Why does $b^{n}=a^{m} \implies \gcd(a,b) > 1$? For $a,b, m, n \in \mathbb{Z} $ ($n,m\neq 0$).
I am failing to see how this implies a common divisor bigger than $1$.
 A: Say a prime $p\mid b$. Then, $p\mid b^n$ and $$p\mid a ^m=a\cdot a\cdot a\cdot\ldots \cdot a.$$ By Euclid’s lemma, $p$ divides at least one factor. This implies $p\mid a$, so that $\gcd(a,b)\geq p >1$.
A: It fails if $\,|a|=1\, (\!\!\iff |b| = 1)\,$ so we must add that hypothesis. We give $4$ proofs below.

With Bezout: $\ {\rm if}\ \ \gcd(a,b)=1\ \ {\rm then}\ \  1^{\phantom{|^{|^{|^|}}}}\!\!\!\!\! = \smash{\overbrace{j\, b^n + k\, a^m = (j\!+\!k)a^m}^{\textstyle b^n = a^m\!\!}}\Rightarrow\, |a| = 1$

With primes: $\,|a|>1\,$ so some prime $\,p\mid a\mid a^m = b^n\,$ so $\,p\mid b^n\,\Rightarrow\, p\mid b,\, $ by existence and uniqueness of prime factorizations, so $\,p\mid a,b,\,$ so $\,\gcd(a,b) > 1$

With gcds: by here $\,1=\gcd(a,b)\,\Rightarrow\, 1 = \gcd(b^n,a^m) = |a|^m,\ $ by $\,b^n = a^m$.

With RRT: wlog $\,m\ge n\,$ so $\,x = b/a\,$ is a root of $\, x^m = b^{m-n}\in\Bbb Z\,$ hence $\,b/a\in \Bbb Z\,$ by RRT = Rational Root Test, so $\,\gcd(a,b)>1\,$ (else reduced $\,b/a\not\in\Bbb Z\,$ by $|a|> 1)$
