square, we must have $$\begin{aligned}
 \label{DD}
\Delta &=&  8t((t-1)u^4-2S_2u^2+(2-t)S_4+S_2^2) ~=~ (\alpha u^2 + \beta u + \gamma)^2, \end{aligned}$$ where $\alpha$, $\beta$, and $\gamma$ belong to $k[x_1,\cdots,x_{n-2}]$. Equating the coefficients of $u^4$ and $u^3$ in (\[DD\]), we see that $\alpha^2 = 8t (t-1) \ne 0$ and $2 \alpha \beta = 0$. Therefore $\beta =0$, and we have $$\begin{aligned}
 \label{DDDD} 
\Delta &=&  8t((t-1)u^4-2S_2u^2+(2-t)S_4+S_2^2) ~=~ (\alpha u^2 + \gamma)^2. \end{aligned}$$ Equating coefficients in (\[DDDD\]), we find that $$\alpha^2 = 8t(t-1),~~\gamma^2 = 8t ((2-t)S_4+S_2^2),~~2 \alpha \gamma = 8t
 (-2S_2).$$ It follows that $(t-1) ((2-t)S_4+S_2^2) =  (-S_2)^2.$ This simplifies into $(2-t)((t-1)S_4  - S_2^2) =  0.$ Since $t \ne 2$, it follows that $S_2^2$ and $S_4$ are linearly dependent over $k$. This happens if and only if $n=3$, in which case $S_4=x_1^4=(x_1^2)^2=S_2^2$, and $t-1=1$. This is the case treated earlier. Thus $\Delta$ is not a square, a contradiction. This completes the proof. $\Box$

\[BBB\] 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 $n \ge 3$. Then $f$ has a linear factor if and only if $n=3$ and $t=2$. In this case, $f$ factors as 