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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 $1$.

Bill Dubuque
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    Well, you need $n,m\neq 0$. If you have that, then let $p$ be a prime dividing $b$ and argue that the equality implies that $p$ divides $a$ as well. – lulu Jan 08 '20 at 19:04
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    You also need $|a|\neq 1$ or $|b|\neq 1$. – Arturo Magidin Jan 08 '20 at 20:18
  • "I am failing to see how this implies a common divisor bigger than 1" Well, what does the largest common divisor being less than or equal to $1$ mean? $\gcd(a,b) = 1$ means that $a$ and $b$ have no factors in common. $\gcd(a,b) > 1$ means they do have a factor in common. And $\gcd(a,b)< 1$ is not possible. So this boils down to: If $a,b$ and integers and $a^n=b^m$ then $a$ and $b$ have factors in common. Can you prove that? – fleablood Jan 08 '20 at 22:33

2 Answers2

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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 $\gcd(a,b)\geq p >1$.

ViHdzP
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nonuser
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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 $|a|> 1)$

Bill Dubuque
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