Prove $ax^2+bx+c=0$ has no rational roots if $a,b,c$ are odd If $a,b,c$ are odd, how can we prove that $ax^2+bx+c=0$ has no rational roots?
I was unable to proceed beyond this: Roots are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and rational numbers are of the form $\frac pq$.

Suppose that $a,b,c$ are odd. $ax^2+bx+c=0$ has rational roots iff the discriminant is the square of an integer. That is, there is an integer $d$ so that $d^2=b^2-4ac$. Since $a,b,c$ are odd, $d$ must also be odd.
Note that the right hand side of
$$
(b-d)(b+d)=4ac\tag{1}
$$
has exactly two factors of $2$. However, since $b$ and $d$ are both odd, $2d\equiv2\pmod{4}$ and so one of $b-d$ and $b+d$ is $0\bmod{4}$ and the other is $2\bmod{4}$. Thus, the left hand side  of $(1)$ has at least three factors of $2$. Contradiction.
