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I don't know how to prove the following result.

Let $p$ be a prime number and let $a,b \in \mathbb Z$ such that $\gcd(a,b)=1$

Then $\gcd (a-b,\frac{a^p-b^p}{a-b}) = 1 $ or $ p $

(gcd should be $1$)

I know that $\dfrac{a^p-b^p}{a-b} = a^{p-1}+a^{p-2}b+\cdots $

I don't know what to do next...

Thanks for your help.

Michael Hardy
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Gabriel Romon
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1 Answers1

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Note that $a^m b^n + (a-b)a^mb^{n-1} = a^{m+1}b^{n-1}$. By applying this repeatedly we get $$\gcd(a-b,a^{p-1}+a^{p-2}b+\cdots+b^{p-1}) = \gcd(a-b,pa^{p-1}).$$ Since $\gcd(a,b)=1$, also $\gcd(a-b,a^{p-1})=1$ and hence $$\gcd(a-b,pa^{p-1}) = \gcd(a-b,p),$$ which is $1$ or $p$, since $p$ is prime.

Abel
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