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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.

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    $\begingroup$ $4!-1$ disproves B. So the only one left is C. $\endgroup$ – Arthur May 3 at 22:08
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Proof of (C): let $N<m<N!$ be the smallest divisor of $(N!-1)$. We argue by contradiction: if $m$ is not prime, then it has some nontrivial divisor $1<m'<m$, and thus $m'$ also divides $(N!-1)$. By minimality of $m$ it follows that $m'\le N$, but no such number can divide $(N!-1)$.

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(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 $\sqrt{40319} < 40318$.

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  • $\begingroup$ 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? $\endgroup$ – Tapi May 4 at 18:27
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    $\begingroup$ 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. $\endgroup$ – Mr. Brooks May 4 at 20:41
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    $\begingroup$ @Mr.Brooks: For the question in your comment, take mod $4$. $\endgroup$ – user21820 May 7 at 16:17
  • $\begingroup$ @user21820 Right, of course. Thank you very much. $\endgroup$ – Mr. Brooks May 9 at 23:10

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