<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] Consider a quadratic equation of the form $a\cdot x^2 + b\cdot x + c = 0$. The only way, it can have rational roots IFF there exist two integers $\alpha$ and $\beta$ such that
$$\alpha \cdot \beta = a\cdot c\tag1$$
$$\alpha + \beta = b\tag2$$
$$
Explanation\left\{ \begin{align}
if\,\alpha\cdot \beta &= a\cdot c,\\
\frac{\alpha}{a} &= \frac{c}{\beta}\\ 
a\cdot x^2 + b\cdot x + c& = a\cdot x^2 + (\alpha + \beta)\cdot x + c\\
& = a\cdot x^2 + \alpha\cdot x + \beta\cdot x + c\\ 
& =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{c}{\beta}))\\ 
& =a\cdot x(x + \frac{\alpha}{a}) + \beta\cdot (x + \frac{\alpha}{a}))\\ 
& =(x+\frac{\alpha}{a})\cdot (a\cdot x + \beta)\\
&\text {As a Quadratic equation has only two roots,}\\
&\text {there would be no other way to factorize the equation}
\end{align}\right.
$$
$$\text{Reason }\alpha,\beta\in\mathbb{Z}\begin{cases}
\text{Given a Ring R, with two operations }\left\{⋅,+\right\},\text{ on }\mathbb{Q}\\
\text{and if } \alpha,\beta \in \mathbb{Q},\alpha\cdot \beta \in \mathbb{Z},\alpha+\beta\in\mathbb{Z}\\ 
\Rightarrow\alpha,\beta \in \mathbb{Z}\\
\text{ where } \mathbb{Z}\subset\mathbb{Q}
\end{cases}
$$
If $(a,b,c)$ are odd then $\alpha \cdot \beta$ is odd and $\alpha + \beta$ is odd, but you cannot have two integers whose product and sum are odd.
So by contradiction we prove that the equation cannot have rational roots 
</s>