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?
 A: 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 $a^z$.
A: 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.
A: Clearly we must have $x, y<z$. Suppose $x\not=y$; WLOG, suppose $x<y$. Then this can be rewritten as $$a^x(1+a^{y-x})=a^xa^{z-x},$$ with $z-x, y-x>0$. This gives $$1+ a^{y-x}=a^{z-x};$$ but the right hand side is divisible by $a$, while the left hand side is not.
So we must have $x=y$. But then this yields $$2a^x=a^z,$$ which in turn yields $$2=a^{z-x}$$ with $z-x>0$. So there are no nontrivial solutions.
A: The equation $a^x+a^y=a^z$  has no solutions in positive integers with $a\gt 2$. For suppose the relation holds. Without loss of generality we may assume $x\le y$. Then
$$1+a^{y-x}=a^{z-x}.$$
This is only possible if $y=x$,  $a=2$, and $z=x+1$.
A: 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 $a=0$.
