TAOCP 4.6.2 Exercise 33
The statement is false.
Section 4.6.2: Factorization of Polynomials
Exercise 33. [M18] True or false: If $u(x) \ne 0$ and the complete factorization of $u(x)$ modulo $p$ is $p_1(x)^{e_1} \cdots p_r(x)^{e_r}$, then $u(x)/\gcd(u(x), u'(x)) = p_1(x) \cdots p_r(x)$.
Verified: yes
Solve time: 1m20s
Solution
The statement is false.
Let
$$ u(x)=\prod_{i=1}^r p_i(x)^{e_i}, $$
where the $p_i(x)$ are distinct monic irreducible polynomials and $e_i\ge1$.
By differentiating,
$$ u'(x) = \sum_{i=1}^r e_i,p_i(x)^{e_i-1}p_i'(x) \prod_{j\ne i} p_j(x)^{e_j}. $$
Fix an index $i$. Since $p_i(x)$ is irreducible, there are two possibilities.
If $p\nmid e_i$, then the coefficient $e_i$ is nonzero modulo $p$. Furthermore, because $p_i(x)$ is irreducible, $p_i(x)\mid p_i'(x)$ would imply $p_i'(x)=0$, which can occur only when every exponent of $p_i(x)$ is divisible by $p$. In that case $p_i(x)$ would be reducible over the algebraic closure, hence cannot occur as a squarefree irreducible factor. Therefore $p_i(x)\nmid p_i'(x)$. Examining the displayed formula for $u'(x)$, the exact power of $p_i(x)$ dividing $u'(x)$ is then $e_i-1$.
If $p\mid e_i$ and $p_i'(x)\ne0$, the same argument shows that the exact power of $p_i(x)$ dividing $u'(x)$ is again $e_i-1$.
However, if $p_i'(x)=0$, then every exponent occurring in $p_i(x)$ is a multiple of $p$, and every term of $u'(x)$ contains at least $p_i(x)^{e_i}$. In this case the exact power of $p_i(x)$ dividing $u'(x)$ is at least $e_i$.
Consequently,
$$ \gcd(u,u') = \prod_{i=1}^r p_i(x)^{f_i}, $$
where
$$ f_i= \begin{cases} e_i-1,& p_i'(x)\ne0,\[2mm] e_i,& p_i'(x)=0. \end{cases} $$
Hence
$$ \frac{u(x)}{\gcd(u(x),u'(x))} = \prod_{p_i'(x)\ne0} p_i(x). $$
The factors satisfying $p_i'(x)=0$ disappear completely. Therefore the quotient need not equal $p_1(x)\cdots p_r(x)$.
A counterexample modulo $2$ is
$$ u(x)=x^2=(x)^2. $$
Here
$$ u'(x)=2x=0, $$
so
$$ \gcd(u,u')=\gcd(x^2,0)=x^2. $$
Therefore
$$ \frac{u(x)}{\gcd(u(x),u'(x))}=1, $$
while
$$ p_1(x)\cdots p_r(x)=x. $$
Since $1\ne x$, the asserted identity fails.
Thus the statement is
$$ \boxed{\text{False}.} $$
∎