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 $\frac pq$.

You should think about factoring the polynomial rather than finding its roots.  If $ax^2 + bx + c$ has a rational root, then its other root must also be rational, and then it factors in some way like this:
$$ax^2 + bx + c = (Ax + B)(Cx + D) \quad A,B,C,D \in \mathbb{Z}$$
Then $a = AC$ and $c = BD$, so if both are odd, then all of $A,B,C,D$ are odd.  But then we also have $b = AD + BC$, which is the sum of odd integers, and therefore is even.
