Problem 1588

1. Let $W(x)$ be a fourth-degree polynomial with integer coefficients, and let $x_1, x_2, x_3, x_4$ be its roots. By Vieta's formulas, we know:
$$x_1 + x_2 + x_3 + x_4 \in \mathbb{Q}$$
$$x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 \in \mathbb{Q}$$
$$x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 \in \mathbb{Q}$$
$$x_1x_2x_3x_4 \in \mathbb{Q}$$

2. Given that $x_3 + x_4 \in \mathbb{Q}$ and $x_3x_4 \in \mathbb{R} \setminus \mathbb{Q}$, we need to show that $x_1 + x_2 = x_3 + x_4$.

3. From Vieta's formulas, we have:
$$x_1 + x_2 + x_3 + x_4 \in \mathbb{Q}$$
Since $x_3 + x_4 \in \mathbb{Q}$, it follows that:
$$x_1 + x_2 = (x_1 + x_2 + x_3 + x_4) - (x_3 + x_4) \in \mathbb{Q}$$

4. Next, consider the product of the roots:
$$x_1x_2x_3x_4 \in \mathbb{Q}$$
Given $x_3x_4 \in \mathbb{R} \setminus \mathbb{Q}$, it follows that:
$$x_1x_2 = \frac{x_1x_2x_3x_4}{x_3x_4} \in \mathbb{R} \setminus \mathbb{Q}$$
because the product of a rational number and an irrational number is irrational.

5. Now, consider the sum of the products of the roots taken two at a time:
$$x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 \in \mathbb{Q}$$
Since $x_3x_4 \in \mathbb{R} \setminus \mathbb{Q}$ and $x_1x_2 \in \mathbb{R} \setminus \mathbb{Q}$, the remaining terms must sum to a rational number:
$$x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 \in \mathbb{Q}$$

6. We also have:
$$(x_1 + x_2)(x_3 + x_4) = x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 + x_1x_2$$
Since $x_1 + x_2 \in \mathbb{Q}$ and $x_3 + x_4 \in \mathbb{Q}$, it follows that:
$$(x_1 + x_2)(x_3 + x_4) \in \mathbb{Q}$$
Therefore:
$$x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 + x_1x_2 \in \mathbb{Q}$$

7. Subtracting the rational terms $x_3x_4 + x_1x_2$ from both sides, we get:
$$x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 \in \mathbb{Q}$$

8. Since $x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 \in \mathbb{Q}$ and $x_1x_2 \in \mathbb{R} \setminus \mathbb{Q}$, the only way for the sum to be rational is if:
$$x_1 + x_2 = x_3 + x_4$$

$\blacksquare$