Rational Solutions to Two Variable Quadratic EquationDate: 11/25/2003 at 01:29:22 From: Sam Subject: Rational solutions to x^2 + y^2 = 2 Find all the rational solutions to x^2 + y^2 = 2. This is under the section on diophantine equations, which I understand, but this one is different since it is asking for all rational solutions, not just the integer ones. I think that the way to do this is similar to how you would do it for the Diopahntine equations which have integral solutions, but I am not sure if I am getting the correct answer. The answer I get is x = 2*r*s, y = r^2-s^2, where r and s are relatively prime, one odd and one even. Date: 11/25/2003 at 17:59:15 From: Doctor Vogler Subject: Re: Rational solutions to x^2 + y^2 = 2 Sam, There's a handy little trick to get all rational solutions to a quadratic equation in two variables, given one. If you have any one solution (x, y) = (a, b), then any *other* solution (r, s) will lie with (a, b) on a line with rational slope (or infinite slope, a vertical line; this is a special case). Furthermore, every line through (a, b) with rational slope will intersect with the quadratic curve in exactly two points. That's because every quadratic equation has either no solutions or two, and we already know that (a, b) is one solution. So to find all solutions, you first look at the vertical line case by substituting x = a and seeing what two solutions you get. One will be y = b, and the other gives a solution (which may be the same solution). Next, you take a line with rational slope m through (a, b), so that x = a + t, y = b + tm, and you substitute this parameterization into your curve, simplify, divide by t (since t = 0 is always a solution), and you end up with a linear equation in t which gives all other solutions in terms of a, b, and m. See if you can take it from there and write back if you need more help. - 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/