SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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

Assume that $a,b$ are two distinct elements of $K$ and $r_1,r_2,\dots ,r_i$ are the roots of $P(x)-a$ with multiplicity $k_1,k_2,\dots ,k_i$, respectively. Then $r_1,r_2,\dots ,r_i$ are also the roots of $Q(x)-a$. Let $t_1,t_2,\dots ,t_j$ be the roots of $P(x)-b$ with multiplicity $s_1,s_2,\dots ,s_j$, respectively. Then $t_1,t_2,\dots ,t_j$ are also the roots of $Q(x)-b$.

Assume that $\mathrm{deg}P(x)\ge \mathrm{deg}Q(x)$, and $R(x)=P(x)-Q(x)$ is not identically zero. Thus, $\mathrm{deg}P(x)=k_1+\cdots +k_i=s_1+\cdots +s_j\ge \mathrm{deg}R(x)\ge i+j$, then we can get $$\begin{array}{rl}\mathrm{deg}{P}^{\prime }(x)& \ge (k_1-1)+\cdots +(k_i-1)+(s_1-1)+\cdots +(s_j-1)\\ & =(k_1+\cdots +k_i)+(s_1+\cdots +s_j)-(i+j)\ge \mathrm{deg}P(x),\end{array}$$ a contradiction. $\square$