Prove that if all $a,b,c$ are odd integers, then $ax^2+bx+c=0$ has no rational roots.

Question

My thinking was that if I divided the entire equation by $a$, then the roots would have to sum to $b/a$ and multiply to $c/a$. We can assume for the case that a divides b and c. This means that $c/a$ would be an odd integer, because an odd divided by an odd is still odd. Therefore, the two roots would have to be odd, and they would sum to an even integer b, which is a contradiction. I am not sure how to prove the case where a doesn't divide b or c.

Answer 1

Suppose $p/q$ is a rational root, with $p,q$ coprime. Now $p\mid c$ and $q\mid a$, i.e., they are odd. Hence $$ ap^2+bpq+cq^2=0 $$ which is impossible, since an odd sum of odd integers is odd.

Answer 2

Hint: Multiply the equation by $a$ and put $y=ax$ to obtain $y^2+by+ac=0$

Now $y$ must be an integer.

Answer 3

If $\frac{x}{y}$ is a rational root in lowest terms, so $x$ and $y$ are not both even, then $$ax^2+bxy+cy^2=0.$$ Show this is impossible since the left side is odd.

Answer 4

The roots are $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, which are rational if and only if $b^2-4ac$ is the square of an integer. However, since $b$ is odd and $ac$ is odd, $$b^2-4ac\equiv 1-4\equiv5\pmod 8$$

And $5$ is not a quadratic residue modulo $8$.