Catalan's ConjectureDate: 06/02/2007 at 21:31:49 From: Thomas Subject: Solve a^b = b^a + 1 Solve a^b = b^a + 1. How many integer solutions are there? Are there an infinite number? How are the solutions distributed? I need some clues as to how to analyze this equation. I noticed that 8 is 2^3 while 9 is 3^2. I started thinking about pairs of consecutive numbers where one is of the form a^b and the other is of the form b^a. I expressed the relationship between a and b in the above equation and tried tried taking the log of both sides, but didn't know how to expand log(b^a + 1) or if this was the best approach. Is there a way to analyze this problem other than random guessing? Date: 06/03/2007 at 08:54:13 From: Doctor Vogler Subject: Re: Solve a^b = b^a + 1 Hi Thomas, Thanks for writing to Dr. Math. The only integer solution with a>1 and b>1 is the one you already found. In fact, more is true; see "Catalan's Conjecture" From MathWorld - A Wolfram Web Resource. http://mathworld.wolfram.com/CatalansConjecture.html Of course, if b = 1, then a = 2, and if a = 1, then b = 0. So these would be the other integer solutions. On the other hand, proving Catalan's Conjecture takes a lot of work, so you could probably come up with a more accessible analysis of your equation, sort of along the lines of Solving the Diophantine Equation x^y - y^x = x + y http://mathforum.org/library/drmath/view/66640.html I guess the way I would attack this problem is by using the function f(x) defined in Solving the Equation x^y = y^x http://mathforum.org/library/drmath/view/66166.html A solution to your equation has a^b = b^a + 1 and therefore a^b > b^a which means that b log a > a log b and therefore (log a)/a > (log b)/b or f(a) > f(b). That means that either a or b is 1 or 2, or both are bigger than e and therefore a < b which means (since a and b are both integers) that a <= b - 1. Next, I would try to prove that for fixed a >= 3, the function g(x) = a^x - x^a is increasing on x > a. (Hint: x > a > e implies a^x > x^a.) That means that g(b) >= g(a+1). Next, I would try to prove that for a >= 3, g(a+1) > 1, perhaps by defining another function h(x) = x^(x+1) - (x+1)^x - 1 and looking at its derivative.... Of course, you also have to check the special cases when a or b is 1 or 2, but these are much easier than the general case I just outlined. 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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