Seems terribly difficult and nice

harazi

Are there infinitely many numbers n such that $2^n-1$ divides $(n+1)!$ ?

**Posted:** Thu Aug 26, 2004 3:20 am

harazi

The answer is no. Schinzel has proved that for $n>12$ the largest prime factor of $2^n-1$ is greater than or equal to 2n+1.

**Posted:** Sat Apr 02, 2005 10:59 pm

nntrkien · Hodge Conjecture Offline Joined: 24 Apr 2004 Posts: 60

Can you post a solution, Harazi ? It is interesting !

**Posted:** Sun Apr 03, 2005 12:41 am

_________________
Nguyen Trung Kien

harazi

I don't have the article where he proved this result, unfortunately. I just read about this result yesterday. Hope that someone can provide the article.

**Posted:** Sun Apr 03, 2005 1:02 am