SAUDI ARABIAN MATHEMATICAL COMPETITIONS

We will show that $a_k=a_{k-1}=a_{k-2}=\cdots =a_0=0$ by induction on $n$, the degree of $P(x)$.

In fact, we may assume that the leading coefficient of $P(x)$ is $1$. For $n=1,2$, the result follows immediately.

Assume that the induction hypothesis is true for every $n<m$, we shall prove it is also true for $n=m$. Denote by $P_m(x)=x^m+\cdots +a_{1}x+a_0$, and for some $k<m$, $a_k=a_{k-1}=0$.

By taking derivative of $P_m(x)$, we obtain $P_m^{\prime }(x)=b_{m-1}{x}^{m-1}+\cdots +b_{1}x+b_0$, for some real numbers $b_{m-1},\dots ,b_0$.

Since $a_k=a_{k-1}=0$, we conclude that $b_{k-1}=b_{k-2}=0$.

This together with $P_m(x)$ has only real roots, implies that $P_m^{\prime }(x)$ also has only real roots.

Hence, by induction hypothesis, we get $b_{k-3}=\cdots =b_0=0$. In other words, $a_{k-2}=\cdots =a_1=0$. It remains to show that $a_0=0$. Assume that $a_0\ne 0$, then if $r_1,\dots ,r_m$ are the roots of $P_m(x)$, then by Vieta's theorem, $$r_1\cdots r_m=(-1{)}^m{a}_0,r_1+r_2+\cdots +r_m=0$$ and $$\sum _{i,j}\frac{1}{r_i{r}_j}=0\Rightarrow \sum _i\frac{1}{r_i^2}=0$$ a contradiction. Therefore, $a_0=0$, the induction process is completed.

Obviously from that, we get $0$ is the root of $P(x)$ with multiplicity at least $k+1$.