Elementary Number Theory Problems 3.2 Solution (David M. Burton's 7th Edition) - Q9

Background

All theorems, corollaries, and definitions listed in the book's order:

I will only use theorems or facts that are proved before this question. So you will not see that I quote theorems or facts from the later chapters.

Question

(a) Prove that if $n \gt 2$, then there exists a prime $p$ satisfying $n \lt p \lt n!$.
[Hint: If $n! - 1$ is not prime, then it has a prime divisor $p$; and $p \leq n$ implies $p \mid n!$, leading to a contradiction.]
(b) For $n \gt 1$, show that every prime divisor of $n! + 1$ is an odd integer that is greater than $n$.

Solution

(a)

Let us consider the integer $n! - 1$.

If $n! - 1$ is a prime, we are done since $n \lt n! - 1 \lt n!$. If $n! - 1$ is not a prime, using Theorem 3.2, $n! - 1$ is divisible by some prime $q$.