Can you just assume without proof for coprime bases?

Question

Let $x,y ∈ \mathbb{Z}$ and $a,b ∈ \mathbb{Z+}$, where $a\ne b$ and a,b are coprime

Let $a^x=b^y$

The only solution is when $x = 0$ and $y = 0$, but is it necessary to prove it?

I've got as far as saying:



$a^x$ has factors $a^n$, where $1 \le n \le x$

Likewise, $b^y$ has factors $b^m$, where $1 \le m \le y$

As $a,b$ are coprime, so are $a^n$ and $b^n$. In order for $a^x=b^y$, the "answer" ($a^x$ and $b^y$) must have:

(factors $a^n$) $\cap$ (factors $b^m$). These are mutually exclusive conditions, so it's impossible

Hence, $a^x$ and $b^y$ both cannot have factors $a$ or $b$. This only happens when $x, y = 0$

Is this sufficient, or indeed necessary? Thanks for any help

Answer 1

That is not correct. You have no reason to assume that $a$ or $b$ is a prime number.

Besides, the statement is clearly false if $a=1$ or $b=1$.

Otherwise, let $p$ be a prime factor of $a$. Then $p$ is also a prime factor of $a^x$ and, if $a^x=b^y$, then $p$ will also be a prime factor of $b^y$, if $y>0$. But then it's a prime factor of $b$. This is impossible, since $a$ and $b$ are coprime. So, $y=0$. By the same argument, if $p$ is a prime factor of $b$ and $a^x=b^y$, then $x=0$. Since both $x$ and $y$ must have prime factors, this proves that $x=y=0$.