I want rational solutions

If $a$ , $b$ and $c$ are all odd integers, prove that $ax^2+bx+c=0$ can't have a rational solution.

We begin by assuming that a rational number $\frac{p}{q}$ is the solution to $ax^2+bx+c=0$ . Without loss of generality, we can assume that $p$ and $q$ are co-prime. This means $p$ and $q$ can't be even. At least one of them has to be odd.

If $\frac{p}{q}$ is indeed a solution to the quadratic, we must have:

$\begin{aligned} a\left(\frac{p}{q}\right)^2+b\frac{p}{q}+c&=0\\ ap^2+bpq+cq^2&=0. \end{aligned}$

Now the right hand side is even, so the left hand side has to be even too. But since $a$ , $b$ , $c$ are all odd, that can happen only if both $p$ and $q$ are even and we made it very clear that they aren't.

So our initial assumption was wrong.

Therefore, a rational number $\frac{p}{q}$ can't be a solution to $ax^2+bx+c=0$ if $a$ , $b$ and $c$ are all odd. $_\square$

Thus the value of $X$ is $0$ and we get $\boxed{0}$ as the answer