Let $P(x)$ be a polynomial with integer coefficient such that $P(1) = 10$ and $P(-1) = 22$.
(a) Give an example of $P(x)$ such that $P(x) = 0$ has an integer root.
(b) Suppose that $P(0) = 4$, prove that $P(x) = 0$ does not have an integer root.
(a) We can take $P(x)=x(16x-6)$ which has a root at $0$ (this is of course motivated by part (b)).
(b) Using $x-y \mid P(x)-P(y)$ you would then get $x \mid 4$ and $x-1 \mid 10$ and $x+1 \mid 22$. Since $x \ne \pm 1$, we get that $x$ must be even and hence $x-1 \mid 5$ and $x+1 \mid 11$.
But it is easy to see that this is not possible unless $x=0$ which of course does not work by assumption.