0

Find all solutions to $1!\times 3! \times \cdots \times (2n -1)!=m!$

I have found $(n,m)=(1,1),(2,3),(3,6),(4,10)$

How can one prove there are no more? Also is it a coincidence that these are the triangle numbers? If not, why is this so? I cannot work it out.

2 Answers2

2

The following is to convince you that there cannot be really large $m$ and $n$:

Let $p$ be the biggest prime on the left hand side; for $p$ sufficiently large, the left hand side will be much bigger than $2p!$(You can verify this inductively by simply noting that $n! (n-2)! (n-4)! > (2n)!$ when $n$ gets too large since left side grows in $O(n^3)$ and right side grows in $O(n^2)$.) However, by Bertrand's postulate, there exists a prime between $p$ and $2p$ bigger than $p$, (i.e. the right hand side has a prime bigger then $p$) but then, the biggest prime on the left cannot be $p$ if this is to be an equality.

You -hopefully- should be able to run a simulation for the small values of $n$ before you enter the "large" region.

As for the triangle numbers showing up, I do not know.

E-A
  • 6,075
  • 1
  • 11
  • 21
1

Following on E-A's answer you have them all. For $n=5$ the left side is close to $15!$, but $15!$ has factors of $11,13$. For $n=6$ the left side is close to $21!$, but $21!$ has factors $13,17,19$. For $n=7$ the left side is close to $28!$ and we are already into the regime that Bertrand settles the issue.

Ross Millikan
  • 383,099
  • Wow, I did not realize it was that close; thanks for completing the answer. – E-A Jun 21 '18 at 21:20
  • 1
    @E-A: If you want large numbers, factorials are a fast way to get them. We are piling up a bunch of factorials on the left. – Ross Millikan Jun 21 '18 at 21:24