Let's assume this equation has a rational root $\frac{m}{n}$ where $r$ and $s$ are relatively prime to each other. Then $$a\Big(\frac{m}{n}\Big)^2 + b\Big(\frac{m}{n}\Big) + c = 0\implies am^2 + bmn + cn^2 = 0$$ Because $a\equiv b\equiv c \equiv 1\pmod{2}$, then $$am^2 + bmn + cn^2\equiv m^2+mn+n^2\pmod{2}$$ Meanwhile, because $m^2 +mn+n^2=0$, we must have $m^2+mn+n^2\equiv 0\pmod{2}$. However, in order to make this relation holds, both $m$ and $n$ must be even. This contradicts to the assumption that $m$ and $n$ are relatively prime.