Belarusian Mathematical Olympiad

Since $abc$ is the decimal representation of a three-digit prime number $p$, we have $a\ne 0$, $c\ne 0$. Suppose, contrary to our claim, that there exists a rational root of the equation $a{x}^2+bx+c=0$. Then the discriminant of this equation is a perfect square, i.e. $b^2-4ac=n^2$, where $n$ is a positive integer number, $n<b$. Multiplying the equality $p=100a+10b+c$ by $4a$ we have $$4np=400a^2+40ub+4ac=400a^2+40ub+b^2-n^2.$$ Thus $4np=(20a+b{)}^2-n^2=(20a+b+n)(20a+b-n)$. The numbers $20a+b+n$ and $20a+b-n$ have the same parity and since their product is divisible by 4 both of $10a+(b+n)/2$ are positive integers. Since their product is divisible by the prime number $10a+(b+n)/2\le 10a+b\le 99<p$. This contradiction proves the required statement.