<s>[INST] 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 $\frac pq$.
 [/INST] $x= \dfrac{-b+ \sqrt{b^2-4ac}}{2a} $ or $\dfrac{-b- \sqrt{b^2-4ac}}{2a}$
If $x=\dfrac{p}{q} \implies b^2-4ac=k^2$, and $k$ is odd. ($odd-even=odd$).
Considering $a,b,c$ odd. 
$k^2 \equiv 1 \mod 8$
$b^2 \equiv 1 \mod 8$
$4ac \equiv 4 \mod 8$
$b^2-4ac=-3 \mod 8 \implies 5 \mod 8$, a contradiction.(Either of $a$ or $c$ has to be even), since $k^2 \equiv 1 \mod 8$
$\therefore$ You don't get rational roots when all are odd.
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