Determine all possible nonzero integers A and B, with A ≠ B, that satisfy this equation.

                                         A^B = B^A-B

Let t be the divisor of A,B such that A=ta and B=tb
and a,b are co-prime to each other,i.e., the common
divisors of a,b are +1,-1 only.
(ta)^(tb)=(tb)^(t(a-b))
=> (ta)^b=(tb)^(a-b)
=> (a^b)/(b^(a-b))=t^(a-2b)
a,b are co-prime to each other, so from above eqn
either a=1 or b=1
Case(1): b=1
a=t^(a-2)
a ≤ -2 and a ≥ 5. there is no solution to above eqn
a=-1 => t =-1 => A=1,B=-1
a=1 =>t=1 => A=1,B=1 (can't be the solution)
a=2 => 2=t^0 => 2=1(contradiction)
a=3=> 3=t => A=9,B=3
a=4 => t=+/-2 => A=8,B=2 , A=-8,B=-2
Case(2): a=1
b^(1-b)=t^(2b-1)
Consider b>1 => |LHS|<1 and |RHS|>1
Consider b<-1 => |LHS|>1 and |RHS|<1
so, b=-1(if b=1,A=B)
1=t^-3 => t=1 => A=1,B=-1
So, the only solutions to the given eqn are
(A,B)=(1,-1),(9,3),(8,2) and (-8,-2)

Posted by Praneeth on 2008-07-21 04:59:01