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.



When $p(x)^2 \mid f(g(x))$?
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 $f\_0^{k+1}$.


