Let $p(x) \in \mathbb{Z}[x], p(2) = p(-2) = 2, p(100) < 0$. Prove that $p(100) < -1000$

Question

I have the following task:

Let $p(x) \in \mathbb{Z}[x], p(2) = p(-2) = 2, p(100) < 0$. Prove that $p(100) < -1000$

I'v tried to make some substitutions, to factorize the polynomial, but didn't succeed and out of ideas now. Can you give me some hints on what to try?

Answer 1

Hint:

Use the fact that for any integers $x,y$ we have $$x-y\mid p(x)-p(y)$$

and mark $p(100)=a$. From a condition you get $a<0$ and ...

Answer 2

.If $q(x) = p(x) - 2$, then $q(-2) = q(2) = 0$, so $q$ is a multiple of both $x-2$ and $x+2$, which are coprime hence $q$ is a multiple of $x^2 - 4$.

Consequently, $q(x) = (x^2 - 4)h(x)$ for some polynomial $h$ with integer coefficients. Substituting, $q(100) = 9996 \times h(100)$, so $p(100) = 2 + 9996 \times h(100)$.

Note that $h(100)$ is an integer, so $p(100) = 2 + 9996k$, for some integer $k$. If $p(100) < 0$, then $2 + 9996k < 0$, so $k < \dfrac{-2}{9996}$. Since $k$ is an integer, $k \leq -1$ , since this is the greatest integer smaller than $\dfrac {-2}{9996}$. Therefore, $p(100) \leq 2 + 9996(-1) \leq -9994$.

It follows that $p(100) \leq -9994 < -1000$.