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.
How to solve $x^3+y^3+z^3=kt^3$?
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diophantine-equations
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0Do you want all solutions or just one (assuming that a solution exists) ? – 2017-01-14
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1All the solutions for it is easy to choose arbitrary $x,y,z$ and simplify this equation. – 2017-01-14
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3Some useful links and equations can be found [here](http://math.stackexchange.com/questions/494424/proving-that-any-rational-number-can-be-represented-as-the-sum-of-the-each-cube) . Ryley's Theorem shows existence of solutions, I do not know if it yields all solutions. – 2017-01-14
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0For some private to write about. http://www.artofproblemsolving.com/community/c3046h1172731_a_cubic_equation_with_the_troika – 2017-01-14
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0@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. – 2017-01-14
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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 $k=1$.