A divisibility problem

Quote:

Originally Posted by eyke

Show that if $(a,b)=1$ and $p$ is an odd prime, then

$( a+b, (a^p+b^p)/a+b ) = 1$ or $p$

$\frac{a^p+b^p}{a+b}=\sum_{j=1}^p (-1)^{j-1} a^{p-j}b^{j-1}= \sum_{j=1}^p (-1)^{j-1} (a+b \ - \ b)^{p-j}b^{j-1} \equiv pb^{p-1} \mod a+b.$ similarly $\frac{a^p+b^p}{a+b} \equiv pa^{p-1} \mod a+b.$ so if $d \mid a+b$ and $d \mid \frac{a^p + b^p}{a+b},$ then $d \mid pa^{p-1}$ and $d \mid pb^{p-1}.$

thus $d \mid p \gcd(a^{p-1},b^{p-1})=p. \ \Box$