$a^m=b^m$ and $a^n=b^n$ imply $a=b$ 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 $a=b$?
 A: I came across a similar problem but with the condition that $m$ and $n$ are positive coprime integers. The following is my solution. 
If $a=0$, then $b^m=a^m=0$. It follows that $b=0=a$ since $D$ is an integral domain.
If $a \ne 0$, since gcd(m,n)=1, $\exists s, t\in \mathbb z $ such that $1=sm+tn$. Because $ m, n>1$, $s,t \in \mathbb z$ and $1 = sm + tn$, we have either $s>0, t<0$ or $t>0, s<0$ (Otherwise we cannot have $1=sm+tn$). 
Let's assume $s>0, t<0$. Then $-t>0$. From $a^m = b^m$, we have $(a^m)^s=(b^m)^s$. From $a^n=b^n$, we have $(a^n)^{-t}=(b^n)^{-t}$. We can factor out as follows.
$$a^{-tn}(a^{sm+tn})=b^{-tn}(b^{sm+tn})$$ 
$$a^{-tn}(a^{sm+tn})=a^{-tn}(b^{sm+tn})$$
$a^{-tn} \ne 0$ since $a \ne 0$ and $a$ is in an integral domain. We can cancel $a^{-tn}$ on both sides.Then we get $a^{sm+tn}=b^{sm+tn}$, i.e. $a^1=b^1$.
The proof for the case where $s<0, t>0$ is similar.
A: 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The pair $(m,n)$ has now been replaced by the pair $(m,n-m)$.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.
A: 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 $a=b$. 
A: Hint $ $ The set $\,S\,$ of naturals $\,k>0\,$ with $\,a^{\large k} = b^{\large k}\,$ is nonempty & closed under positive subtraction, i.e. $\, j>k \in S\,\Rightarrow\,j-k\in S\,$ (by cancelling $\,a^{\large k} = b^{\large k}\,$ from $a^{\large j} = b^{\large j}).$ Thus a fundamental lemma implies the least element $\,\ell\in S\,$ divides every element of $\,S.\,$ In particular, $\,\ell\,$ is a common divisor of the coprimes $\,m,n\in S,\,$ thus $\,\ell = 1.\,$ Hence $\,1\in S,\,$ i.e. $\, a = b.\ \ $ QED
A: Let $a^m = b^m$ $\rightarrow$ Implies that: 
$$O(a)=m \text{ and } O(b)=n \tag{1}$$       
Let $a^m = b^m$ $\rightarrow$ Implies that:
$$O(a)=n \text{ and } O(b)=n  \tag{2}$$
From (1) and (2)
$$O(a) = m = n \text{ and } O(b) = m = n$$
$\rightarrow$
Implies that:   $a^{m=n} = b^{m=n}$
Which is $a^{(m|n)} = b^{(m|n)}$ by relatively prime 
Implies that $a^1 = b^1$
Hence proved which is $a = b$. 
A: What's the necessity of $D$ to be a domain?
As $(m,n) = 1$ we have two integers $x$ and $y$ such that $mx+ny = 1$
so, $a^{mx} = b^{mx}$ and $a^{ny} = b^{ny}$ which implies, $a^{mx+ny} = b^{mx+ny} = a = b$
