Proving that, for $l$, $m$, $n$ odd integers, $\sqrt{l^2-4mn}$ is never rational

Question

So the question I was solving was:

Prove that if $l$,$m$ & $n$ are odd then the line $lx+my+n=0$ will never intersect the parabola $y=x^2$ at a rational point.

I tried to solve the question and arrived at the condition that for this to be true $\sqrt{l^2-4mn}$ should never be rational. How do I prove this?

Answer 1

Since $l^2-4mn$ would have to be an odd square, i.e. $1$ more than a multiple of $8$ just like $l^2$, $mn$ would be even, a contradiction.

Answer 2

The system of equations $$\begin{cases} my + lx + n = 0 \\ y = x^2 \end{cases}$$ Can be simplified as $$mx^2 + lx + n = 0$$

As the right hand side is divisible by 2, the left hand side must be too.

But $$mx^2 + lx + n \equiv x^2 + x + 1 \equiv x + x + 1 \equiv 1 \pmod{2}$$ as $m,l,n$ are odd.