# Zero sum of four cubes in K[X]

Good evening,

I'm trying to solve the equation

$A^3+B^3+C^3+D^3=0$

in the ring of complex polynomials; indeed, I don't know if I can find all solutions, prove there aren't non constant solutions or exhibit one.

One should certainly add a condition such as $\gcd(A,B,C,D)=1$ in order to avoid other kind of trivial solutions like $(A,B,-A,-B)$.

I tried to use the Mason-Stothers theorem, since it may be used to show Fermat's Last Theorem for polynomials or Catalan's conjecture for rational functions, but I could not get a strong enough condition on degrees to get a solution or a contradiction.

May anybody help me?

Thanks!

• Solve under what assumptions? For example $A=B=-C=-D$ will always satisfy the equation, but that's likely not what you are after. – dxiv Feb 22 '17 at 20:18
• You're right, one should assume that the polynomials are coprime. – Tig la Pomme Feb 22 '17 at 20:23
• That is significant information which you should edit into the question itself, rather than leave in a comment. – dxiv Feb 22 '17 at 20:55
• You're right. Thank you for your interest in this topic. – Tig la Pomme Feb 22 '17 at 21:13
• math.stackexchange.com/questions/107570/… – individ Feb 23 '17 at 6:32

Examples can be found based on known parametrizations of the solutions to the diophantine equation $A^3+B^3=C^3+D^3\,$ (see 3.2.2 here). For example, Ramanjuan's identity:

$$(a^2+7ab-9b^2)^3+(2a^2-4ab+12b^2)^3=(2a^2+10b^2)^3+(a^2-9ab-b^2)^3$$

For $b=1$ the above gives:

$$(x^2+7x-9)^3+(2x^2-4x+12)^3+(-2x^2-10)^3+(-x^2+9x+1)^3 = 0$$

It can be easily verified that the $4$ polynomials are mutually coprime.

• Thank you! Though it's interesting, I would be interested in a constructive way to get such formulas from the initial problem. – Tig la Pomme Feb 23 '17 at 8:49
• The above answers the prove there aren't non constant solutions or exhibit one part of the question. There are several known polynomial parametrizations to the taxicab solutions, which can be used to generate many such examples. For some of the theory behind them, see for example Silverman's Taxicabs and Sums of Two Cubes. – dxiv Feb 23 '17 at 19:54
• @dxiv: There are in fact infinitely many Ramanujan-type quadratic parameterizations of the taxicab problem. Kindly see this post. – Tito Piezas III Feb 12 '18 at 12:37
• @TitoPiezasIII Neat. Thank you for the pointer. – dxiv Feb 12 '18 at 17:23