Finding Integer Solutions to a^b = b^aDate: 02/02/2005 at 12:31:23 From: jim Subject: Number Theory Find all positive integers a and b such that a^b = b^a. I can think of some examples of numbers that will work (a = 2 and b = 4), but I don't know how to generalize the equation. I need to have a general proof for this problem. If I take the log of each side I can get b log a = a log b, then divide by ab to get (log a)/a = (log b)/b. From there I don't know where to go. Date: 02/02/2005 at 19:00:11 From: Doctor Vogler Subject: Re: Number Theory Hi Jim, Thanks for writing to Dr. Math. The short answer to your question is that all solutions in positive integers are a = b and (2, 4) or (4, 2). The long answer, of course, is a proof of the same. Before going there, however, I would like to point out that your idea is a terrific idea for finding all *real* solutions. Allow me to elaborate. Since the function f(x) = (log x)/x increases from x=0 to x=e and then decreases from x=e to infinity, if 0 < a < b <= e or if e <= b < a, then we have f(a) < f(b) and therefore (log a)/a < (log b)/b b log a < a log b a^b < b^a. But if we have a < e < b, then that is not enough information to tell whether a^b or b^a is bigger. And for each a < e there is exactly one b > e where f(a) = f(b) and therefore a^b = b^a, and for smaller values of b (but still bigger than e) you will have a^b < b^a while for larger values of b you will have a^b > b^a. But this idea of using continuous functions becomes less useful when dealing with integer or rational solutions, since the Intermediate Value Theorem doesn't apply. So we need to try a different idea. Try this: Substitute t = b/a into your equation. If a and b are integers, or if a and b are rational numbers, then t will be rational. Substituting a*t for b, we get t a a (a ) = (at) . Now, taking the (positive real) a'th root of both sides of the equation, we get t a = at and therefore t-1 a = t and 1/(t-1) a = t . In fact, we can get a parametrized form for all rational solutions from this equation. Let n = 1/(t-1), so that a = (1 + 1/n)^n b = (1 + 1/n)^(n+1). These are rational solutions for all (nonzero) integers n. It takes a little more work to show that these (along with a=b) are *all* rational solutions. And that work begins with proving the lemma: If r/s and c/d are rational numbers in lowest terms, and (r/s)^(c/d) is also rational, then r and s are both d'th powers of integers. But you wanted integer solutions. And I'm done with my tangents. The first step is to prove that t is an integer. Then all that's left is a simple induction argument. Of course, t is not an integer if b=2 and a=4. But if we permit ourselves to switch a with b (note that this doesn't change the equation) so that a <= b then we can prove that t will be an integer. You see, if a <= b, then a b a divides a . But a^b = b^a, which means that a a a divides b , and this is enough to imply that a divides b (think about the previous statement in terms of the prime factorizations of a and b), and therefore t=b/a is an integer. And if a and b are positive, then t is also positive. Now t=1 gives the solutions with a=b. And if t=2, then t-1 a = t gives a = 2 (and therefore b = 4). And a=1 quickly implies b=1. But if a >= 2, then t-1 t-1 a >= 2 , and we can prove by induction that t-1 2 > t for all integers t >= 3. This is easily verified for t=3. Now suppose that t-2 2 > t-1 and then we have t-1 t-2 2 = 2 * 2 > 2(t-1) = t + (t-2) > t. Therefore, there are no solutions with t > 2, and so we have found all of the integer solutions, namely (k,k), (2,4), and (4,2). 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: 08/01/2011 From: Nathan Subject: Simpler solution? I was thinking about this same problem (integer solutions to a^b=b^a), Googled it and found this page. I read the beginning, up to f(x) = (log x)/x increases from x=0 to x=e and then decreases from x=e to infinity and decided that I knew exactly where it would go from there... The solutions should be pairs from the sets (1, e) and (e, \inf). Since there is clearly a useful bijection between them, and there's only one integer, 2, in the first set, a=2, b=4 is the only solution. |
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