a^b = b^a for integers a,b > 0

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Pathway
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a^b = b^a for integers a,b > 0

Postby Pathway » Sun Nov 30, 2008 4:37 am UTC

So I think I have a correct answer to the question of which distinct positive integers (a,b) satisfy ab = ba. Did I miss anything here?

Spoiler:
First, neither a nor b can be 1: assuming wlog b =1 since (a,b) is a solution iff (b,a) is a solution, a1=1b gives a = 1. Therefore if one of {a,b} is 1 then they both are and hence aren't distinct.

With the provision that neither a nor b is =1, and that they are both positive, we have

a/log(a) = b/log(b). [log(x) is the for natural logarithm.]

This indicates to us that the problem is equivalent to finding a, b positive integers such that for f(x)=x/log(x), f(a)=f(b).

Now f'(x) = 1/log(x) - 1/(log(x))2. If x > e, then log(x) > 1 and 0 < 1/log(x) < 1. This means that for x > e, 1 > 1/log(x) > 1/log(x)2 > 0. Hence f'(x) > 0 for x > e, i.e. f is strictly increasing for x > e.

Since f is strictly increasing for x > e, there are no a, b both greater than e such that f(a) = f(b) [edit: unless a=b, which we don't care about so assume wlog at all times a!=b]. This means there are no pairs (a,b) of integers satisfying the original equation with a, b both greater than 2. Hence one of {a,b} must be less than or equal to 2, and since neither can be 1, one of {a,b} must be 2.

Note at this point that 24 = 42 = 16, and that (2,3) does not satisfy our equation. Hence any k for which (2,k) might be a valid pair is greater than 4. Then we have that f(2) = f(4) and that f(x) increases between 4 and k, so f(k) > f(2) and hence k > 4 implies that (2,k) is not a valid pair. By symmetry (k,2) for k > 4 is also invalid. Hence the only pairs containing 2 are (2,4) and (4,2).

Since every valid pair must contain 2 and since the only pairs containing 2 are (2,4) and (4,2), these are the only distinct integers for which ab = ba.
Last edited by Pathway on Sun Nov 30, 2008 8:22 am UTC, edited 1 time in total.
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Re: a^b = b^a for integers a,b > 0

Postby jestingrabbit » Sun Nov 30, 2008 5:01 am UTC

looks pretty tight though you can shorten the argument a little by saying that

Spoiler:
f(x) is "U shaped" on (1,\infty) (this can be made formal) so for each y!=e there exists a unique x such that x!=y and f(x)=y, so given that one of them must be 2, and 4 pairs with 2, the only pairs are (2,4) and (4,2).
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Re: a^b = b^a for integers a,b > 0

Postby t0rajir0u » Mon Dec 01, 2008 2:33 am UTC

The integer argument can be phrased in a way that avoids any analysis, and the benefit of this is that the same method generalizes immediately to tell you what the rational solutions are. The first step is to note the following

Lemma: If positive integers a, b, c, d satisfy a^b = c^d, then there exists a positive integer e such that a, c are both powers of e.


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