All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Unequally Yoked (Posted on 2004-02-26)
Find all pairs of unequal integers a, b such that a^b = b^a.

 Submitted by Aaron Rating: 3.6250 (8 votes) Solution: (Hide) Suppose first that m and n are both positive. Assume m > n. Then, we can put m = n + k with k > 0, hence (1 + k/n)^n = n^k. But for x > 1, we have 1 + x < e^x (the derivative of f(x) = e^x - x - 1 is positive and f(0) = 0) and hence (1 + k/n)^n < e^k. So, there are no solutions for n > 2. If n = 1, then n^m = 1 and hence m = 1, contradicting the fact that m and n are unequal. If n = 2, then m must be a power of 2. Suppose m = 2^h. Then we find h = 1 or 2. h = 1 is invalid (because m and n are unequal), so m = 4. There is also the corresponding solution with m < n (m = 2, n = 4). If n < 0 and m > 0, then n^m = 1/m-n. So m divides 1 and hence m = 1. But m must be even for this to make n^m positive. Contradiction. So there are no solutions of this type. If m and n are both negative, then -m, -n is a solution, so the only possibilities are (-2, -4) and (-4, -2), and it is readily checked that these are indeed solutions. So, the only four pairs are (2, 4), (4, 2), (-2, -4), (-4, -2).

 Subject Author Date Solution Math Man 2015-03-14 08:41:40 Puzzle Solution K Sengupta 2008-05-12 16:26:22 answer K Sengupta 2008-01-16 11:55:49 Me thinks....... abhishek 2004-04-07 07:26:28 i once set this challenge to my teacher, who didnt get it sassy 2004-03-16 12:47:09 re: such effort Penny 2004-02-26 23:57:01 such effort DJ 2004-02-26 15:18:45 re(2): Solution Federico Kereki 2004-02-26 13:50:13 re: Solution Frank Riddle 2004-02-26 12:59:32 Solution e.g. 2004-02-26 10:40:21 Not Many Juggler 2004-02-26 10:06:00 No Subject Ady TZIDON 2004-02-26 09:52:11

 Search: Search body:
Forums (0)