Let $f,h \in \mathbb{C}[t]$ be two polynomials satisfying the following conditions:
(1) $h$ is separable, namely, $\gcd(h,h')=1$.
(2) $f+f'$ is a multiple of $h$.
(3) $\gcd(f,h)=1$. Moreover, $h'f=1+Ah$, for some $A \in \mathbb{C}[t]$.
(4) $\gcd(f',h)=1$. Moreover, $h'f'=-1+Bh$, for some $B \in \mathbb{C}[t]$.
Should $h$ be linear? I do not think this should be true, although in the few examples I have found, it is.
Non-example: $f=x^3+x^2+x+1$. $f'=3x^2+2x+1$. $f+f'=h=x^3+4x^2+3x+2$. $h'=3x^2+8x+3$. But $h'f\neq 1+Ah$, since $h'f=3x^5+11x^4+14x^3+14x^2+11x+3=h(3x^2-x+9)+(-25x^2-14x-15)$.
Example: $f=x^3+x^2-x+1$. $f'=3x^2+2x-1$. $f+f'=x^3+4x^2+x$. $h=x$. $h'=1$. All conditions are satisfied, and $h$ is linear.
Thank you very much!
No, try $h = t^2-t$ and $f = -4t^3 + 5t^2 + t - 1$.
If I have calculated this right, then $h$ is separable, $$f+f' = h \cdot (-4t-11),$$ $f$ and $f'$ do not have roots at $0$ or $1$ and so are coprime to $h$, and
$$\begin{align*} h'f &= 1 + h \cdot (-8t^2 + 6t + 3), \\ h'f' &= -1 + h \cdot (-24t+8). \end{align*}$$