this is to complete my previous work by considering all possible real values of r

let a=b^r  with a,b integers and r a real number

(b^r)^(b^2)=b^(b^r)
b^(rb^2)=b^(b^r)
rb^2=b^r
r=b^(r-2)

now there are 3 possibilities for r
(1) r is integer
(2) r is rational (non-integer)
(3) r is irrational

I already took care of case 1 in previous posts

as for case 2 if r is rational then r is equal to b raised to a rational power.  But since b is an integer then b raised to a rational power would either be an integer or irrational thus not being able to be equal to r.  Thus r can not be rational

as for r being irrational we have as above that
r=b^(r-2)=(b^r)/(b^2)=a/(b^2)
but the right side makes r rational and thus we have a contradiction and thus r can only be integer in value and in my other post I showed the only possible solutions with r being integer.

Posted by Daniel on 2006-07-24 22:32:13