Prove fact about polynomial with coefficients from $\mathbb{Z}$ Let $f \in \mathbb{Z}[x]$. 
And for more than 3 (i.e $\geqslant 4$) distinct $a \in \mathbb{Z}\ f(a) = 1$. 
Prove that $\forall a \in \mathbb{Z} \ f(a)\neq -1$.
I have clearly no idea how to tackle this. 
The only thing I've noticed (I'm pretty sure it is an obvious fact for the most of you) is that since polynomial is continuous function we have to prove that $f(a) > -1\  \forall a \in \mathbb{Z}$.

Let 
$$
f(x)=(x-1)(x-3)(x-5)(x-7)+1.
$$
Then $f(a)=1$ for $a=1,3,5,7$. Nevertheless we have $f(6)=-14<-1$.
As for the first claim, compare with this question, for an idea how to tackle this.
