1

Problem: Find the sum of all positive integers $n$ such that $n!+2$ divides $(2n)!$.

My approach: At first, I tried if at some point $n!+2 > (2n)!$. But $(2n)!$ grows massively. Then, I tried some values and got $2$ and $3$ works. Claim: There is no $n>3$ works. I tried to prove it but couldn't. Induction shouldn't work here because $1$ doesn't work and by domino effect, $2$ wouldn't work too.$n!|(2n)!$ and $n!+2|(2n)!$. GCD($n!,n!+2$)=$2$. So, $n!(n!+2)|2(2n)!$. I tried to bound it but it didn't work.

Conclusion: I think my claim is correct but I can't prove it. How can I? Any suggestions are appreciated.

Geometry99
  • 377

1 Answers1

2

Since $(1! + 2) \nmid 2!$, we know that $n! + 2$ is even for all such $n$. As you noted, $\gcd(n!+2,n!) = 2$ then. And in the comments it was mentioned that $(n!)^2 \mid (2n)!$. Thus it follows that \begin{equation} \frac{n!+2}{2} \mathrel{\bigg\vert} \binom{2n}{n}\,, \end{equation} in particular \begin{equation} \frac{n!+2}{2} \leqslant \binom{2n}{n} < 2^{2n}\,. \end{equation} It is easy to obtain a small upper bound for $n$ from this inequality, and one quickly finds the sum of all such $n$ to be $5 = 2+3$.

Dermot Craddock
  • 2,341
  • 3
  • 15
  • What should be the intuition? – Geometry99 Jun 20 '25 at 21:03
  • The intuition for what? The bound? The intuition for that is that $n!$ grows much faster than $4^n$, so the left hand side will soon overtake the right hand side. That leaves you with only a handful of $n$ to check. Quite possibly you have already checked them all when you came to conjecture that no $n > 3$ works. – Dermot Craddock Jun 20 '25 at 21:09
  • Intuition for the whole solution. How did you struggle and what should be the intuition to get $(n!)^2|(2n)!$? I mean, I know the proof but it has to come to my mind. I always try to bound in these types of math. But never used binomial coefficients here. In other words, how do I know I have to use binomial coefficients? – Geometry99 Jun 21 '25 at 07:32