5

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 $b<2c$.

IraeVid
  • 3,216
Old Peter
  • 1,357
  • 1
    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 06 '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 06 '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 09 '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
  • @Adam Bailey Thank you for your interest. Yes, I’m interested in negative integers, but on reflection perhaps I would have been better to split up this question. It’s precision that’s the problem with my search, not time, although my gut feeling is that the known primitive solution is unique. For the all positive case, I’ve made the substitution $a=qb$ where $q$ is rational giving $$w^3=b^6(2q^2)(q+1)^2(q^2-2q+3)$$ If $q$ is integer, I only find solutions of $q = 2$ and $q = 3$. However, proving $q$ to be integer is beyond my skills. – Old Peter Jun 12 '16 at 15:45
  • @Adam Bailey My apologies for failing to understand how valuable your comment that, for the “all variables positive” case, “$a$ must be divisible by 2, and the highest power of 2 that divides $a$ must be of the form $3n+1$”. I put $a=2c$ to obtain
    $$W^3= 8 (c^2) (b+2c)^2 (3b^2-4bc+4c^2)$$ with ($c$ odd, or the highest power of 2 that divides $c$ must be of the form $3n$) and $b<2c$. Hence, I can extract the prime factors (until I find a non-cube) of $W^3$ from much smaller numbers than previously.     – Old Peter Jun 16 '16 at 15:32
  • Further to my previous comment, I’ve now checked $c^2 (b+2c)^2 (3b^2-4bc+4c^2)$ to be a cube up to $c=200001$, finding no solutions other than $b=c$. The program uses just one processer and can currently handle 3500 $c$ values an hour, clearly slowing as $c$ increases. I would welcome any views on ploughing on with this, or any ideas for a different approach. – Old Peter Jun 20 '16 at 18:31
  • 1
    @AdamBailey It seems OldPeter's 2016 post was only answered this year. But this related 2014 post may be of interest. – Tito Piezas III Dec 08 '23 at 15:05

1 Answers1

3

A. Genocchi proved that (3,4,5,6) is indeed the only solution to your question. Reference: Atti Accad. Pont. Nuovi Lincei, 19, 1865-6, 49; Annali di Mat., 7, 1865, 157; French transl., Jour. de. Math., (2), 11, 1866, 185-7.

Source: Dickson, History of the Theory of Numbers, Vol.2, pp.613.

Mu-Tsun Tsai
  • 146