Problem 986

To prove the following statement: If $f$ can be factored into the product of two integer-coefficient polynomials $\varphi$ and $\psi$
$$f(x) = \varphi(x) \psi(x),$$
then one of $\varphi$ and $\psi$ has a degree $\geqslant m$, and this polynomial (for the same $p$ and $m$) satisfies similar conditions (i), (ii), and (iii).

Let us prove this statement. Suppose
$$\begin{array}{l} \varphi(x) = a_{r} x^{r} + a_{r-1} x^{r-1} + \cdots + a_{0}, \\ \psi(x) = b_{s} x^{s} + b_{s-1} x^{s-1} + \cdots + b_{0}. \end{array}$$

Since $p \mid c_{0} = a_{0} b_{0}$ and $p^{2} \nmid c_{0} = a_{0} b_{0}$, we can assume without loss of generality that $p \mid a_{0}$, $p^{2} \nmid a_{0}$, and $p \nmid b_{0}$. Also, since $p \nmid c_{n} = a_{r} b_{s}$ and $p \nmid a_{r}$, we can assume without loss of generality that $a_{k}$ is the first among $a_{0}, a_{1}, \cdots, a_{r}$ that is not divisible by $p$. Consider
$$c_{k} = a_{k} b_{0} + a_{k-1} b_{1} + \cdots + a_{0} b_{k}.$$

We know that $p \nmid c_{k}$, hence $k \geqslant m$. We confirm that the polynomial $\varphi$ satisfies the following conditions:
(i) $p \nmid a_{r}$,
(ii) $p \mid a_{j} \ (j=0,1, \cdots, m-1)$,
(iii) $p^{2} \nmid a_{0}$.

If $\varphi$ is irreducible, then the conclusion of the theorem is already proven. Otherwise, we can repeat a similar discussion until we obtain an irreducible factor with a degree $\geqslant m$.