Determine all possible positive integral solutions of a^bē=b^a.

not quite complete but its a start on an analytical solution

a^(b^2)=b^a

either a=b,a>b,a<b

if a=b then
a^(a^2)=a^a thus
a^2=a thus
a=0 or a=1
so this gives us the solutions (0,0) and (1,1)
depending on your perspective you may wish to discard (0,0)

if a>b or a<b  we can find a real number r such that a=b^r
then we have
(b^r)^(b^2)=b^(b^r)
b^(r*b*b)=b^[b*b^(r-1)]
(b^b)^(r*b)=(b^b)^[b^(r-1)]
thus
r*b=b^(r-1)
r=b^(r-2)
b=r^(1/(r-2))
thus we want to find real r such that both
b=r^(1/(r-2)) and
a=r^(r/(r-2)) are both integers

if r=1
b=1^(-1)=1
a=1

if r=3
b=3^(1/1)=1
a=3^(3/1)=27

if r=4
b=4^(1/2)=2
a=4^(2)=16

now if r>4 it would make 1<b<2 and thus
no further solutions can be found

now if r=2 we have a=b^2 and thus
(b^2)^(b^2)=b^(b^2)
b^(2*b*b)=b^(b*b)
(b^b)^(2b)=(b^b)^b
2b=b
b=0 then a=0 which we already found


so it would seem
(0,0),(1,1),(16,2),(27,3) are the only
solutions

Posted by Daniel on 2006-07-23 19:08:38