Prove that this polynomial is irreducible over $\mathbb Z$ I want to prove that the following polynomial is irreducible: $$x^3 - x^2 - x + 3$$
My question gives the hint to apply the substitution $x \mapsto x+1$ but I've tried this and when multiplied out I'm getting $x^3 + x^2 +2$. I tried this mod 2 but came out with a reducible polynomial ($x^3 + x^2$ which can be written $(x^2)(x+1)$). 
Can anyone tell me what I'm missing? I know that irreducible mod prime number implies irreducible in $\mathbb Z$, but am I wrong in thinking it works the other way i.e. reducible mod prime means reducible in $\mathbb Z$?
 A: The thing is that with $f(x)=x^3-x^2-x+3$ you get
$$
f(x+1)=x^3+2x^2+2
$$
and we are good...
A: You didn't try enough finite fields. 
If you passes $x^3 + x^2 +2$ in the field $F_3$ you get a polynommial which doesn't have any root, hence is irreducible.
Since it's irreducible in $F_3$ it must be irreducible also in $\mathbb Z$.
Addressing the second part of the question: is not generally true that if a polynomial is reducible $\text{mod }p$ then is reducible in $\mathbb Z$. 
By the way it's true the following if $f$ can be factor in a $\mathbb Z/p \mathbb Z$ with enough big $p$ then $f$ factors in $\mathbb Z$.
Here by enough big we mean that $p$ must be such that if $g,h$ are the polynomial such that $f=gh$ in $\mathbb Z/p \mathbb Z[X]$ then it must be that $p$ is bigger than any sum of the product of the coefficients of $g$ and $h$. That's is basically saying that coefficient of the polynomials doesn't get truncated $\text{mod } p$ while performing the product of the polynomial in $\mathbb Z/p \mathbb Z$ and so the factorization works in $\mathbb Z$ as well.
