Sum of the cubes of a Pythagorean triple equal a cube.

Apart from (3, 4, 5, 6) are there any more primitive solutions to $x^3+y^3+z^3=w^3$ where $x^2+y^2=z^2$ ?

I’ve noted that if gcd(x ,y ,z) = k, then k divides w, so non-primitive Pythagorean triples are irrelevant.

The set of primitive Pythagorean triples is given by, $$\pm x= a^2-b^2$$ $$\pm y=2ab$$ $$\pm z= a^2+b^2$$ where $gcd(a,b)=1$ and $a, b$ are of opposite parity (one odd, one even).

So w is given by one of the following $$w^3=2a^2(a^4+4ab^3+3b^4)$$ $$w^3=2a^2(-a^4+4ab^3-3b^4)$$ $$w^3=2a^2(a^4-4ab^3+3b^4)$$ $$w^3=2b^2(3a^4+4ab^3+b^4)$$ $$w^3=2b^2(-3a^4+4ab^3-b^4)$$ $$w^3=2b^2(3a^4-4ab^3+b^4)$$

I’ve written a program to generate (x,y,z) from (a,b) with b < a and of opposite parity, and then run it, with all the sign options, to a = 41282.

Edit 17 July 2016

If nobody objects, I should now like to restrict this question to positive integer values of $x, y, z$. I consider my original idea of allowing signed integers resulted in too many sub-problems, thus discouraging further progress.

Hence, the equation to be solved is $$w^3=2a^2(a^4+4ab^3+3b^4)$$

After the substitutions $a=2c$ and $w=2v$ this is
$$v^3= c^2 (b+2c)^2 (3b^2-4bc+4c^2)$$ with ($c$ odd, or even with the highest power of 2 that divides $c$ of the form $3n$) $b$ odd and $b<2c$.

• There seem to be also complete characterization of solutions of $x^3+y^3+z^3=w^3$ is known, see this question or Wikipedia. However, it does not seem very likely that this would give something simple. – Martin Sleziak Jun 6 '16 at 10:05
• @Martin Sleziak. Thank you for your kind comment, and for tidying my first attempt at using this mark-up language. I’ve read all the info you provided, but I’m still unsure in which direction to proceed. To be honest, I expected to find this problem had already been solved and that I was just looking in the wrong places for a solution. Thanks again for your help. – Old Peter Jun 6 '16 at 19:11
• @Martin Sleziak 1. I apologise for omitting a vital part of the primitive Pythagorean triples parametric solution in my OP; a, b must be of opposite parity (one odd, one even). 2. I’ve investigated Choudhry's On Equal Sums of Cubes (1998), but found the parametric solutions unsuitable for me to make further progress in this direction. 3. I’m working on a possible method to move nearer to a solution. Should I post this as a comment or an answer? – Old Peter Jun 9 '16 at 18:41
• A Pythagorean triple is usually considered to be a triple of positive integers, so doesn't it suffice to specify $a > b > 0$ and consider just the first of your six formulae for $w^3$? Or are you interested in solutions including negative integers? – Adam Bailey Jun 12 '16 at 10:42
• Considering the first of the six formulae, since $a$ and $b$ have opposite parity $a^4 + 4ab^3 + 3b^4$ must be odd, so to make the whole formula equal a cube, $a$ must be divisible by 2, and the highest power of 2 that divides $a$ must be of the form $3n+1$. This narrows the possibilities to be considered in any search. – Adam Bailey Jun 12 '16 at 10:50