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

Posted:

**Sun Nov 30, 2008 4:37 am UTC**by

**Pathway**So I think I have a correct answer to the question of which distinct positive integers (a,b) satisfy a

^{b}= b^{a}. 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, a

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))

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 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 a

^{1}=1^{b}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 2

^{4}= 4^{2}= 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 a

^{b}= b^{a}.