```Author: bandsmer (Michael Bandsmer)
Date: 2003-Oct-02 18:31:58

Find all integer solutions to the equation

x^y = y^x .

Here, "^" is the exponentiation operator.

Author: tibberio (Paolo Andreoni)
Date: 2003-Oct-13 19:01:28

I would like to propose another simplier solution

m^n=n^m  and take m>n, m-n=k, m,n,k are integer(1,n,n+1), so

m^n     m^n        1
1 = ----- = ----- * --------
n^m     n^n     n^(m-n)

so

n^k = (1+ k/n)^n , n^k is integer too, so k=R*n,with R integer, so:

n^(R*n)=(1+R)^n ---> n=(1+R)^(1/R)=f(R)

there is only one solution for these rules, because
n is integer and f(R) have solutions in [2,1[ and decreases:
R=1, n=2, m=4

that's all (I hope so ... I'm not a mathematician)

Author: bandsmer (Michael Bandsmer)
Date: 2003-Oct-14 18:46:14

This is great, tibberio!  Your observation that R must be an integer is
very helpful, since it allows us to easily show that f(R) is decreasing.
(Although you didn't formally prove that f(R) is decreasing, the calculus
to show it is fairly straightforward, now that we know R is an integer).
Good job!

Author: tibberio (Paolo Andreoni)
Date: 2003-Oct-16 09:31:44

Ok , it's too easy say that (1+R)^(1/R) is decreasing.
I think now I'm able to demostrate it.
We can look for which S (in R) is true:

f(R)=(1+R)^(1/R) > (1+S*R)^(1/S*R) ...

we can bound from below

... >= (1+R)^(1/S*R)

(1+R) >=(1+R)^(1/S)  and so S>= 1

Now  take  sequences s(i) >= 1 for all i (in N)
and R(i) (in N) where R(i+1)>=R(i)

f( R(1) ) >= f( s(1)*R(1) )
f( R(2) ) >= f( s(2)*R(2) )
.............
f( R(n) ) >= f( s(n)*R(n) )

but if I take s(i)=[R(i)+1]/R(i) we obtain that s(1)*R(1)=R(2)
and so

f( R(1) ) >= f( R(2) ) ........... >= f( R(n) )

so f(R) is or flat or like a scale
(could I think that probably it decrease ? :)).

In the last post I have found that in  m^n=n^n m>n, one solution
is n=2
so

2^m=m^2 and so m=2*t, t in N t>1 because m>n

so

2^(2*t)=(2*t)^2 ---> 2^t=2*t

t=2 is one solution, but it's the only, because for t>2
2^t is more powerfull then 2*t. [m=4]

That' all (folks!), I hope

sorry s(i)=[R(i+1)+1]/R(1)

another error s(i)=../R(i) :P

sorry, this is the last,  is a  promise!

S(i)=[R(i)+1]/R(i)=R(i+1)/R(i) OK (I hope)

I feel a little anti-zeronism here,
why is (0,0) excluded from the (x,x) solutions?

(0,0) is excluded because 0^0 is undefined.

> (0,0) is excluded because 0^0 is undefined.

I strongly disagree.
0^0 is very defined here, more precisely as being 1.
Do they leave it undefined in your college/university? Strange.

Author: bandsmer (Michael Bandsmer)
Date: 2003-Oct-17 02:29:30

Hmm... I stand corrected.  It would seem that 0^0 is normally defined as 1,
although it depends on the textbook and/or application where it's used.
See <http://mathforum.org/dr.math/faq/faq.0.to.0.power.html> for a brief
discussion on the issue, which concludes that "Consensus has recently been
built around setting the value of 0^0 = 1".  Thanks for pointing this out.
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