## Composite factorial plus one

How many positive integers $$n$$ are there such that $$n! + 1$$ is composite?

[SOLVED]

#### Tanny Libman solved this puzzle:

By Wilson's theorem, $$(p - 1)! quiv -1 \pmod p$$ for any prime $$p$$. If $$n = p - 1$$, then we get $$n! + 1 quiv 0 \pmod p$$. Since $$n! + 1 \gt n + 1 = p$$ for any $$n \gt 2$$, we'll always have $$n! + 1 = pk$$ for some integer $$k \gt 1$$. That means $$n! + 1$$ is composite. Since there are infinitely many primes $$p$$, there are infinitely many values for $$p - 1 = n$$ such that $$n! + 1$$ is composite.

#### Rino Raj solved this puzzle:

For any prime $$p$$, we have $$(p - 1)! \equiv -1 \pmod p$$ by Wilson's theorem, so $$p \mid (p-1)! + 1$$. Further for $$p \geq 5$$, $$(p - 1)! + 1 \gt p$$, so $$(p - 1)! + 1$$ is composite. As the number of primes is infinite, the number of such composite cases is also infinite.

#### Mark Brader solved this puzzle:

Infinitely many, or to be more specific, $$\aleph_0$$.

Wilson's theorem states that $$p$$ is prime if and only if $$(p - 1)! + 1$$ is divisible by $$p$$. If $$x \gt 3$$, then $$(x - 1)! + 1 \gt x$$. Therefore the desired property holds for all numbers $$n \gt 4$$ where $$n - 1$$ is prime, and there are $$\aleph_0$$ of those.

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