TITLE: Sequence involving primes of form $n^2 + n+1$
QUESTION [5 upvotes]: Looking at prime numbers $p_i $ of the form $n^2+n+1$ and the derived expression
$$1 -  \prod_{i=1}^{j}\frac{(p_i-1)}{p_i}$$ 
it seems (I do not claim it and do not see why it should be true) that the numerators of these numbers (in lowest terms) beyond 1 may contain only prime factors that are themselves factors of numbers of the form $n^2+n+1.$ It is true for
$$1, 3, 43, 1381, 8689, 642937, 7812553,$$ 
and I cannot verify it for the next element, 1,655,913,643. (Note: the sequence $p_i$ may be finite.) Am only asking in case there is an elementary reason for this that I do not see. Thanks.  
Thanks.

REPLY [8 votes]: A prime $p$ is a factor of a number of the form $n^2+n+1$ if and only if $p \neq 2 \pmod 3$ :
$3$ is a factor of $1^2+1+1 = 3$
If $p \neq 3$, then $(n^2+n+1)(n-1)=n^3-1$, thus $p$ is a factor of some $n^2+n+1$ if and only if it is a factor of $n^3-1$ where $n \neq 1 \pmod p$, which is equivalent to the existence of a nontrivial cube root of $1$ in $\Bbb Z/p \Bbb Z$. Since $(\Bbb Z/p \Bbb Z)^*$ is a cyclic group of order $p-1$, there is one if and only if $p-1 \equiv 0 \pmod 3$.
Since $773 \equiv2 \pmod 3$, it is not a factor of a number of the form $n^2+n+1$