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Number theory problem #1

MountEvariste

Well-known member
Jun 29, 2017
79
1. For $(a,b) = 1$ and prime $p\ne 2$, prove that $\displaystyle \left(a+b, \frac{a^p+b^p}{a+b}\right) = 1$ or $p$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
1. For $(a,b) = 1$ and prime $p\ne 2$, prove that $\displaystyle \left(a+b, \frac{a^p+b^p}{a+b}\right) = 1$ or $p$.
If $q$ is a prime divisor of $a+b$ then $q$ cannot divide $a$ or $b$ (because $a$ and $b$ are coprime).

If $q$ is also a divisor of $\dfrac{a^p+b^p}{a+b}$ then $q$ divides $$\begin{aligned}\dfrac{a^p+b^p}{a+b} &= a^{p-1} - a^{p-2}b + a^{p-3}b^2 - \ldots + b^{p-1} \\ &= (a+b)\bigl(a^{p-2} -2a^{p-3}b + 3a^{p-4}b^2 - \ldots - (p-1)b^{p-2}\bigr) + pb^{p-1} .\end{aligned}$$ Therefore $q$ divides $pb^{p-1}$. But $q$ dnes not divide $b$, so it follows that $q$ divides $p$. Hence the only possible prime divisors of $a+b$ and $\dfrac{a^p+b^p}{a+b}$ (and consequently the only possible common divisors) are $1$ and $p$.