<s>[INST] $x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
 [/INST] Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$
Then we proceed simply this way:

$ x^y = y^x $       
$ x = y^{\frac x y } $         
$ \frac x y = y^{\frac x y -1} $       
$ \frac x y -1 = y^{\frac x y -1} - 1 $       

Now we expand the rhs into its well-known exponential-series      

$ \frac x y -1 =  \ln(y)*(\frac x y -1) + \frac {((\ln(y)*(\frac x y -1))^2}{2!} + ... $       

Here by the definition x>y the lhs is positive, so if $ \ln(y) $ >=1 we had lhs $\lt$ rhs
Thus $ \ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.

[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.    
We ask for $x=4^{1+\delta}   ,\delta > 0 $ inserting the value 2 for y:    

$ 4^{(1+\delta)*2}=2^{4^{(1+\delta)}} $      

Take log to base 2:    

$ (1+\delta)*4=4^{(1+\delta)} $      
$ \delta =4^{\delta} - 1 $      
$ \delta = \ln(4)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      
$ 0 = (\ln(4)-1)*\delta  + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $      

Because $ \ln(4)-1 >0 $ this can only be satisfied if $ \delta =0 $    
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
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