Bounding Mixed Exponential-Algebraic Diophantine EquationsDate: 10/21/2010 at 06:46:32 From: Nick Subject: solve in natural numbers the equation 2^x+x^2=3^y+y^3. Dear Doctor Math, I'm trying to solve this system of equations, where k is an integer: 2^x - 3^y = k y^3 - x^2 = k The obvious solution is (x,y) = (3,2). I think that is the only one -- but I can't prove it. I know papers that discuss such equations invoke Baker's method and lower bounds of the forms |a^x - b^y| |y^3 - x^2| Date: 10/23/2010 at 21:20:34 From: Doctor Vogler Subject: Re: solve in natural numbers the equation 2^x+x^2=3^y+y^3. Hi Nick, Thanks for writing to Dr. Math. That looks like a challenging Diophantine equation to solve, particularly because it is mixed exponential (2^x, 3^y) and algebraic (x^2, y^3). I suspect that modular arithmetic won't get you very far on this one, which is usually the easy way to solve these kinds of equations. Perhaps Baker's method is indeed your best bet. If you haven't used it before, then I would recommend a book by Nigel Smart, entitled "The Algorithmic Resolution of Diophantine Equations." The basic idea is that there are various theorems which will give you a lower bound on the quantity ... x log 2 - y log 3 ... which depends on the size of x and y. The bounds generally have this form, for some constant C: log | x log 2 - y log 3 | > -C * log max(|x|, |y|) We use this theorem like so. Suppose 2^x + x^2 = 3^y + y^3. Then it follows that ... 2^x * 3^-y = 1 + (3^-y)(y^3 - x^2) ... so that x log 2 - y log 3 = log(1 + (3^-y)(y^3 - x^2)). The left side of the equation is our linear form in logarithms. The right side of the equation is very close (when y is large) to the quantity (3^-y)(y^3 - x^2) Using our inequality from Baker Theorem, we get log | log(1 + (3^-y)(y^3 - x^2)) | > -C * log max(|x|, |y|) This is pretty close to ... log | (3^-y)(y^3 - x^2) | > -C * log max(|x|, |y|) ... and that's the same as ... -y log 3 + log|y^3 - x^2| > -C * log max(|x|, |y|) ... or log|y^3 - x^2| + C * log max(|x|, |y|) > y log 3 Notice that the left side of this equation involves logs of y, while the right side does not. So you can use this equation to show that y can't be too large and still solve your Diophantine equations. Unfortunately, even for your case -- which is as simple as you can get, with two variables and integer operands -- a typical theorem that gives a lower bound on a linear form in logarithms will have a constant that is in the millions or billions or so. So your upper bound on y is generally pretty big -- too big to actually test all of the y values up to the bound. Here, a technique such as lattice reduction, which uses the actual values of the logarithms (to a finite precision) and a large bound on y, would reduce it to something more reasonable, or at least small enough that a computer could then try all of the y values up to the given bound. It's all a lot of work, but it's pretty straight-forward, and the end result is that Baker's Theorem gives a way to completely solve your Diophantine equations. 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: 10/25/2010 at 04:52:36 From: Nick Subject: Thank you (solve in natural numbers the equation 2^x+x^2=3^y+y^3.) Thank you! Let me study what you wrote and try to solve the equation. If I need some more help, I'll write back. Yours sincerely, Nick |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/