Integer Proof Using Diophantine Equation
Date: 01/10/2005 at 10:38:58 From: Eugene Subject: Diophantine equations?? The integer 26 is the only integer preceded by a a square and followed by a cube. What is the proof of this?
Date: 01/10/2005 at 11:32:06 From: Doctor Vogler Subject: Re: Diophantine equations?? Hi Eugene, Thanks for writing to Dr Math. I'm surprised you recognized this as a Diophantine equation, but you're absolutely right. You are looking for integer solutions to n + 1 = x^3 n - 1 = y^2, which, of course, is the same as solving y^2 + 1 = x^3 - 1 or y^2 = x^3 - 2. This last is, in fact, an elliptic curve, and so there are methods to find all rational solutions as well. But you wanted all of the integer solutions. For this, you can use the theory of number fields. This does, unfortunately, require some background knowledge. I shall give a description of the solution using some theory from number fields, but you might find some gaps in it if you are unfamiliar with the subject. First we change the equation to x^3 = y^2 + 2 and we wish to factor the right side. Of course, the right side does not factor over the rational numbers (or the real numbers), but it does in the field of rational numbers with sqrt(-2), which I shall call a = sqrt(-2), so that the field is Q(a). Then we get x^3 = y^2 + 2 = (y - a)(y + a). We want to show that the two factors on the right are relatively prime. If we can do that, then the knowledge that the ring of integers in Q(a) is a PID and therefore a UFD will tell us that y - a and y + a are both cubes times some unit. But since Q(a) is an imaginary quadratic field, the Dirichlet Unit Theorem tells us that the only units are +1 and -1, so that y - a and y + a are both cubes. So let's show that y - a and y + a are relatively prime. Clearly, their GCD must divide their difference, which is -2a = a^3. Since a is a prime number in the ring of integers of Q(a), the GCD must be a power of a. Now, if y is even, then 4 divides y^2, so x^3 = y^2 + 2 must be even but not divisible by 4, which is impossible for the cube of an integer. Therefore we may conclude that y is odd. But the prime a lies over the rational prime 2, which means that a does not divide y, and therefore a does not divide y - a or y + a. Therefore, their GCD must be 1. Now we know that y - a and y + a are both cubes of algebraic integers in Q(a). But the algebraic integers in Q(a) are the numbers r + s*a where r and s are (normal) integers. So that means that y + a = (r + s*a)^3 = r^3 + 3r^2*sa + 3r(sa)^2 + (sa)^3 = r^3 + 3r^2*sa + 3rs^2*(-2) + s^3*(-2)a = (r^3 - 6rs^2) + (3sr^2 - 2s^3)a. But if the second part, s(3r^2 - 2s^2) = 1, then that means that s is +1 or -1, and therefore 3r^2 - 2 = +1 or -1 (in fact, equals s), and therefore r is +1 or -1 as well. So that means that y + a = (-1 + a)^3 or y + a = (1 + a)^3. Solving these gives y = +5 and y = -5, respectively. So these are the only solutions, and they both give x^3 = 27 and therefore x = 3. You'll see where I used some facts from algebraic number theory--that is, the theory of number fields. To understand those parts better, I might recommend reading a book about number fields and learning the subject, which I found to be very interesting. I learned from "Number Fields" by D.A. Marcus (which you can find by searching amazon.com for "number fields"), and I thought it was a good book, but there are also many other books available which teach the same kinds of things. 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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