Is $n^2 + n + 1$ prime for all n? I recently stumbled across this question in a test.

Paul says that "$n^2+n+1$ is prime $\forall\:n\in \mathbb{N}$".  
  
  
*
  
*Paul is correct, because...
  
*Paul is wrong, because...
  

The answers sheet says that "all answers providing a valid counterexample, e.g. $n=4$, are valid; however, I'm interested in understanding why - is there some sort of theorem which can predict whether a polynomial is prime or not?
What I tried was:

$n^2+n+1$ can be factorized as $(n+\frac{1+\sqrt{3}i}{2})(n+\frac{1-\sqrt{3}i}{2})$. Neither of these factors is a suitable factor (i.e. is natural, different from $1$, and different from $n^2+n+1$) for any $n\in\mathbb{N}$, therefore $n^2+n+1$ must be prime $\forall\:n\in\mathbb{N}$.
  Instead, $n^2-1$ can be factorized as $(n+1)(n-1)$. The factor $(n+1)$ is a suitable factor (i.e. natural, different from $1$, and different from $n^2-1$) for $n>2$; therefore, $n^2-1$ is not prime for $n>2$.

However, the above reasoning is clearly wrong, because "$n^2+n+1$ is prime $\forall\:n\in\mathbb{N}$" doesn't hold for the case $n=4$.
What is wrong in the reasoning? Can one tell whether a polynomial $p(n)$ is prime for all $n$ a priori?

No. For $n=43$, $n^2+n+1=1893$ which is divisible by $3$.
