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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?

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 $k=2$, $n=4$.

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Here's my thoughts for the case $k=2$:

For a polynomial $p(x)$ let $\tilde p(x)=x^{\deg p}p(1/x)$ denote the polynomial with coefficients in reverse order. Your polynomial is $f(x)=(x-1)^n+xg(x)$ where $\deg g=n-2$. We have $\tilde f(x) = (-1)^n(x-1)^n+x\tilde g(x)$. Note that $f$ has two roots with product $1$ iff $f$ and $\tilde f$ have a root in common (well, he have to be careful if that common root is $1$). The general tool for this is to check if the resultant $\operatorname{res}(f(x),\tilde f(x))$ is zero. Expressed in terms of the coefficients of $g$ (i.e., the $(-1)^ia_i$), this resultant is a polynomial $Q\in\Bbb Z[a_1,\ldots,a_{n-1}]$.

Trivially, $Q$ is zero if $g=(-1)^n\tilde g$. For $n$ odd this trivial case amounts to $a_1+a_{n-1}=a_2+a_{n-2}=\ldots=0$ and for $n$ even to $a_1-a_{n-1}=a_2-a_{n-2}=\ldots=0$ (where notably $a_{n/2}$ does not occur). This can be summarized by writing $n=2m+r$ with $r\in \{0,1\}$ and then the condition becomes that $a_k-(-1)^na_{n-k}=0$ concurrently for $1\le k\le m$.

Another trivial case - but for an undesired solution - is when $f(1)=0$, i.e., $g(1)=0$, i.e., $a_n-a_{n-1}\pm\ldots+(-1)^{n-1}a_1=0$. We conclude that $$ Q=(a_n-a_{n-1}\pm\ldots+(-1)^{n-1}a_1)\sum_{k=1}^m (a_k-(-1)^na_{n-k})P_k(a_1,\ldots,a_{n-1})$$ and that the second factor, the sum, is our desired $P$. A linear term can occur only if one of the $P_k$ has a constant term. So if for such $k$ (with $2k<n$) we substitute $a_k=T$ and $a_i=0$ otherwise, we should obtain that $Q=Q(T)=cT^2+O(T^3)$ with $c\ne 0$. So what can we say about the lower degree parts of $\operatorname{res}((x-1)^n+Tx^k,(-1)^n(x-1)^n+Tx^ {n-k})$?

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