Odd zero

Let the roots be $\alpha$ , $\alpha + 2$ . Using Vieta's relation

$\begin{cases} b = (\alpha) + (\alpha + 2) \\ a = (\alpha)(\alpha + 2) \end{cases}$

\[\begin{align} b^2 - 4a &= (2\alpha + 2)^2 - 4(\alpha)(\alpha + 2) \\ &= 4\alpha^2 + 8\alpha +4 - ( 4\alpha^2 + 8 \alpha) \\ &= \boxed{4}

\end{align}\]

Regardless of whether $\alpha$ is odd or not does not matter. Thus $b^2 - 4a = \boxed{4}$ is true for any two roots which have positive difference 2

The two roots are $\dfrac {b \pm \sqrt{b^2-4a}}2$ . Since the two roots have a difference of $2$ , then $\sqrt{b^2-4a} = 2 \implies b^2 - 4a = \boxed 4$ .

Let the roots of the equation be $2n-1$ and $2n+1$ respectively, where $n$ is an integer. Then

$2n-1+2n+1=b\implies n=\dfrac{b}{4}$ .

And $(2n-1)(2n+1)=a\implies 4n^2-1=a$

$\implies 4\times \dfrac{b^2}{16}=a+1$

$\implies b^2-4a=\boxed 4$

The same result appears for two consecutive even roots also

For the two roots to be consecutive integers, the relation is $b^2-4a=1$ .