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Ajay Chauhan Reasoning Book Pdf In Hindi Free 43
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Jawaharlal Nehru University, New Delhi, India.Q:
Number of prime divisors of $p^n+1$
Let $p$ be prime and let $n \in \mathbb{N}$.
Find the number of prime divisors of $p^n+1$.
I thought to divide $p^n+1$ by $p-1$ and apply Lucas’ Theorem, but I don’t think that helps me.
A:
There are only 2 types of primes $p$ satisfying the condition $p^n \equiv 1 \mod p-1$ for a natural $n$.
1, $p^n \equiv 1 \mod p-1 \Rightarrow p^{n+1} \equiv 1 \mod p-1$,
$p^{n+1} – 1 \equiv 0 \mod p-1 \Rightarrow p^{n+1} \equiv 1 \mod p-1$,
where $\gcd(p^{n+1} – 1, p-1) = 1$,
2, $p^n \equiv -1 \mod p-1$ and $p \equiv 2 \mod 4$,
$p^n + 1 \equiv -1 \mod p-1 \Rightarrow p^{n+1} + 1 \equiv 1 \mod p-1$,
$p^{n+1} + 1 \equiv 2 \mod p-1 \Rightarrow p^{n+1} \equiv 2 \mod p-1$,
where $\gcd(p^{n+1} + 1, p-1) = 2$.
Assuming $n$ is not a multiple of $3$, we get $5$ primes of the second kind.
5, $n = 2$,
$p = 2^3, 5^2 + 1 = 3^3 + 1 = (2^3 – 1)(2^3 + 2)$
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