monic irreducible polynomial in $\mathbb Z[x]$ have a multiple root in $F_p$ over $F_p[x]$ for some prime $p$ Could you help me with the following problem which I cannot solve ?

Let $f\in \mathbb Z[x]$ be a monic irreducible polynomial with $deg(f)>1$.
Prove or disprove that there is always a prime $p$ satisfying the following condition:
$\overline{f}$ have a multiple root in $F_p$
where $\overline{f}$ is a polynomial in $F_{p}[x]$, whose coefficients are the reduction of those of $f$ modulo $p$.


From Minkowski's bound, |disc$(f)|>1$.
So we can take a prime $p$ dividing  |disc$(f)|$.
Then $\overline{f}$  have a multiple root in algebraic closure of $F_p$.
However, this does not mean $\overline{f}$ have a multiple root in $F_p$.
For example, $f=x^4 + 2x^3 + 3x^2 + 2x + 6$ and $p=5$.
In this case, disc$(f)$$=35600$ and $\overline{f}=(x^2+x+1)^2$ over $F_p$.
(On the other hand,$\overline{f}=x^2(x+1)^2$ over $F_2$)

The statement is actually false.
Consider the polynomial $f=\Phi_{12}=X^4-X^2+1$. It's straightforward to see that $\Phi_{12}$ divides $X^{12}-1$. The latter polynomial is coprime with its derivative in $\mathbb{F}_p$ for $p > 3$. So we need to check that $\Phi_{12}$ has no double root in $\mathbb{F}_p$ for $p=2$ and $p=3$.
For $p=2$, $\overline{f}(x) = (x^2+x+1)^2$ so we're done. For $p=3$, $\overline{f}(x)=(x^2+1)^2$ and we're also done.
