Solving Multivariable Diophantine Equations
Date: 05/03/98 at 17:32:06 From: Jin Subject: Diophantine equations How would I go about solving this multivariable diophantine equation? x^2 + y^2 - 2xz = 0 For example, I see that (1,3,5) works, but I'd like to know how to solve it.
Date: 05/07/98 at 10:54:26 From: Doctor Nick Subject: Re: Diophantine equations Hi Jin - This is a nice problem. Suppose x = 0. Then y^2 = 0, so y = 0, and z can be anything. Now we can suppose that x is not 0. The first thing I noticed about your equation is that it would be easy to solve for z. So, I did: z = (x^2 + y^2)/(2x) Now, z is supposed to be an integer. That means that x^2 + y^2 has to be a multiple of 2x. Suppose first that x is odd. Then x^2 is odd, and since x^2 + y^2 = 2x, that's even, and so y must be odd. Now (x^2 + y^2)/2x = (x^2)/(2x) + (y^2)/(2x) = x/2 + (y^2)/(2x) has to be an integer. But x/2 is not an integer -- it's a whole (odd) multiple of 1/2. So, (y^2)/(2x) has to be a whole multiple of 1/2 -- that means that x divides y^2. And if y is odd, and x divides y^2, then we get a solution. Now, suppose x is even. Then x^2 is even, so y^2 and y must be even. Then z = (x^2 + y^2)/(2x) = x/2 + (y^2)/(2x) has to be an integer. But that would mean x/2 is an integer, so 2x must divide y^2. And if y is even, and 2x divides y^2, then we get a solution. That's it: pick an x that's not zero. If x is odd, pick an odd y such that x divides y^2, and you get z automatically (remember that z = (x^2 + y^2)/(2x)). If x is even, pick an even y such that 2x divides y^2, and you get z automatically. These are all the solutions. There are lots of them. For example: x = 1, y = 1, z = 1 x = 25, y = 35, z = 37 x = 16, y = 8, z = 10 x = -5, y = 15, z = -25 Notice that x = y = z always works (like x = y = z = 1, x = y = z = 2, and so on). Have fun, -Doctor Nick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 05/17/98 at 11:53:26 From: Jin Subject: Diophantine equations Thanks for your response to my last question. I have another multivariable diophantine equation I was hoping you could solve. It's slightly more complicated. How can you find the solutions to this, other than letting one of the variables equal to zero? x^4 + y^4 - 4z(x^3) + 6(z^2)(x^2) - 4(z^3)x = 0
Date: 05/19/98 at 13:36:39 From: Doctor Nick Subject: Re: Diophantine equations Hi Jin - The left hand side of your equation factors to: (x - z)^4 - z^4 + y^4 So your equation can be re-written: z^4 = y^4 + (x - z)^4 and if we let w = x - z, we have: z^4 = y^4 + w^4 which ought to look familiar -- it's a case of Fermat's Last Theorem. The proof that there are no non-trivial solutions (i.e., a solution with no zeros) was given by Fermat, using his Method of Infinite Descent. For a discussion, take a look at: http://www.math.toronto.edu/mathnet/plain/questionCorner/fermat4.html If you haven't already, I recommend you get a book on number theory. A really good book is Hardy and Wright's _An Introduction to the Theory of Numbers_. It's a super classic, and covers a lot of stuff, including your equation. It might seem kind of sophisticated at first, but from the sound of it, you're well on your way to "growing into" it. Another book that's probably easier to dip into is Kenneth Rosen's _Introduction to Number Theory_, which I know covers this equation. If you have more trouble with this equation, let me know, and definitely write if you have more diophantine difficulties. Check out some books to learn some techniques. When you write, it might help us if you include some idea of why you're trying to solve a particular equation. Feel free to elaborate, and say what you've tried as methods of solution. Have fun, -Doctor Nick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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