Problem 1835

First we see that if the polynomial $P(x)-a$ has a root $\alpha$ with multiplicity $k$, then $P'(x)$ also has the root $\alpha$ with multiplicity $k-1$.

Assume that $a, b$ are two distinct elements of $K$ and $r_{1}, r_{2}, \ldots, r_{i}$ are the roots of $P(x)-a$ with multiplicity $k_{1}, k_{2}, \ldots, k_{i}$, respectively.
Then $r_{1}, r_{2}, \ldots, r_{i}$ are also the roots of $Q(x)-a$.
Let $t_{1}, t_{2}, \ldots, t_{j}$ be the roots of $P(x)-b$ with multiplicity $s_{1}, s_{2}, \ldots, s_{j}$, respectively.
Then $t_{1}, t_{2}, \ldots, t_{j}$ are also the roots of $Q(x)-b$.

Assume that $\deg P(x) \geq \deg Q(x)$, and $R(x) = P(x) - Q(x)$ is not identically zero. Thus, $\deg P(x) = k_{1} + \cdots + k_{i} = s_{1} + \cdots + s_{j} \geq \deg R(x) \geq i + j$, then we can get
$$\begin{aligned} \deg P'(x) & \geq (k_{1} - 1) + \cdots + (k_{i} - 1) + (s_{1} - 1) + \cdots + (s_{j} - 1) \\ & = (k_{1} + \cdots + k_{i}) + (s_{1} + \cdots + s_{j}) - (i + j) \geq \deg P(x), \end{aligned}$$
a contradiction. $\square$