we obtain $Uu=a$, $Uv+Vu=b$, and $Vv=c$. Therefore $\Delta = b^2 - 4ac =  (Uv+Vu)^2 - 4 (Uu)(Vv) =(Uv-Vu)^2$, which is a square, as desired.

To prove (ii), suppose that $\Delta = b^2 - 4ac$ is a square, say $\Delta = \delta^2$, and that $a$ is a unit. Letting $\alpha = \left(-b+\delta\right)\left(2a\right)^{-1}$ and $\beta = \left(-b-\delta\right)\left(2a\right)^{-1}$, we can easily see that $H=a(x-\alpha)(x-\beta),$ and hence is reducible, as desired.

To prove (iii), suppose that $H$ is a square, say $H=(ux+v)^2$, where $u, v \in A$. Then $a = u^2$, $b=2uv$, and $c=v^2$. Thus $b^2-4ac=0$.

To prove (iv), suppose that $2a$ is a unit and that $a$ is a square, say $a=u^2$, and $b^2-4ac=0$. Then it is easy to check that $H=\left(ux+ b (2u)^{-1} \right)^2,$ a square. $\Box$

[ *The assumption that $2a$ is a unit in (ii) and (iv) above cannot be replaced by the weaker assumption that $a$ is a unit. For (ii), consider the example $A = {\mathbb{Z}}[\sqrt{5}]$ and $H=x^2+\sqrt{5} x + 1$. Then $a=\Delta = 1$, but $H$ is irreducible. In fact, if $H=(x+s)(x+t)$, then $st=1$, $s+t=\sqrt{5}$, and $(s-t)^2 = (s+t)^2-4st=1,$ and hence $s-t = \pm 1$. In view of the fact that $s+t=\sqrt{5}$, we obtain $2s=\sqrt{5}\pm 1$, contradicting 