$a^m=b^m$ and $a^n=b^n$ imply $a=b$

Question

Let $D$ be an integral domain and let $a^m=b^m$ and $a^n=b^n$ where $m$ and $n$ are relatively prime integers, $a,b \in D$.

How do I show $a=b$?

Answer 1

No generality is lost by supposing $m < n$. So $a^n=b^n$ implies $a^{m+(n-m)}=b^{m+(n-m)}$, or $a^ma^{n-m}=b^mb^{n-m}$. In integral domains, there's a cancellation property, so $a^{n-m}=b^{n-m}$.

The pair $(m,n)$ has now been replaced by the pair $(m,n-m)$. If you iterate that process, replacing the pair you've got with the pair consisting of the larger of the two and the difference---the larger minus the smaller, that's Euclid's algorithm. It ends when you reach the gcd.

Answer 2

Since $m$ and $n$ are coprime then $x m - y n=1$ for some $x,y \in \mathbb Z$. The equality $a^m=b^m$ implies that $a^{xm}=b^{xm}$ so $a^{1+yn}=b^{1+yn}$ which implies that : $$(*)\;\;\; \;a a^{yn}=b b^{yn}$$ Since $a^{n}= b^{n}$ then $a^{yn}=b^{yn}\not = 0$ so we can cancel $a^{yn}$ and $b^{yn}$ from both sides of $(*)$ since we are working in an integral domain, and then we get $a=b$.