Problem 5

Suppose $f$ is a proper factor of $P(x)-p$, so $P(x)-p=f(x)g(x)$ with $f,g$ non-constant. By Gauss's lemma there are integer-coefficient $f_1,g_1$, scalar multiples of $f,g$, with $P(x)-p=f_1(x)g_1(x)$; since $f_1(0)g_1(0)=P(0)-p=-p$ and $p$ is prime, we may take $f_1(0)=\pm1$. If every root of $f_1$ had absolute value greater than $1$, then $|f_1(0)|$ would equal the leading coefficient of $f_1$ (an integer of absolute value $\ge1$) times the product of root magnitudes, which exceeds $1$; this contradicts $|f_1(0)|=1$. So $f_1$, and hence $P(x)-p$, has a root $r$ with $|r|\le1$.