TITLE: When $p(x)^2 \mid f(g(x))$?
QUESTION [5 upvotes]: Let $f(x),g(x),p(x)$ be non-constant polynomials with rational coefficients.
Is it true that for all $f$ exist $g,p$ such that $p(x)^2 \mid f(g(x))$?
Partial results:
$f(g(x))$ is divisible by square iff the discriminant of $f(g(x))$ is zero.
For variables $z_i$, write $g_0(x)=\sum_{i=0}^n z_i x^i$.
Then the discriminant of $f(g_0(x))$ is polynomial $D$ in variables $z_i$.
Solution $D=0$ will find $p(x)$, but finding points on variety is hard.
If we allow $g,p$ to be with coefficients algebraic integers, the solution
is easy: in $D$ fix all but one variables say $z_0$ to be integers and work in the
number field with defining polynomial $D(z_0)$.
As corollary to positive solution, we have $f(g(x))$ reducible.

REPLY [15 votes]: Yes, it is true.
Let $f_0$ be an irreducible divisor of $f$. It suffices to find $g$ such that $f_0^2$ divides $f_0(g(x))$ (which, in turn, divides, $f(g(x))$).
Try to choose $g(x)=x+h(x)f_0(x)$. Then $f_0(g(x))=f_0(x+h(x)f_0(x))\equiv f_0(x)+f_0'(x)h(x)f_0(x) \pmod {f_0^2(x)}$, and we need $1+f_0'(x)h(x)$ to be divisible by $f_0$. Since $f_0'$ and $f_0$ are coprime (as $f_0$ is irreducible and $0\leqslant \deg f_0'<\deg f_0$), such $h$ exists.
Actually by the same reasoning once we found $g_k$ for which $f_0^k$ divides $f_0(g_k)$, where $k\geqslant 1$, we may find $h$ such that $f_0(g_k+f_0^kh)$ is divisible by $f_0^{k+1}$.