and $2 \ne 0$.

If (char ($k$) $\ne 2$, and) $n = 1$, then $h$ factors into $h = (a_0 + x_1)(-a_0 + x_1)$.

If char ($k$) $\ne 2$, and $n = 2$, then comparing the coefficients of $x_1$, we obtain $-2 a_0 a_1 = 0$, and hence $a_0 = 0$, since $a_1^2 = -t_1 \ne 0$ and $2 \ne 0$. Therefore $h$ factors into $h = (a_1 x_1 + x_2)(-a_1 x_1 + x_2).$ This completes the proof. $\Box$

It follows, for example, that the polynomial $x_1^2 + \cdots + x_n^2$ is reducible over the field ${\mathbb{C}}$ of complex numbers if and only if $n=1$ or $n=2$. In particular, $x^2+y^2+z^2$ is irreducible over ${\mathbb{C}}$, a fact that appears as a problem in [@AMATYC].

In view of the lemma above, we may exclude the case $t=0$ in (\[g\]). Also, it is obvious that if ${\mbox{char~}}k = 2$, then $g = (1-t) (a+x_1+\cdots+x_n)^4$. That is why the lemmas and theorems below assume that $t\ne 0$ and ${\mbox{char~}}k \ne 2$.

\[AAA\] Let $k$ be a field with ${\mbox{char~}}k \ne 2$, and let $$\begin{aligned}
f &=& (x_1^2+\cdots+x_n^2)^2-t(x_1^4+ \cdots +x_n^4) \in k[x_1,\cdots,x_n], \end{aligned}$$ where $t \in k$ and $t \ne 0$, and where $n \ge 