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 $P_{m}'(x) = b_{m-1} x^{m-1} + \cdots + b_{1} x + b_{0}$, for some real numbers $b_{m-1}, \ldots, b_{0}$.

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 $P_{m}'(x)$ 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 $a_{0} \neq 0$, then if $r_{1}, \ldots, r_{m}$ are the roots of $P_{m}(x)$, then by Vieta's theorem,
$$r_{1} \cdots r_{m} = (-1)^{m} a_{0}, \quad r_{1} + r_{2} + \cdots + r_{m} = 0$$
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 $k+1$.