# sum of cubes of two rationals

How to find two rational numbers $x,y$ such that $$x^3+y^3=6$$ I know that $x=17/21,y=37/21$ is a solution but I am interested in a method how is achieved and does exists other solutions

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That equation defines an elliptic curve. The problem of finding all rational points on an elliptic curve can be solved for many a curve (or class of curves), but the math is very non-trivial in general. There may be only finitely many solution or the solutions can all be generated via the secant-tangent method starting from a carefully chosen set of solutions. – Jyrki Lahtonen Jan 27 '13 at 10:52
Finding a rational solution of $x^3+y^3=6$ was set as a puzzle by Dudeney in Amusements in Mathematics, about $100$ years ago. I'm sure his method was educated trial-and-error. Once you have found one solution, you can often find others, but the denominators tend to grow very fast. The line tangent to the graph of $x^3+y^3=6$ at $(17/21,37/21)$ hits the curve at a point which will have rational coordinates. – Gerry Myerson Jan 27 '13 at 11:36
Magma says this curve has a minimal model $y^2 = x^3-243$ and is rank 1, in case others are interested. I am not familiar with Magma so I don't know what the explicit isomorphism between these curves is. – user27126 Jan 27 '13 at 11:45
This is relevant. – P.. Jan 27 '13 at 12:51
@GerryMyerson By the way, that solution that you described is: $$(x,y) = \left(\frac{-1805723}{960540},\frac{2237723}{960540}\right)$$ – Rustyn Jan 27 '13 at 19:44

Solutions $z$ of the diophantine equation $x^3 + y^3 = 6z^3$ are tabulated at the Online Encyclopedia of Integer Sequences. Only $4$ are given (though infinitely many exist): $21$, $960540$, $16418498901144294337512360$, and $436066841882071117095002459324085167366543342937477344818646196279385$ $305441506861017701946929489111120$.

See also this mathforum post, and the article, The £$450$ question, by J. H. E. Cohn, Mathematics Magazine 73, No. 3 (Jun., 2000) 220-226.

EDIT: Indeed, Cohn gives a solution not in the OEIS, and smaller than that last solution: $$z=1097408669115641639274297227729214734500292503382977739220$$ It's a very nice paper.

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I find that an easy way to write special characters is to copy them from unicodeforyou.appspot.com, e.g. pound. – Rahul Jan 29 '13 at 0:03
If you want to generate more, they give you the maple code @ that link also. but by 7th or so $z$, we are up to 10's of thousands of digits. – Rustyn Jan 29 '13 at 1:17

Using the maple syntax from this site,
I have here $6$ $z$ such that: $$x^3 + y^3 =6z^3$$ I have excluded the other $z$'s for the $7^{\text{th}}$ is nearly $30,000$ digits long.

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The "really cool solution" comes from the 4th solution in the oeis list given in my answer. – Gerry Myerson Jan 29 '13 at 2:41

I have used Microsoft Solver Foundation to find a (different) solution:

SolverContext context = SolverContext.GetContext();

Decision a = new Decision(Domain.IntegerNonnegative, "A");
Decision b = new Decision(Domain.IntegerNonnegative, "B");
Decision c = new Decision(Domain.IntegerNonnegative, "C");
Decision d = new Decision(Domain.IntegerNonnegative, "D");

Model model = context.CreateModel();

Term a3 = a * a * a;
Term b3 = b * b * b;
Term c3 = c * c * c;
Term d3 = d * d * d;

Term res = a3 * d3 + c3 * b3 - 6 * b3 * d3;

// model.AddConstraint("a3", a > c);  //  symmetry breaking

model.AddConstraint("b3", b != 21);   //  want something different!

Solution solution = context.Solve();

Console.WriteLine("a={0} b={1} c={2} d={3}", a, b, c, d);


The solver re-discovers your solution in a couple of seconds but is unable to find a different one with numbers below 1000000.

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Try running it to $3,000,000$ to see whether it finds the solution in the comment by @Rustyn. – Gerry Myerson Jan 28 '13 at 23:06
3,000,000 does not work in the MSF solver. This might be caused by the fact that 3,000,000^3 requires a number representation above 64 bits. – Axel Kemper Jan 28 '13 at 23:12

This is an old question, but anyway. Given an initial solution $x,y,z$, to,

$$ax^3+by^3 = cz^3$$

then a new one can be derived as,

$$a(-bxy^3-cxz^3)^3 + b(ax^3y+cyz^3)^3 = c(-ax^3z+by^3z)^3\tag{0}$$

For example, given the OP's,

$$x^3+y^3 = 6z^3$$

starting with initial,

$$x,y,z = 17, 37, 21\tag{1}$$

using $(0)$, we find a second,

$$x,y,z = -1805723,\, 2237723,\, 960540\tag{2}$$

which is the point given by Myerson and Yazdanpour. Using $(2)$, we can find a third and so on, ad infinitum.

P.S. 1. Presumably, a positive $x,y,z$ will appear after every few iterations. 2. For some reason, the solution given by Kohn is skipped by this process.

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