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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8That 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 solutions, 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
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2Finding 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
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2Magma 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. – Jan 27 '13 at 11:45
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5This is relevant. – P.. Jan 27 '13 at 12:51
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3@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
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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.
Aeolian
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Gerry Myerson
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I find that an easy way to write special characters is to copy them from http://unicodeforyou.appspot.com/, e.g. pound. – Jan 29 '13 at 00:03
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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 01:17
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The solution given by Cohn is from 6P which is the 6th multiple of the point $P=(17,37,21)$ of the curve $x^3+y^3=6z^3$. – Somos Dec 30 '21 at 16:13
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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.
Tito Piezas III
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Is there an explanation/proof anywhere of how this derivation is obtained/derived?
Are there equivalents for higher powers?
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1@KierenMacMillan: Ultimately this can be reduced to an elliptic curve $ax^3+by^3=cz^2$. This has a 4th power analogue $ax^4+by^4=cz^2$ but CANNOT be generalized to the homogeneous form $ax^4+by^4=c\color{red}{z^4}$. – Tito Piezas III Mar 21 '20 at 09:26
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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.
link
Rustyn
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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 02:41
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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();
model.AddDecisions(a, b, c, d);
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("eq", res == 0);
model.AddConstraint("a1", a < 1000000);
model.AddConstraint("b1", b < 1000000);
model.AddConstraint("c1", c < 1000000);
model.AddConstraint("d1", d < 1000000);
model.AddConstraint("a2", a >= 1);
model.AddConstraint("b2", b >= 1);
model.AddConstraint("c2", c >= 1);
model.AddConstraint("d2", d >= 1);
// 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.
Axel Kemper
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1Try 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
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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
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Note the danger that most solutions found by the Julia program will reduce to 17/21 and 37/21 for a and b or the reverse. However If this program uses BigInt for the integers it will find another unique solution in a week to two. You also can test and skip any a/c that reduces to 17/21 with the test
if a//c == 17/21; continue; end
before the print statement and declare a,b,c BigInt and increase the range of c by another decade.
