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If $p$ is an odd prime and $a,b$ are relatively prime integers, prove that : $$\text{gcd}\bigg(a+b, \frac{a^p+b^p}{a+b}\bigg)=1 \ \text{or} \ p$$

Since it's obligatory to show the attempts to avoid the closing of the posts, here's mine :

First of all I tried to simplify :

Factoring gives : $$\frac{a^p+b^p}{a+b}=a^{p-1}-a^{p-2}+a^{p-3}-a^{p-4}+\dots-a+1$$ To calculate $\text{gcd}$, we can reduce it, using the fact that $p$ is an odd prime, since it is odd, $p-1$ is even, hence : \begin{align} \frac{a^p+b^p}{a+b}&=a^{p-1}-a^{p-2}+\dots+a^{2x}-a^{2x-1}+\dots-a+1 \\\ &=(-1)^{p-1}-(-1)^{p-2}+\dots+(-1)^{2x}-a+1 \ (\text{mod} \ a+1) \\\ &=1+1+1\dots+1 \equiv p(\text{mod} \ a+1) \end{align} Thus by the Euclidean algorithm : $$\text{gcd}\bigg(a+b, \frac{a^p+b^p}{a+b}\bigg)=\text{gcd}(a+1,p)$$

And here where I'm stuck, I thought that I'm supposed to rearrange it but, unfortunately I got nothing, Any suggestions ? I'll be really thankful !

11PkA_Fon
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  • https://math.stackexchange.com/questions/3100848/if-gcda-b-1-then-gcd-big-fracapbpab-ab-big-1-unless-pab?noredirect=1&lq=1 https://math.stackexchange.com/questions/247146/show-that-gcd-left-fracan-bna-b-a-b-right-gcdn-dn-1-a-b – lab bhattacharjee May 11 '20 at 17:13

1 Answers1

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Hint :

Use a similar factorization : $$\frac{a^p+b^p}{a+b}=a^{p-1}-a^{p-2}b+a^{p-3}b^2+\dots -ab^{p-2}+b^{p-1}$$

euler_med
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