# Factorials and prime numbers

Let $$N$$ be a positive integer not equal to $$1$$. Then note that none of the numbers $$2, 3, \ldots, N$$ is a divisor of $$N! - 1$$. From this we can conclude that:

(A) $$N! – 1$$ is a prime number;

(B) at least one of the numbers $$N + 1, N + 2, \ldots, N! – 2$$ is a divisor of $$N! - 1$$;

(C) the smallest number between $$N$$ and $$N!$$ which is a divisor of $$N! - 1$$, is a prime number.

My working: $$N! - 1$$ is not necessarily a prime, as $$5! - 1 = 119$$ and $$7 \mid 119$$. I cannot proceed further. Hints are appreciated.

• $4!-1$ disproves B. So the only one left is C. – Arthur May 3 at 22:08

Proof of (C): let $$N be the smallest divisor of $$(N!-1)$$. We argue by contradiction: if $$m$$ is not prime, then it has some nontrivial divisor $$1, and thus $$m'$$ also divides $$(N!-1)$$. By minimality of $$m$$ it follows that $$m'\le N$$, but no such number can divide $$(N!-1)$$.
(A) is a premature conclusion. If $$N! - 1$$ is composite, its least prime factor could be as large as $$\sqrt{N! - 1}$$. And if $$N > 4$$, then $$\sqrt{N! - 1} > N$$, as in, for example, $$\sqrt{119} \approx 10.9 > 5$$.
(B) contradicts (A), but it seems like it could be a more useful conclusion if we adjust it to exclude the "factorial primes". However, since $$N! - 2 > \sqrt{N! - 1}$$ by a rapidly widening margin, it turns out to be overkill, e.g., $$8! - 1 = 40319 = 23 \times 1753$$, and clearly $$\sqrt{40319} < 40318$$.
• You're right about (B). However, if $N! - 1$ is composite, shouldn't its greatest prime factor be as large as (N! - 1)^1/2? I also didn't get what the second sentence under (A) is trying to imply. Please explain? – Tapi May 4 at 18:27
• Forgive the butting in, Mr. Soupe. If $N! - 1$ is composite, its least prime factor is less than or equal to $\sqrt{N! - 1}$, and its greatest prime factor is greater than $\sqrt{N! - 1}$. $5! - 1$ is a good example of this: $119 = 7 \times 17$ and $7 < \sqrt{119} < 17$. Though I admit I'm not certain that there are numbers of the form $N! - 1$ that are also perfect squares. – Mr. Brooks May 4 at 20:41
• @Mr.Brooks: For the question in your comment, take mod $4$. – user21820 May 7 at 16:17