China Southeastern Mathematical Olympiad

If $\overline{abc} = p$ is a prime number, and the roots of the equation $f(x) = ax^2 + bx + c = 0$ are rational numbers, then $b^2 - 4ac$ is a square number, $x_1, x_2$ are negative, and $f(x) = a(x - x_1)(x - x_2)$.

So $p = f(10) = a(10 - x_1)(10 - x_2)$, and we get $4ap = (20a - 2ax_1)(20a - 2ax_2)$.

Since $x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we see that $20a - 2ax_1$, $20a - 2ax_2$ are integers. Now $p \mid 20a - 2ax_1$ or $p \mid 20a - 2ax_2$, and we can suppose that $p \mid 20a - 2ax_1$; then $p \le 20a - 2ax_1$, and $4a \ge 20a - 2ax_2$, which is a contradiction to $x_2$ being negative.

We prove by contradiction. If $abc = p$ is a prime number, the rational root of quadratic equation $f(x) = ax^2 + bx + c = 0$ is $x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Obviously, $b^2 - 4ac$ is a perfect square number, and $x_1, x_2$ are all negative, and

$$f(x) = a(x - x_1)(x - x_2).$$

Thus,
$$p = f(10) = a(10 - x_1)(10 - x_2).$$
So,
$$4ap = (20a - 2ax_1)(20a - 2ax_2).$$
It is easy to see that $(20a - 2ax_1)$ and $(20a - 2ax_2)$ are all positive integers. Consequently, $p \mid (20a - 2ax_1)$ or $p \mid (20a - 2ax_2)$. If $p \mid (20a - 2ax_1)$, then $p \le 20a - 2ax_1$, so, $80 - 8x_1 - 10x_2 + x_1x_2 \le 0$, which contradicts to $x_1, x_2 < 0$. Similarly, $p \mid (20a - 2ax_1)$ is not true. $\square$