From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: interesting problem
Date: 29 Oct 2001 16:18:52 GMT
Newsgroups: sci.math,alt.math.undergrad,alt.math.recreational
Summary: Parameterizing the rational/integral solutions to a^(b^2) = b^a

In article <b2b2562d.0110290537.41da1d8e@posting.google.com>,
Dorgale <try_josua@hotmail.com> wrote:
>Find all pairs (a,b) of integers a>1, b>1 that satisfy the equation: 
>
>a^(b^2) = b^a

This is a FAQ here except for the "2", but the method can be extended.

Given positive integers (or indeed rational numbers)  a, b  satisfying the 
given equation, we may let  t = a/b^2. Then we have
  (b^2*t)^(b^2)=b^(b^2*t)
which expands to
  ((b)^(b^2))^2*t^(b^2)=(b^(b^2))^(t)
Divide both sides by  ((b)^(b^2))^2  to get
  t^(b^2)=(b^(b^2))^(t-2) = (b^(t-2))^(b^2)
Extract  b^2-th roots:
  t=(b)^(t-2)
Solve for  b, and then for  a = b^2 * t:
  b = t ^ (1/(t-2))
  a = t ^ (t/(t-2))

Conversely, for any rational  t (except  t=2)  we find that this pair
of expressions satisfies the desired equation. Thus we have a complete
parameterization of the positive rational solutions of the initial equation.
(We can use this parameterization to pick out all the positive real
solutions, too.)

A fractional power of a rational number is never an integer
unless the base is an integer (and the exponent positive), or the
base is the reciprocal of an integer (and the exponent negative), or
the exponent is zero.  So our expression for  b  shows that either
t  is an integer greater than  2, or  1/t  is a positive integer.

In the first case we need the integer  t  to be the  (t-2)th power of
an integer, and in particular  t >= 2^(t-2). This happens only for t <= 4,
and so we have the solutions with t=3 and t=4 (respectively,
a=27, b=3  and  a=16, b=2). 

In the second case let t = 1/n  with  n  a positive integer. Then
b=n^(n/(2n-1)), a=n^(1/(2n-1))  and again for  n>1  this means  n
is at least  2^(2n-1), which is a contradition, giving us only a=b=1,
which was excluded in the problem statement.

dave