Prove that if $p$ is a prime $p^p – 1$ has a prime factor $\equiv1(\mod p)$.
My approach was to write the congruence $p^p\equiv 1\pmod q$$(1)$ for some prime factor of $p^p-1$ $q$ and I noticed that $gcd(p,q) = 1$. Let $o =$ order $p$ of a modulo $m$. From $(1)$ $o\in\{1,p\}$. And here I don't know how to proceed. I see that it sufficies to prove that there exist a prime $q$ for which $o = p$ and we know that $o = p$ divides $\phi(q) = q-1$ so $q\equiv 1\pmod p$. Any help appreciated.
$p^p – 1$ has a prime factor of the form $lp+1$
elementary-number-theory
Best Answer
First work modulo $p-1$ so $N:=\sum_{j=0}^{p-1}p^j=p=1$, and fix a prime factor $q$ of $N\ge3$ (and hence of $p^p-1$), which won't be $p$ or a prime factor of $p-1$. Now work modulo $q$: since $p\ne1$ but by Fermat's little theorem $p^{\gcd\{p,\,q-1\}}=1$, the GCD is $p$ as required.