Problem 1388

1. Assume all roots are rational:
Let the polynomial $ax^3 + bx^2 + cx + d$ have all rational roots, denoted by $\alpha_1, \alpha_2, \alpha_3$.

2. Apply Vieta's formulas:
By Vieta's formulas, the relationships between the coefficients and the roots are:
$$\alpha_1 + \alpha_2 + \alpha_3 = -\frac{b}{a}$$
$$\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = \frac{c}{a}$$
$$\alpha_1 \alpha_2 \alpha_3 = -\frac{d}{a}$$

3. Analyze the product of the roots:
Given $ad$ is odd, both $a$ and $d$ must be odd (since the product of two integers is odd if and only if both integers are odd). Therefore:
$$\alpha_1 \alpha_2 \alpha_3 = -\frac{d}{a}$$
Since $a$ and $d$ are odd, $-\frac{d}{a}$ is a rational number that is also odd.

4. Analyze the sum of the roots:
$$\alpha_1 + \alpha_2 + \alpha_3 = -\frac{b}{a}$$
For this to be rational, $b$ must be an integer. Given $a$ is odd, for $-\frac{b}{a}$ to be rational, $b$ must be an integer. However, we need to determine the parity of $b$.

5. Analyze the sum of the products of the roots taken two at a time:
$$\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = \frac{c}{a}$$
For this to be rational, $c$ must be an integer. Given $a$ is odd, for $\frac{c}{a}$ to be rational, $c$ must be an integer. However, we need to determine the parity of $c$.

6. **Determine the parity of $b$ and $c$:**
Since $ad$ is odd, both $a$ and $d$ are odd. From the given condition $bc$ is even, at least one of $b$ or $c$ must be even.

7. Derive a contradiction:
If all roots are rational, then:
- $\alpha_1 \alpha_2 \alpha_3 = -\frac{d}{a}$ implies $\alpha_1, \alpha_2, \alpha_3$ are rational.
- Given $ad$ is odd, both $a$ and $d$ are odd.
- From Vieta's formulas, $\alpha_1 + \alpha_2 + \alpha_3 = -\frac{b}{a}$ implies $b$ must be odd.
- From Vieta's formulas, $\alpha_1 \alpha_2 + \alpha_2 \alpha_3 + \alpha_3 \alpha_1 = \frac{c}{a}$ implies $c$ must be odd.

Therefore, both $b$ and $c$ must be odd, which contradicts the given condition that $bc$ is even.

8. Conclusion:
Since assuming all roots are rational leads to a contradiction, not all roots can be rational.

$\blacksquare$