IMC 2012 · Day 2 · P8

Consider a positive integer $n$ with $n! + 1 \mid (2012n)!$. It is well-known that for arbitrary nonnegative integers $a_1, \dots, a_k$, the number $(a_1 + \dots + a_k)!$ is divisible by $a_1! \cdot \dots \cdot a_k!$. (The number of sequences consisting of $a_1$ digits $1$, …, $a_k$ digits $k$, is $\frac{(a_1 + \dots + a_k)!}{a_1! \cdot \dots \cdot a_k!}$.) In particular, $(n!)^{2012}$ divides $(2012n)!$.

Since $n! + 1$ is co-prime with $(n!)^{2012}$, their product $(n! + 1)(n!)^{2012}$ also divides $(2012n)!$, and therefore $$(n! + 1) \cdot (n!)^{2012} \le (2012n)!.$$ By the known inequalities $\left( \frac{n+1}{e} \right)^n < n! \le n^n$, we get $$\left( \frac{n}{e} \right)^{2013n} < (n!)^{2013} < (n! + 1) \cdot (n!)^{2012} \le (2012n)! < (2012n)^{2012n}$$ $$n < 2012^{2012} e^{2013}.$$ Therefore, there are only finitely many such integers $n$.

Remark. Instead of the estimate $\left( \frac{n+1}{e} \right)^n < n!$, we may apply the Multinomial theorem: $$(x_1 + \dots + x_\ell)^N = \sum_{k_1 + \dots + k_\ell = N} \frac{N!}{k_1! \cdot \dots \cdot k_\ell!} x_1^{k_1} \dots x_\ell^{k_\ell}.$$ Applying this to $N = 2012n$, $\ell = 2012$ and $x_1 = \dots = x_\ell = 1$, $$\frac{(2012n)!}{(n!)^{2012}} < (\underbrace{1 + 1 + \dots + 1}_{2012})^{2012n} = 2012^{2012n},$$ $$n! < n! + 1 \le \frac{(2012n)!}{(n!)^{2012}} < 2012^{2012n}.$$ On the right-hand side we have a geometric progression which increases slower than the factorial on the left-hand side, so this is true only for finitely many $n$.

Assume that $n > 2012$ is an integer with $n! + 1 \mid (2012n)!$. Notice that all prime divisors of $n! + 1$ are greater than $n$, and all prime divisors of $(2012n)!$ are smaller than $2012n$.

Consider a prime $p$ with $n < p < 2012n$. Among $1, 2, \dots, 2012n$ there are $\left\lfloor \frac{2012n}{p} \right\rfloor < 2012$ numbers divisible by $p$; by $p^2 > n^2 > 2012n$, none of them is divisible by $p^2$. Therefore, the exponent of $p$ in the prime factorization of $(2012n)!$ is at most 2011. Hence, $$n! + 1 = \gcd \bigl( n! + 1, (2012n)! \bigr) < \prod_{n < p < 2012p} p^{2011}.$$ Applying the inequality $\prod\limits_{p \le X} p < 4^X$, $$\tag{2} n! < \prod_{n < p < 2012p} p^{2011} < \left( \prod_{p < 2012n} p \right)^{2011} < \bigl( 4^{2012n} \bigr)^{2011} = \bigl( 4^{2012 \cdot 2011} \bigr)^n.$$ Again, we have a factorial on the left-and side and a geometric progression on the right-hand side.