Opposite of Fermat's Last Theorem? So Wiles' proof showed that no three positive integers $a$, $b$, and $c$ can solve the equation $a^n+b^n=c^n$ for any integer value of $n$ greater than $2$. Now what about the opposite?

What does this mean for any $a$ greater than $2$, and $x$, $y$ and $z$ are positive integers in the equation $a^x+a^y=a^z$. Is there any relation? Is it solvable?
Here's a less algebraic way to interpret your question: Look at things in base $a$. Then the equation $a^x+a^y=a^z$ looks like $1 \ldots 0 + 1 \ldots 0 = 1\ldots0$. (where the number of zeros is $x$, $y$, or $z$) This can only work if we're in base $2$ and the addition carries to produce a $1$ in the next digit.
It is clear that $1+a^{y-x}=a^{z-x}$ where $a=1$,$x=y=z$ is an obvious solution. On the other hand the minimun difference of powers of integers is of the forme $2k+1$ which corresponds to squares.So, (making abstraction of the previous division by zero!) this would gives $a=0$.