3

I am almost certain it is a duplicate question but I am looking for a reference regarding how solve the diophantine equation $$x^3+y^3+z^3=kt^3$$ where $x,y,z,k$ are pairwise co-prime. Please help me find a reference or with any hints. Thanks.

  • Do you want all solutions or just one (assuming that a solution exists) ? – Peter Jan 14 '17 at 11:48
  • 1
    All the solutions for it is easy to choose arbitrary $x,y,z$ and simplify this equation. –  Jan 14 '17 at 11:50
  • 3
    Some useful links and equations can be found here . Ryley's Theorem shows existence of solutions, I do not know if it yields all solutions. – lulu Jan 14 '17 at 12:15
  • For some private to write about. http://www.artofproblemsolving.com/community/c3046h1172731_a_cubic_equation_with_the_troika – individ Jan 14 '17 at 12:29
  • @individ, your methods are quite special. I cannot verify whether or not they actually yield true or accurate results. I am quite sure there must exist small integer solutions. Anyways, thanks for your input. –  Jan 14 '17 at 21:05

1 Answers1

3

Ryley's Theorem states that, $$x_1^3+x_2^3+x_3^3 = k\tag1$$ is solvable in the rationals for all $k$. His original formulation is, $$(p^3+qr)^3 + (-p^3+pr)^3 + (-qr)^3 = N (6Nvp^2)^3$$ where, $$(p,q,r)= (N^2+3v^3,\; N^2-3v^3,\; 36N^2v^3)$$ However, this is not complete. While his was a sextic in $k$, a cubic in $k$ is also possible, $$(27m^3-n^9)^3 + (-27m^3+9mn^6+n^9)^3 + (27m^2n^3+9mn^6)^3 = m(27m^2n^2 +9mn^5+3n^8)^3$$ P.S. It seems the complete rational solution to $(1)$ is known only for $k=1$.

Tito Piezas III
  • 60,745