TITLE: Opposite of Fermat's Last Theorem?
QUESTION [14 upvotes]: 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?

REPLY [7 votes]: Informally, but intuitively:
Let's say $a = 3$. Then, we're looking at the sequence $\{1, 3, 9, 27, 81\}$. The numbers are clearly too far apart for adding two of them to reach the next one.
This leads us to a slightly more formal proof:
Let's say that $x \leq y$. Then, $$a^x + a^y \leq a^y + a^y = 2a^y < a(a^y) = a^{y+1}$$ So $z$ must be between $y$ and $y + 1$, so there's no possible solution.
We can also note that $$\sum_{k=1}^{k=z-1}{a^k} = {a^z-1\over a-1} < a^z$$
Meaning, even if we added all the exponents of $a$ less than $z$, we still wouldn't reach $a^z$.