Let $a \in - 2 + 5\mathbb{Z}$. Show that the polynomial $f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) - a$ doesn't have zeroes in $\mathbb{Z}$.
Hint says to prove directly, or by using projection $\mathbb{Z} \to {\mathbb{Z}_5}$ of the polynomial coefficients.
I must admit that I have no idea how to show that, either directly, or via projection. Resulting polynomial is $g\left( x \right) = {x^5} - x - 2$, and it obviously doesn't have zeroes in $\mathbb{Z}$, but I don't know how to, from that fact, justify that $f$ doesn't have any either.
Can this problem be solved within the framework of the field extension theory? Even if not, I'd appreciate a more elaborate hint.