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 $f(a) > -1\ \forall a \in \mathbb{Z}$.
Supose $f$ is a polynomial with coefficients in $\mathbb{Z}$ such that $f(a_1) = f(a_2) = f(a_3) = f(a_4) = 1$ where $a_1 < \dots < a_4 \in \mathbb{Z}$. Assume that $f(c) = -1$ for some integer $c$.
Set $g(x) = f(x) - 1$, then $g(a_i) = 0$, and also $g(c) = -2$.
Consequently we may factor $g(x) = (x-a_1)(x-a_2)(x-a_3)(x-a_4)h(x)$ where $h$ is another polynomial. Now $h$ has integer coefficients as well, since it is obtained by repeated applications of the Euclidean Algorithm. Substitute $x = c$ to obtain $$ -2 = (c-a_1)(c-a_2)(c-a_3)(c-a_4)h(c) $$ and all factors are integers. The first four factors are all different. That is impossible since their product would be at least 4 in absolute value.
Note that $f(x) = x^3 - 7x^2 + 14x - 7$ indeed satisfies $f(1) = f(2) = f(4) = 1$ and $f(3) = -1$. So there have to be at least four $a_i$ with $f(a_i) = 1$.
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.
