$a + b$, $b^{p - 1}$ coprime when $a$, $b$ coprime for odd prime $p$

Question

I was reading a proof of the theorem $$\gcd\bigg(\frac{a^p + b^p}{a + b}, a + b\bigg) \in \{1, p\}$$ where $a$, $b$ are coprime integers and $p$ is an odd prime. Using long division, we get $$\gcd(a + b, pb^{p - 1})$$ which equals $\gcd(a + b, p) \in \{1, p\}$ because $a + b$ and $b^{p - 1}$ are coprime.

I don't understand how that's true though. I tried using the Euclidean algorithm but still don't see how $\gcd(a + b, b^{p - 1}) = 1$.

$(a+b,b)=(a,b)=1$ so $(a+b,b^n)=1$ by iterating Euclid's Lemma, or by coprimes (to $a+b)$ are closed under products, or by the gcd Freshman's Dream etc as in the linked dupes . — Bill Dubuque, Dec 31 '23 at 00:07

score 1 · Accepted Answer · answered Dec 30 '23 at 23:45

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$\gcd(a+b,b^p)|\gcd(a+b,b)^p=1^p=1$

answered Dec 30 '23 at 23:45

Eric

6,348

Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque Dec 30 '23 at 23:53
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Agreed - didn’t realize it was a dupe (though I probably should have checked given the simplicity). Also, he wasn’t asking that question - he was asking for clarification on a solution. – Eric Dec 30 '23 at 23:58
We have hundreds of dupes of this. It is best not to add more (beware that continuing to answer dupes can lead to account suspension, cf. above-linked policy). – Bill Dubuque Dec 31 '23 at 00:08

$a + b$, $b^{p - 1}$ coprime when $a$, $b$ coprime for odd prime $p$

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