China Southeastern Mathematical Olympiad

We prove by contradiction. If $abc=p$ is a prime number, the rational root of quadratic equation $f(x)=a{x}^2+bx+c=0$ 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, $$4ap=(20a-2a{x}_1)(20a-2a{x}_2).$$ It is easy to see that $(20a-2a{x}_1)$ and $(20a-2a{x}_2)$ are all positive integers. Consequently, $p\mid (20a-2a{x}_1)$ or $p\mid (20a-2a{x}_2)$. If $p\mid (20a-2a{x}_1)$, then $p\le 20a-2a{x}_1$, so, $80-8x_1-10x_2+x_1{x}_2\le 0$, which contradicts to $x_1,x_2<0$. Similarly, $p\mid (20a-2a{x}_1)$ is not true. $\square$