I thought since my solution wound up spread across several posts I would repost it so it could be viewed all at once
we have that
(1) a^(b^2)=b^a with a,b positive integers
now since a,b are positive integers we can find a real number r such that a=b^r
substituting this into (1) gives us
(b^r)^(b^2)=b^(b^r)
b^(r*b^2)=b^(b^r)
thus
r*b^2=b^r
r=b^(r-2)
now there are 3 possibilities for r
1) irrational
2) rational (non integer)
3) integer
now if case 1 is true then we have that
b=r^(1/(r-2))
now b can either be rational or irrational
if it is irrational then we have a contridiction and r can't be irrational
if b is rational then a=b^r and thus a is a rational number raised to a irrational power and thus is irrational and once again we reach a contridiction and thus we can say that r can not be irrational
now if case 2 is true then we have that b^(r-2) is an integer raised to a rational power. A integer raised to a rational power is either going to be a integer or irrational either case it can not be equal to r (a rational number). Thus r can not be a strictly rational number.
That only leaves r to be an integer.
r can not be negative because that would give a=1/(b^(-r)) thus contradicting a being integer
if r=0 then
a=b^0=1
putting this into (1) we get
1^(b^2)=b^1
b=1
thus r=0 gives the solution a=1 b=1
if r=1 then
a=b
putting this into (1) we get
a^(a^2)=a^(a)
a^2=a
a^2-a=0
a=0 or a=1
we already found a=1 thus we shall use a=0
if a=0 then b=0 now this solution can be agrued based on the fact that I have consistantly seen conflicting views on this site as to weather or not 0 is a positive integer. I will leave that to discussion
if r=2 then
a=b^2
(b^2)^(b^2)=b^(b^2)
b^(2b^2)=b^(b^2)
2b^2=b^2
2=1 contradiction thus r=2 gives no solution
if r=3 then
a=b^3
(b^3)^(b^2)=b^(b^3)
3b^2=b^3
b=3 and a=27
if r=4 then
a=b^4
(b^4)^(b^2)=b^(b^4)
4b^2=b^4
b^2=4
b=2 and a=16
now we will consider all r>=5
a=b^r
(b^r)^(b^2)=b^(b^r)
r*b^2=b^r
b^(r-2)=r
(2) b=r^(1/(r-2))
now for r>=5 the right hand side of (2) is decreasing and for r>=5 we have b<=5^(1/3) and since b is integer b<=1 but we have already considered all possible values for positive b<=1 and thus all r>=5 will produce no more further unique answers.
thus
a b
0 0
1 1
27 3
16 2
are the only solutions with (0,0) being arguable based on your definition of "positive" integers
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Posted by Daniel
on 2006-07-25 19:15:34 |
analytical solution (repost)
