Solving a Diophantine Equation by Use of Elliptic CurvesDate: 03/23/2008 at 17:36:33 From: Wayne Subject: source for solution to particular Diophantine equation Please help me to find a source for solving the equation (or similar) 4u^3 - v^2 = 3. It has solutions: (1,-1), (1,1), (7,-37), and (7,37). My guess is that these are all the solutions, but I can't find any exposition on solving these types of equations. I've tried factoring in Z[omega], the ring of algebraic number of the quadratic number field obtained by adjoining the square root of negative 3 to the rationals. I have not been able to make any headway. Thanks. Date: 04/03/2008 at 20:15:29 From: Doctor Vogler Subject: Re: source for solution to particular diophantine equation Hi Wayne, Thanks for writing to Dr. Math. I'm impressed at your trying to work in Z[omega]. This is actually a useful method, and I do some of this in this answer: Solving a Nonlinear Diophantine Equation http://mathforum.org/library/drmath/view/68414.html But often the end result is a thing known as a Thue Equation, which takes some additional work to solve. Either that, or you need to use linear forms in logarithms, which are also pretty complicated. There are, in fact, many expositions on solving these types of equations. The trick is to search for the phrase "elliptic curve." Your equation v^2 = 4u^3 - 3 is equivalent to y^2 = x^3 - 48 where y = 4v and x = 4u. This last equation is an elliptic curve in standard (Weierstrass) form. Many elliptic curves (in fact, about half of them) have only finitely many rational solutions. And finding the rational solutions is generally much easier than finding the integer solutions. So if there are only finitely many rational solutions, then you just check which ones are integer solutions. I describe how to find the rational solutions in these answers: Solving System of Equations Using Elliptic Curves http://mathforum.org/library/drmath/view/70511.html Using Elliptical Curves to Solve an Arithmetic Sequence http://mathforum.org/library/drmath/view/69827.html Unfortunately for you, there are infinitely many rational solutions, since your curve has trivial torsion and rank 1. That is, all integer multiples of the point (x, y) = (4, 4) are rational points on your curve (and all rational points are integer multiples of that point). These multiples include . . . -3(4, 4) = (73/9, -595/27) -2(4, 4) = (28, 148) -1(4, 4) = (4, -4) 0(4, 4) = 0 1(4, 4) = (4, 4) 2(4, 4) = (28, -148) 3(4, 4) = (73/9, 595/27) 4(4, 4) = (9772/1369, -899996/50653) 5(4, 4) = (1184884/32041, 1289162164/5735339) 6(4, 4) = (48833569/12744900, -130710951503/45499293000) . . . and so on. (I computed the rank in mwrank, and the above points in Pari, as described in the above answers.) The (u, v) values you can get by dividing x and y by 4. But every elliptic curve has only finitely many integer solutions (this is known as Siegel's Theorem), so when there are infinitely many rational points, it can be hard to be sure that you have found all of the integer ones. There are three general methods for doing this, all of them described in detail in the book "The Algorithmic Resolution of Diophantine Equations" by Nigel P Smart. But all three methods are quite complicated and take a significant amount of computation, enough to make anyone do it on a computer. But you can find all integer solutions and *prove* that you have them all. Of course, if you don't need a complete proof, then you use the fact that the number of digits in the numerators *and* denominators of the x and y coordinates of the point n(4, 4) is roughly a multiple of n^2 (and this becomes less rough as n gets large), which means that you're not likely to get a denominator of 1 (i.e. an integer solution) when n is remotely large. So you check the first small integers like the ones I already did, and when you see the denominator get big and keep growing, you can be pretty confident that you aren't going to find any more integers. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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