# Polynomials with some roots whose product is 1

I asked this question in this post but have not got a full answer. So I post it again on MO.

Consider the complex coefficient polynomial equation \begin{eqnarray} x^n-\left(a_1+\binom{n}{1}\right)x^{n-1}+\cdots+(-1)^k\left(a_k+\binom{n}{k}\right)x^{n-k}+\cdots+(-1)^{n-1}\left(a_{n-1}+\binom{n}{n-1}\right)x+(-1)^n=0 \end{eqnarray} By Vieta Theorem, the product of its roots is 1. If we impose the condition that, among the $n$ roots, there exist $k$ roots (counted with multiplicity) whose product is 1, then $a_1, \cdots, a_{n-1}$ have to satisfy a polynomial equation $P(a_1, \cdots, a_{n-1})=0$, where $P\in\mathbb{C}[a_1, \cdots, a_{n-1}]$ has 0 as the constant term.

Question: Under what condition does $P$ has nonzero linear term?

Equivalently, I am asking for condition under which the affine hypersurface $V(P)\subset \mathbb{A}^{n-1}$ is smooth at the point $(0, 0, \cdots, 0)$.

The following are some easy examples I have worked out.

If $k=1$, then that means one of the roots is 1. Plugging $x=1$ to the original polynomial equation yields that $P$ can be \begin{eqnarray} \sum_{i=1}^{n-1} (-1)^ia_i \end{eqnarray} whose linear term is nonzero. For $k=2$, $n=3$ or $4$, $P$ also has nonzero linear term.

If $k=2$ and $n=5$, then the original polynomial equation can be factored as

\begin{eqnarray} (x^3+px^2+qx-1)(x^2+rx+1)=0 \end{eqnarray} Comparing coefficients, we have \begin{align*} a_1&=-p-r-5\\ a_2&=pr+q-9\\ a_3&=-p-qr-9\\ a_4&=q-r-5 \end{align*} According to Mathematica, $P$ is, up to a constant multiple, \begin{eqnarray} -a_1^3+a_3 a_1^2+a_4 a_1^2+a_1^2+a_4^2 a_1-2 a_2 a_1+2 a_3 a_1-a_2 a_4 a_1-a_3 a_4 a_1-2 a_4 a_1-a_4^3+a_2^2+a_3^2+a_2 a_4^2+a_4^2-2 a_2 a_3+2 a_2 a_4-2 a_3 a_4\end{eqnarray} which does not have nonzero linear terms.

For $k=3$, $n=6$, $P$ is \begin{eqnarray} a_1^3-a_4 a_1^2+3 a_1^2-4 a_2 a_1+6 a_3 a_1-12 a_4 a_1+a_3 a_5 a_1+22 a_5 a_1+a_5^3-a_3^2-a_2 a_5^2+3 a_5^2+4 a_2 a_4-12 a_2 a_5+6 a_3 a_5-4 a_4 a_5\end{eqnarray} which again does not have nonzero linear terms. My guess is that, if $k\geq 2$ and $n-k\geq$ 2, then $P$ does not have nonzero linear term, except the case $k=2$, $n=4$.