TITLE: Find all integers $x,y,z,t$ such that $x^y+y^z+z^t = x^{2005}$
QUESTION [5 upvotes]: First I tried taking logarithm on both sides, but that just made it messy. 
Next I tried putting $x=y=z=t=k(say)$. That made the problem : 
$k^k + k^k + k^k = k^{2005}$ 
$=>3k^k=k^{2005}$
$=>3 = k^{2005-k}$ 
How can I proceed?
Even by taking every variable equal, I am not able to solve the problem.
How to solve the problem without making unnecessay assumptions ?

REPLY [2 votes]: The comments have illustrated a range of possible solutions, without anyone really getting close to demonstrating that we have covered all possible solutions. 
For a solution in positive integers, we can set $y=t=1$ reducing the equality to:
$x^1+1^z+z^1 = x^{2005} \implies x+1+z = x^{2005} $
and we have the solutions $(x,y,z,t) = (x, 1, x^{2005}-x-1, 1)$ which will be all positive for $x>1$.

Capturing other families of solutions suggested in comments:
@Math_QED : $(x,y,z,t)=(x,2005,-2005,-2005)$  
@DejanGovc : $(x,y,z,t)=(0,a^k,-a,-ka)$ ($a,k>0$ and odd)
@DejanGovc + : $(x,y,z,t)=(\pm1,a^k,-a,-ka)$ with $a,k>0$ and odd
@Dylan : $(x,y,z,t)=(\{0,1\},y,-y,-y)$