China Southeastern Mathematical Olympiad

If $\overline{abc}=p$ is a prime number, and the roots of the equation $f(x)=a{x}^2+bx+c=0$ 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 $4ap=(20a-2a{x}_1)(20a-2a{x}_2)$.

Since $x_1,x_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we see that $20a-2a{x}_1$, $20a-2a{x}_2$ are integers. Now $p\mid 20a-2a{x}_1$ or $p\mid 20a-2a{x}_2$, and we can suppose that $p\mid 20a-2a{x}_1$; then $p\le 20a-2a{x}_1$, and $4a\ge 20a-2a{x}_2$, which is a contradiction to $x_2$ being negative.