TITLE: Is this number prime?
QUESTION [14 upvotes]: Prove of disprove the following:  If $m,n$ are positive integers, where $n=m^2+m+1$, and $n$ has no prime divisors smaller than $m+1-\sqrt{m}$, then $n$ is prime.  

I've been playing around with this problem for quite some time now.  I'm convinced there is a very large counterexample.  A basic number theory fact is that if a number $n$ has no prime divisors smaller than $\sqrt{n}$ then it must be prime.  For this problem $\sqrt{n}=\sqrt{m^2+m+1}$.  Now there is a gap between $m+1-\sqrt{m}$ and $\sqrt{m^2+m+1}$ which is small for small $m$ but becomes larger for bigger $m$.  This is why I believe there is a large counter example.  Therefore, I need to find primes $x_1,x_2, \cdots x_n \in \mathbb{N}$ such that $$\prod_{i=1}^n x_i=m^2+m+1$$ for some $m \in \mathbb{N}$ and, 
$$x_i \in (m+1-\sqrt{m} , \sqrt{m^2+m+1})$$
Or I need to prove that no such numbers can exist in which I am unsure how to proceed.
Any help or suggestions are appreciated, thanks!
EDIT:  I've consulted with my professor while he would not give any hints he did confirm that the interval should in fact be an open interval.

REPLY [5 votes]: If you let $m=a^2$ then $n = (m+1)^2 - m = (a^2+1)^2 - a^2 = (a^2-a+1)(a^2+a+1).$
There are lots of choices for $a$ which make these last two factors simultaneously prime.  E.g., $a= 2, 3, 6, 15, 21, \ldots.$  Both factors are in your range, if the range is a closed interval.  For $a=6$, $m=36$ and $n=31\cdot 43.$  The factor $31$ is right on the boundary of your interval.