Prove the divisibility of the polynomial $x f'(x) - n f(x)$

Question

Given an $n$-th order polynomial $f(x)$, prove or falsify that there exists an $n$-th order polynomial $g(x)$ such that (n is integer) $$ [xf'(x)-mf(x)]\ |\ g(f(x)) $$ where $m$ is also an integer. This statement is correct when $m,n<4$, for example with $f(x)=ax^2+bx+c$ and $m=1$, $$ xf'(x)-mf(x)=ax^2-c $$ while for $g(x)=x^2+4ax+b^2-4ac$, $$ g(f(x))=(ax^2-c)(ax^2+2bx+3c+4a+\frac{b^2}a) $$ but I still don't know if it could be extended for all integers.

Answer 1

Let the roots of $p(x) = x f'(x) - m f(x)$ be $x_j \big|_{j=1,2,\ldots,k}$ with multiplicities $n_j$ so that $\sum_{j=1}^k n_j = n$. Define the $n^{th}$ degree polynomial $g(x) = \prod_{j=1}^k \big(x - f(x_j)\big)^{n_j}$, then $g\big(f(x)\big) = \prod_{j=1}^k \big(f(x) - f(x_j)\big)^{n_j}$ has $x_j$ as a root with multiplicity $\ge n_j$, so $p(x) \mid g\big(f(x)\big)$.

While the above only proves the existence, $g(x)$ is symmetric in the roots of $p(x)$, so it can be effectively constructed in terms of the coefficients of $p(x)$ i.e. without determining the actual roots.

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[ EDIT ] $\;$For a fully worked out example, suppose $f(x) = x^2+bx+c$ and $m=1$, then $p(x) = x f'(x) - m f(x)$ $= x^2 - c$. The roots of $p(x)$ are $x_{1,2} = \pm\sqrt{c}$ of course, but we'll pretend not to know that and only use the symmetric functions $\,x_1+x_2=0\,$ and $x_1x_2=-c$. Then:

$$ \begin{align} g(x) &= \big(x-f(x_1)\big)\big(x-f(x_2)\big) \\ &= \big(x-x_1^2-bx_1-c\big)\big(x-x_2^2-bx_2-c\big) \\ &= x^2 -(x_1^2+bx_1+c+x_2^2+bx_2+c) x \\ &\quad\quad + (x_1^2+bx_1+c)(x_2^2+bx_2+c) \\ &= x^2 -\big((x_1+x_2)^2 - 2x_1x_2 + b(x_1+x_2) + 2c\big) \\ &\quad\quad + x_1^2x_2^2+bx_1x_2(x_1+x_2)+c(x_1^2+x_2^2)+b^2x_1x_2+bc(x_1+x_2)+c^2 \\ &= x^2 - (2c + 2c)x + c^2 + 2c^2-b^2c+c^2 \\ &= x^2 - 4 c x + 4c^2 - b^2c \end{align} $$

It can be easily verified that:

$$ \begin{align} g\big(f(x)\big) &= (x^2+bx+c)^2 - 4c(x^2+bx+c)+ 4c^2 - b^2c \\ &= x^4 + 2 b x^3 + (b^2 - 2 c)x^2 - 2 b c x - b^2 c + c^2 \\ &= (x^2-c)(x^2 + 2 b x + b^2 - c) \end{align} $$