the fact that $s \in A$. For (iv), let $D = \{ f(x) = c_0+c_1 x+\cdots +c_n x^n\in {\mathbb{Z}}[x] : c_1 \mbox{~is even}\},$ and let $H = x^2 + 2x + 1$. Then $\Delta = 0$, and $H$ is not a square in $D$.*]{}

The main results {#333}
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In this section, we establish, in Theorems \[CCDD\], \[DDD\], and \[CM\], the main irreducibility theorems of this paper. Lemmas \[AAA\], \[BBB\], and \[CCC\] are needed in the proofs. Lemma \[CBA\] disposes of the case when $t=0$ in (\[g\]).

\[CBA\] Let $k$ be a field, and let $g=c_0 + c_1x_1^2+\cdots + c_n x_n^2
\in k[x_1,\cdots, x_n]$, where $n \ge 1$ and $c_1,\cdots,c_n \in k \setminus \{0\}$. For $0 \le i \le n$, let $t_i = c_i/c_n$, and let $h=t_0 + t_1x_1^2+\cdots + t_n x_n^2.$ Then $h$ (and hence $g$) is reducible if and only if $-t_j$ is a square for all $j$, $0 \le j \le n-1$ and if

-   either ${\mbox{char~}}k =2$, in which case $$\begin{aligned}
        h&=&\left(\sqrt{t_0} + \sqrt{t_1} x_1+ \cdots +
    \sqrt{t_{n-1}} x_{n-1} +  x_{n}          \right)^2,\end{aligned}$$

-   or ${\mbox{char~}}k \ne 2$ and $n=1$, in which case, $$\begin{aligned}
        h&=&\left(x_1 - \sqrt{-t_0}\right)\left(x_1 + \sqrt{-t_0}\right),
        \end{aligned}$$

-   or ${\mbox{char~}}k \ne 2$, $n=2$, and 