xy + v₂y²)³ + (bx² + v₃xy + v₄y²)³ + (cx² + v₅xy + v₆y²)³ + (dx² + v₇xy + v₈y²)³ = (a³ + b³ + c³ + d³)(x² + v₉xy + v₁₀y²)³

where, with the simplifying substitution a = p+s, b = p-s, c = q-r, d = q+r,

v₁, v₃ (free variables)

v₅ (2pqr) = (qr(q-r)+ps(p-s))v₁ + (qr(q-r)-ps(p+s))v₃

v₇ (2pqr) = (qr(q+r)-ps(p-s))v₁ + (qr(q+r)+ps(p+s))v₃

v₉ = (v₁+v₃)/(2p)

v₁²-4av₂ = f₁, v₃²-4bv₄ = f₁, v₅²-4cv₆ = f₂, v₇²-4dv₈ = f₂, v₉²-4v₁₀ = f₃

f₁ = f₃(p³-qr²-ps²)/p, f₂ = f₃(q³-qr²-ps²)/q, f₃ = –(pv₁-sv₁-pv₃-sv₃)²/(4pqr²)

and,

p = (a+b)/2, s = (a-b)/2, q = (c+d)/2, r = (-c+d)/2

This is a much more satisfying version as this can be applicable to cubic n-tuples without conditions on (a,b,c,d) and dependence on opposite parity of middle terms, other than the old assumption that pairs of quadratic forms on the left hand side, (q₁, q₂) and (q₃, q₄), share the same discriminants. And it also answers one question in the conclusion of [1]: whether S₂, or the system of equations we get by expanding (eq.2) plus two more equating the discriminants, always has a linear solution. So the answer indeed is yes.

--End--

© 2005

Titus Piezas III

Sept 17, 2005

http://www.oocities.org/titus_piezas/ramanujan.html ® Click here for an index

tpiezasIII@uap.edu.ph ® (Remove “III” for email)

Reference:

1. Piezas, T., “Ramanujan and The Cubic Equation 3³ + 4³ + 5³ = 6³”, http://www.oocities.org/titus_piezas/ramanujan_page9.html