Why Isn't The Answer 10000?

Let the polynomial $f(x) \in \mathbb{Z}[x]$ be such that $f(x) = 2$ has $k$ (which here is $N_f$ ) integer roots, or equivalently $f(x) - 2 = (x-a_1)(x-a_2) \cdots (x-a_k)$ with $a_i \in \mathbb{Z}$ .

Now let $G(x) = f(x) + 2 = (x-a_1)(x-a_2) \cdots (x-a_k) + 4 = 0$ for some $x \in \mathbb{Z}$ . Then $(x-a_1)(x-a_2) \cdots (x-a_k) = -4$ .

Each of $x-a_i$ is integral (and distinct), whose product is $-4$ . Clearly to maximize $N_f \times M_f$ , we can take 2 strategies: (letting none to be zero)

• For $k = N_f$ to be maximum, we need as many factors of $-4$ as possible, which can at most be $\{ \pm 1, \pm 2\}$ , so the number of linear terms can be atmost 4. For a particular $x$ , we can get these factors one from each linear term, say for $x=1$ and $f(x)=-x(x-2)(x-3)(x+1)$ . But this having fixed, $f$ is not satisfied by any other integer. The same is true if we maximize $M_f$ .

• If $k=1$ , by similar reasoning $M_f \leq 4$ . If $k=2$ , say $f= x(x-5)$ , $G$ is a degree $2$ polynomial having atmost $2$ roots. Here they are $1$ and $4$ , or equivalently $M_f \leq 2$ . If $N_f=3$ , say $f= (x-a_1)(x-a_2)(x-a_3)= (1)(-1)(4) = (1)(2)(-2)$ . Order being immaterial here, in both cases, $G$ has atmost one root, which doesn't give a maximum case.

Thus we conclude $N_f \times M_f \leq \boxed{4}$ .