Check if polynomial has double root and find $\sum _{k=1}^{4024}\:\frac{1}{x_k}$ Consider the polynomial $P \in \mathbb R[X[, \: P(x) = (X^2 + X -1)^n - X, \: n\in \mathbb N, n\ge 2$
 1)Which is the number of values of $n$ so that $P$ has double roots ?
I have found that $(X^2 -1) \:| \:P$ for $n = 2k +1, k \in \mathbb N^*$ if that helps in any way.
Also, I think the conditions should be  $P(a) = 0, P'(a) = 0, P''(0) \neq 0$, where $a$ is the double root.
 2) If $n = 2012$ and $x_1, x_2, ..., x_{4024} \in \mathbb C$ are the roots of $P$ then 
$$\sum _{k=1}^{4024}\:\frac{1}{x_k} = ?$$
 A: Solution to Part (1)
If $P(a)=0$ and $P'(a)=0$, then $\left(a^2+a-1\right)^n=a$ and $n\,(2a+1)\,\left(a^2+a-1\right)^{n-1}=1$.  Consequently, $$n\,(2a+1)\,a=n\,(2a+1)\,\left(a^2+a-1\right)^n=1\cdot \left(a^2+a-1\right)\,.$$  That is,
$$(2n-1)\,a^2+(n-1)\,a+1=0\,.$$
Since $\left(a^2+a-1\right)^n=a$, we see that $a$ is an algebraic integer.  Therefore, the monic minimal polynomial of $a$ over $\mathbb{Q}$ is in $\mathbb{Z}[x]$.  Hence, there is a monic polynomial $f(x)\in\mathbb{Z}[x]$ dividing the quadratic polynomial $g(x):=(2n-1)\,x^2+(n-1)\,x+1$.  Consequently, either that the leading coefficient $2n-1$ of $g(x)$ divides its constant term $1$ (leading to $n=1$), or that $f(x)$ is linear (so $a\in\mathbb{Z}$, yielding $a\in\{-1,+1\}$).  In the latter case, we have $$
1=n\,(2a+1)\,\left(a^2+a-1\right)^{n-1}=\left\{
\begin{array}{ll}
3n\,,&\text{if }a=+1\,,\\
(-1)^n\,n\,,&\text{if }a=-1\,,
\end{array}\right.$$
which is absurd.  Thus, $n=1$ is the only possibility, but then $P(X)=X^2-1$ has no double roots.
A: Let $f(X)=X^m+a_{m-1}X^{m-1}+\cdots+a_1X+a_0$ be a monic polynomial, and
$x_1,\ldots,x_m$ be its zeros. Then $\sum 1/x_j$ is $-a_1/a_0$. This is routine elementary symmetric function manipulation. Here $a_0=(-1)^n$
and $a_1=n(-1)^{n-1}-1$.
