For every positive integer $n$, $n^2 + n +19$ is prime I'm trying to prove that for every positive integer $n, n^2 + n +19$ is prime. 
I tried to disapprove it saying that is is not prime.
If it's not prime, then $n^2 + n +19$ has to have at least two factors which are greater than $1$. I'm stuck here. 
I'm not sure if I'm going in the right direction.
 A: Let's try some values:
$n=1$
$$1^2+1+19=21=3*7$$
$n=2$
$$2^2+2+19=25=5*5$$
$n=5$
$$5^2+5+19=49=7*7$$
Even one counter example was enough to prove that the statement is false.
A: Are you sure you didn't forget to mention some other condition that makes this a much more difficult problem, such as for example that $n \not\equiv 1 \pmod 3$?
If $n \equiv 1 \pmod 3$ then $n^2 \equiv 1 \pmod 3$, and since $19$ also satisfies that congruence, $n^2 + n + 19$ is a multiple of $3$.
Spend a few minutes looking at this page from FactorDB.com.
A: I answered a similar question once.
"An Introduction To The Theory Of Numbers" by Hardy, Theorem 21, page 18.

THEOREM 21. No polynomial $f(n)$ with integral coefficients, not a
  constant, can be prime for all $n$, or for all sufficiently large $n$.

A: This proposition is not true, and more generally it is not true if 19 is replaced by any positive integer $k$.  For $k>1$ we can obtain a counterexample by putting $n=k-1$ so that:
$$n^2 + n + k = (k-1)^2+(k-1)+k= (k^2-2k+1)+(k-1)+k=k^2$$
For $k=1$ a counterexample is $n=4$ implying $n^2+n+1 = 21 = 3*7$.
