# 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?

• Is it necessary that $x, y$ and $z$ are different?
– mopy
Sep 2, 2015 at 23:32
• "Opposite of Fermat's Last Theorem" ... Fermat's first theorem? Sep 3, 2015 at 0:47
• @Au101 Tamref's first theorem? Sep 3, 2015 at 5:16
• Did you see this in the book "Gödel, Escher, Bach"? Sep 3, 2015 at 10:10
• @Lucas He calls it Fourmi's Well-Tested Conjecture (that no nontrivial solutions exist), in the chapter Ant Fugue: "I have discovered a truly marvelous proof of this statement, which, unfortunately is so small that it would be well-nigh invisible if written in the margin." Sep 3, 2015 at 16:23

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.

• Brilliant method. Sep 3, 2015 at 21:43

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.

• Except in the case of $a=2$, in which case $z=x+1$. This is simply the fact that twice a power of two is the next power of two. Sep 3, 2015 at 4:21
• I consider $a=2$ a trivial solution. Sep 3, 2015 at 4:21
• The question explicitly asks "What does this mean for any a greater than 2".
– user81060
Sep 3, 2015 at 12:42
• @MarkBannister Which is why I consider $a=2$ a trivial solution. Sep 3, 2015 at 15:49
• @MarkBannister What if we include real (or complex) numbers as basis? Ie, which $x\in \mathbb R \ [\mathbb C]$ are solutions to $x^m - x^n = 1$ for some integer $m$, $n >2$? Does it seem like a reasonable question? Sep 4, 2015 at 11:56

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 $z=x+1$.

• It doesn't even have a solution if $a = 2$, you'd always have $1 + 1 = 1$ which is obviously wrong. Only possible solution would be the trivial case $a = 0$.
– mopy
Sep 2, 2015 at 23:36
• @Aldon Careful: $2+2=4$ :P (Or, more generally, if $x=y$, $z=x+1$, and $a=2$, then $a^x+a^y=a^z$.) Sep 2, 2015 at 23:37
• It works if a=x=y=2 and z=3 Sep 2, 2015 at 23:38
• No sorry I was meant to say $a = 1$ that's why I wrote down $1+1=1$. Sorry for the typo.
– mopy
Sep 2, 2015 at 23:44

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 $a^z$.

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 $a=0$.

• $a=1$ doesn't work, since $1+1\not=1$. Sep 3, 2015 at 2:36