Lemma Let $a(x) = b(x)c(x)$, where $a(x), b(x)$, and $c(x)$ have integer coefficients. If a prime number p divides every coefficient of $a(x)$, it either divides every coefficient of $b(x)$ or every coefficient of $c(x)$.
PROOF: If this is not the case, let $b_r$ be the first coefficient of $b(x)$ not divisible by $p$, and let $c_t$ be the first coefficient of $c(x)$ not divisible by $p$. Now, $a(x) = b(x) c(x)$, so
$\qquad a_{r + t} = b_0c_{r + t} +\cdots+ b_rc_t + \cdots + b_{r + t}c_0$
Each term on the right, except $b_rc_t$ is a product $b_ic_j$ where either $i > r$ or $j > t$. By our choice of $b_r$ and $c_t$, $\color{red}{\text{if }i > r\text{ then }p\,|\,b_i,\text{ and }j > t\text{ then }p\,|\,c_j}$. Thus, $p$ is a factor of every term on the right with the possible exception of $b_rc_n$ but $p$ is also a factor of $a_{r + t}$. Thus, $p$ must be a factor of $b_rc_n$ hence of either $b_r$ or $c_n$ and this is impossible.
Is there a typo at where it is highlighted with red in the proof?
The greater than sign ">" should be replaced with less than sign "<" to become:
$\qquad$ "if $i < r$ then $p\,|\,b_i$, and $j < t$ then $p\,|\,c_j$"
Correct?