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Unequally Yoked (Posted on 2004-02-26) Difficulty: 3 of 5
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).

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionMath Man2015-03-14 08:41:40
SolutionPuzzle SolutionK Sengupta2008-05-12 16:26:22
answerK Sengupta2008-01-16 11:55:49
Hints/TipsMe thinks.......abhishek2004-04-07 07:26:28
Solutioni once set this challenge to my teacher, who didnt get itsassy2004-03-16 12:47:09
Some Thoughtsre: such effortPenny2004-02-26 23:57:01
such effortDJ2004-02-26 15:18:45
Some Thoughtsre(2): SolutionFederico Kereki2004-02-26 13:50:13
re: SolutionFrank Riddle2004-02-26 12:59:32
SolutionSolutione.g.2004-02-26 10:40:21
Some ThoughtsNot ManyJuggler2004-02-26 10:06:00
SolutionNo SubjectAdy TZIDON2004-02-26 09:52:11
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