Saudi Arabian Mathematical Competitions

Denote $x_{1} < x_{2} < \ldots < x_{n}$ as the roots of $P(x)$. Let $d = x_{n} - x_{n-1} = \cdots = x_{2} - x_{1} > 0$. Since $P(x)$ has rational coefficients then by applying Vieta's theorem, we have
$$\sum_{i=1}^{n} x_{i} \in \mathbb{Q} \text{ and } \sum_{1 \leq i < j \leq n} x_{i} x_{j} \in \mathbb{Q} .$$
Note that $\sum_{i=1}^{n} x_{i} = \frac{n(x_{1} + x_{n})}{2}$ so $x_{1} + x_{n} \in \mathbb{Q}$. On the other hand,
$$\sum_{i=1}^{n} x_{i}^{2} = \left(\sum_{i=1}^{n} x_{i}\right)^{2} - 2 \sum_{1 \leq i < j \leq n} x_{i} x_{j} \in \mathbb{Q}$$
and
$$\begin{aligned} \sum_{i=1}^{n} x_{i}^{2} & = \sum_{i=1}^{n} \left(x_{1} + (i-1)d\right)^{2} = n x_{1}^{2} + n(n-1) x_{1} d + \frac{n(n-1)(2n-1)}{6} d^{2} \\ & = n\left(x_{1} + \frac{(n-1)d}{2}\right)^{2} + \frac{n^{2}(n^{2}-1)}{12} d^{2} \\ & = \frac{n(x_{1} + x_{n})^{2}}{4} + \frac{n(n^{2}-1)}{12} d^{2} \end{aligned}$$
From these, we can conclude that $\frac{n(n^{2}-1)}{12} d^{2} \in \mathbb{Q}$ so $d^{2} \in \mathbb{Q}$. Thus
$$x_{1} x_{n} = \frac{(x_{1} + x_{n})^{2} - (x_{n} - x_{1})^{2}}{4} = \frac{(x_{1} + x_{n})^{2} - (n-1)^{2} d^{2}}{4} \in \mathbb{Q} .$$
Therefore, $x_{1} + x_{n}, x_{1} x_{n} \in \mathbb{Q}$ which implies that two numbers $x_{1}, x_{n}$ satisfy.