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        riddles >> easy >> a^b = b^a
        
(Message started by: NickH on Feb 8^th, 2003, 2:52pm)

Title: a^b = b^a
Post by NickH on Feb 8^th, 2003, 2:52pm

Find all rational solutions, with a < b, to a^b = b^a.

Title: Re: a^b = b^a
Post by Chronos on Feb 8^th, 2003, 4:24pm

Was that meant to be "all real solutions" rather than "all rational solutions"? I don't think that there are any rational solutions.

Title: Re: a^b = b^a
Post by NickH on Feb 8^th, 2003, 4:30pm

I definitely mean rational, though real would be a challenging extension. (I'm not sure how the solutions would be listed.) The simplest solution is a = 2, b = 4, but there are others...

<edit>(If not clear, I mean a and b to be rational, not necessarily a^b.)</edit>

Title: Re: a^b = b^a
Post by NickH on Feb 20^th, 2003, 1:23pm

I almost have a solution to this one; I'm just short on one part of the proof. See hidden text below...

[hide]Let b = ra, where r > 1, and necessarily rational.
Then a^ra = (ra)^a = r^aa^a.
So a^a(r-1) = r^a, and a^r-1 = r.
Hence a = r^1/(r-1).

Now, I'm pretty sure, and can almost prove, that r^1/(r-1) is irrational unless 1/(r-1) is an integer. So I have a countable infinity of solutions, and an incomplete proof.[/hide]

Title: Re: a^b = b^a
Post by Icarus on Feb 20^th, 2003, 4:09pm

To complete Nick's proof:[hide]

If x, y, and k are positive integers with k>1, then (x+y)^k - x^k = kxy^k-1+ ... > k. Thus the difference between any two distinct k^th powers > k.

Now for any positive integers k,l,m,n, with (k,l)=1, (m,n)=1, (m/n)^k/l is rational iff both m and n are perfect l^th powers.

Let r = m/n. 1/(r-1) = n/(m-n). So for r^1/(r-1) to be rational, both m and n must be perfect m-n powers. But then, their difference would have to be greater than m-n unless m-n = 1 (m > n since r > 1).

So the solution is r = (n+1)/n, a=rⁿ, b=rⁿ⁺¹ for integers n > 0.[/hide]