Cubic Diophantine Equation in Three VariablesDate: 09/10/2004 at 05:42:16 From: Juan Pablo Subject: Number theory I want to find integer solutions of the equation x^3 + y^3 = 31z^3. I know the fundamental solution is (137, -65, 42), but want to have all the values positive. I know also that there is an arithmetic procedure (doubling in the group) to obtain further solutions from the fundamental one, but I do not know the details of this procedure. Could you explain these details, and how to get the first positive solution, in this case and in other cases with fundamental non-positive solutions? Thanks in advance. Date: 10/19/2004 at 15:10:00 From: Doctor Vogler Subject: Re: Number theory Hi Juan, Thanks for writing to Dr. Math. First of all, there is a relationship between solutions in integers and solutions in rational numbers. You see, if (x, y, z) is any one solution, then (kx, ky, kz) is also a solution for any number k. So if you have a rational solution, then you can just let k be the least common denominator and get an integer solution. In the group law, all of these solutions--(kx, ky, kz) for all rational k--are considered to be the same. So if z is nonzero then you can think of "the" solution as being (x/z, y/z, 1). (This is called converting to affine coordinates.) The group law you referred to comes from the theory of Elliptic Curves, and it is probably most easily understood in affine coordinates. So let's assume that z is nonzero. (All the z=0 solutions are (x, -x, 0) for any x, so let's look at other solutions.) Then our equation (x/z)^3 + (y/z)^3 = 31 can be written as X^3 + Y^3 = 31. Let's suppose that we have a line Y = mX + b and we want to know where this line intersects our curve. So we substitute X^3 + (mX + b)^3 = 31 and expand this out into a cubic equation in X. A cubic equation might have no rational solutions, or it might have one, or it might have three, but it certainly can't have only two. So if we start with a line that we already know passes through two points on our curve, then we will most assuredly get a third point on the curve. This gives us a way to take two points and "multiply" them together to get a third point. That might get you excited, but it does not, unfortunately, give you a group, because it does not satisfy the law of associativity. Fortunately, it will satisfy the associative law if we make one small change. (Although proving that it satisfies the law is still not easy.) First we fix any single point on the curve, such as (1, -1, 0) and we call this our "zero" point, usually written as the letter O. Now to add a point A to B, we first do the above construction--we draw the line through A and B, and we get the third point C on our curve that also lies on the line. Then we do the same construction with C and O--we draw the line through C and O and get the third point D on the curve that is also on the line. Then D is the sum A+B. In fact, C is the negative of D, so that the three points of our curve on any single line sum to the identity of the group, which is O. Notice that O is called the "point at infinity" because when we try to convert to affine coordinates, we divide by zero. Lastly, for the "doubling" you might wonder how we add a point to itself. If we only have one point, how do we make a line that goes through just that one point? Well, the answer is that it has to go through that point doubly. That is, when we substitute into our equation and get a cubic in X, the point we know about has to be a double root. To get that, we need to use the TANGENT line to our curve at that point. Now I will do an example of doubling your point. Your point is (x, y, z) = (137, -65, 42) = (137/42, -65/42, 1). We want the tangent line to x^3 + y^3 = 31z^3 at this point, so we take the derivatives and find that the tangent line at the point (a, b, c) is a^2*x + b^2*y = 31c^2*z or a^2*X + b^2*Y = 31c^2 in affine coordinates. So the tangent at the point (-137, -65, 42) = (137/42, -65/42, 1) is (137)^2*X + (-65)^2*Y = 31(42)^2 which is the same as (137/42)^2*X + (-65/42)^2*Y = 31. Now we solve for Y and substitute into the equation X^3 + Y^3 = 31 and get Y = (31*42^2 - 137^2 * X)/65^2 X^3 + (31*42^2 - 137^2 * X)^3/65^6 = 31 which we can simplify as (X - 137/42)^2 * (X - 277028111/119531076) = 0 and we get Y from the line above, Y = (31*42^2 - 137^2 * X)/65^2, and we now have our new point, X = 277028111/119531076 Y = 316425265/119531076 or x = 277028111 y = 316425265 z = 119531076. Since we haven't yet done the line with our "point at infinity," this point is (277028111, 316425265, 119531076) = -2(137, -65, 42). Next we do the line with our point at infinity, (1, -1, 0). Any line through this point at infinity (1, -1, 0) is a line with slope -1. So our line would be Y = -(X - 277028111/119531076) + 316425265/119531076. The reason for the slope condition is that if we write this line as y = -(x - z*277028111/119531076) + z*316425265/119531076, then you'll find that it goes through the point (1, -1, 0). Now we substitute this into our equation X^3 + (-(X - 277028111/119531076) + 316425265/119531076)^3 = 31 and we simplify this into (X - 277028111/119531076) * (X - 316425265/119531076) = 0 and we get our double point X = 316425265/119531076 Y = 277028111/119531076 or x = 316425265 y = 277028111 z = 119531076. So (316425265, 277028111, 119531076) = 2(137, -65, 42). Isn't it curious that in this case we had -(a, b, c) = (b, a, c) ? Can you prove that this is always true? To do so, all you need to do is show that (a, b, c), (b, a, c), and (1, -1, 0) are all on the same line, and they are all on our curve. Then, if you want, you can add this last point to our original point to get 3(137, -65, 42) and then either add the result to the original point again, or double our double point, and both should get the same result for 4(137, -65, 42). Of course, you'll find that the numbers get really big really fast. That has to do with the "height" of the point. If they really do keep getting bigger (which they do), then you'll never repeat points, which means that there are infinitely many different solutions. You might also wonder if you can get all rational points on this curve this way. In elliptic-curve-speak, that question is, "Does my elliptic curve have rank one?" You can answer these questions with some deeper math in elliptic curves, but I can't teach it to you in email; you'd need a book and/or a teacher (preferably both). I learned from Knuth's book Elliptic Curves, but there are many other books on the subject. 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/ Date: 10/20/2004 at 08:44:15 From: Juan Pablo Subject: Thank you (Number theory) Dear Dr. Vogler. Thank you very much for your detailed answer. It solves a problem that I failed to solve by asking other specialists. Actually my problem was to convert the original equation into the Weierstrass standard form and, at the same time, to convert the solution in integers of the first into the corresponding solution for the second. Now, thanks to your clear advice, the problem is solved not only in the given example but in general. I thank you again very much. - Juan Pablo |
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