Solving Diophantine EquationsDate: 04/18/2008 at 14:26:34 From: Garrett Subject: Solutions to 7x^2 + 1 = y^3 over integers (possibly reals). I have recently delved into the world of Diophantine equations, elliptic curves, and number fields. Looking at the equation 7x^2 + 1 = y^3 I have computationally found integer solutions (x>=0): (x,y) = (0,1), (1,2), (3,4), and (39,22) but hit round-off issues around x = 1.05*10^6. Still, being able to search beyond 10^6 makes me question if these are the only integer solutions. Could you please answer the questions as to the number of integer solutions to this equation and either provide an upper bound for finite solutions (or find all solutions) or how to generate infinite solutions? Could you also address this question in x,y over reals? I have only begun reading about elliptic curves and the different methods used to solve these types of equations. I am hoping another solution of this type of question will help me to better understand how to go about solving these types of questions. Thanks for the help! Date: 04/18/2008 at 21:23:57 From: Doctor Vogler Subject: Re: Solutions to 7x^2 + 1 = y^3 over integers (possibly reals). Hi Garrett, Thanks for writing to Dr. Math. You are correct that you have an elliptic curve. Finding the real points is easy. For every real number x, you can set y = (7x^2 + 1)^(1/3). Alternately, y has to be at least 1, since 7x^2 >= 0, but whenever y >= 1, there are two solutions for x, namely sqrt((y^3 - 1)/7) and -sqrt((y^3 - 1)/7). You can plot this curve on a graphing calculator or computer graphing program to see what it looks like. Finding all of the integer points is *much* harder. There are often easy methods to prove that there are no integer points on such an equation, but these methods don't work when there are some. Methods for finding all integer points on an elliptic curve use much of the structure of an elliptic curve. If you want to use programs to help you with your curve, then you should put it in standard (Weierstrass) form. If you set v = 49x u = 7y and you multiply your equation by 7^3 = 343, then you get v^2 = u^3 - 343 which is an elliptic curve in standard 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 rank 1, and torsion of order 2. That is, all integer multiples of the point (u, v) = (14, 49), along with all such points plus the torsion point (7, 0) are rational points on your curve (and this gives you all rational points). These multiples include . . . -5(14, 49) = (3397814/477481, -1374504341/329939371) -5(14, 49) + (7, 0) = (10082548/7921, 32015140437/704969) -4(14, 49) = (6512/169, -523921/2197) -4(14, 49) + (7, 0) = (62146/5329, 13715247/389017) -3(14, 49) = (763/9, 21070/27) -3(14, 49) + (7, 0) = (889/100, -18963/1000) -2(14, 49) = (8, 13) -2(14, 49) + (7, 0) = (154, -1911) -1(14, 49) = (14, -49) -1(14, 49) + (7, 0) = (28, 147) 0(14, 49) = 0 0(14, 49) + (7, 0) = (7, 0) 1(14, 49) = (14, 49) 1(14, 49) + (7, 0) = (28, -147) 2(14, 49) = (8, -13) 2(14, 49) + (7, 0) = (154, 1911) 3(14, 49) = (763/9, -21070/27) 3(14, 49) + (7, 0) = (889/100, 18963/1000) 4(14, 49) = (6512/169, 523921/2197) 4(14, 49) + (7, 0) = (62146/5329, -13715247/389017) 5(14, 49) = (3397814/477481, 1374504341/329939371) 5(14, 49) + (7, 0) = (10082548/7921, -32015140437/704969) . . . and so on. (I computed the rank in mwrank, and the above points in Pari, as described in the above answers I linked.) The (x, y) values you can get by dividing v by 49 and u by 7. 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/