I am given that $x^3+px+q\in\mathbb Z[x]$ is irreducible, and I need to show that the discriminant, which I'm given is equal to $-4p^3-27q^2$, cannot be $0$ or $\pm 1$.
Now, since it's irreducible, the discriminant can't be $0$ since irreducible polynomials over $\mathbb Z$ have distinct complex roots. I can rule out $D=-1$ by looking mod $4$, since it reduces to $q^2$. Similarly, I can rule out $D=1$ by looking mod $9$, since this forces $p^3\equiv 2$ mod $9$, which is again impossible. However, since this problem was from the section talking about the Galois Groups of the cubic and quartic, I was expecting the fact that $D$ was a square to be involved. Is there another way to rule out either of the nonzero cases?
if $|D| = 1$ then the extension obtained by adding the roots of $P$ is unramified except possibly at infinity. But every extension of $\Bbb Q$ is ramified at some finite prime, so $P$ has to split over $\Bbb Z$.
Moreover, you can't find three integers that are distance $1$ apart with each other, so it is impossible to have $|-4p^3-27q^2| = 1$ with $p,q \in \Bbb Z$