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Let $\operatorname{char}\mathbb K = 0$. Consider the polynomial $$q(x) := (x - a_1)^{n_1}...(x - a_m)^{n_m} \in \mathbb K[x]$$ and the polynomial $$Q(x) := q(x)(x - a_1)...(x - a_m) \in \mathbb K[x],$$ where $n_i \in \mathbb N \setminus\{0\}$ and $a_i \neq a_j, i \neq j$. Now consider $$f(x) := \frac{Q^{'}}{q} \in \mathbb K[x].$$ Is it true that all roots of $f$ are simple(of multiplicity 1)?

I can show that this is true for $m \leq 3$. I'm doing this by directly writing down $f(x)$ and calculating it's discriminant.

Any help is welcome

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By my reckoning, this is true only when $\mathbb K$ is an ordered field. If you can give nonzero numbers $w_j\in\mathbb K$ with $\sum \beta_jw_j=\sum\beta_jw_j^2=0$ (here $\beta_j$ are integers $\ge 2$), then you will have an example where it fails. You can achieve this, for example, by taking $\beta_j=2$ for all $j=1,2,3$ and $w_j$ the cube roots of unity in $\Bbb C$. [So, I don't believe you have a proof for $m=3$.] However, in an ordered field (e.g., $\Bbb R$), there can be only the trivial solution of the second equation.

Here's how you get to this. Set $q(x) = \prod_{i=1}^m (x-a_i)^{\gamma_i-1}$ (with $\gamma_i\ge 0$ integers) and $Q(x) = \prod_{i=1}^m (x-a_i)^{\gamma_i}$. You set $f = Q'/q$. By logarithmic differentiation (or not), $$\frac{Q'(x)}{Q(x)} = \sum \frac{\gamma_i}{x-a_i},$$ so $$f(x)=\sum \gamma_i \prod_{j\ne i} (x-a_j) = \underbrace{\prod_j (x-a_j)}_{g(x)} \underbrace{\sum \frac{\gamma_i}{x-a_i}}_{h(x)}.$$ Now, suppose $f(w)=0$. We want to know if we can also have $f'(w)=0$. Note that $g(w)\ne 0$, since $w$ cannot be one of the $a_i$, and so we must have $h(w)=0$. By the product rule, $f'(w)=g(w)h'(w)$, so $f'(w)=0 \iff h'(w)=0$. Note that $$h(w) = \sum\frac{\gamma_i}{w-a_i} \quad\text{and}\quad h'(w)=-\sum\frac{\gamma_i}{(w-a_i)^2}.$$ Since $\gamma_i>0$, there can be no solutions of $h'(w)=0$ in an ordered field, but, as I pointed out at the beginning, we can certainly have nontrivial solutions to both equations, e.g., in $\Bbb C$.

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