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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

$$\sum _{k=1}^{4024}\:\frac{1}{x_k} = ?$$

Liviu
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    If $P'(x)=0$, then so is $(X^2+X-1)P'(x)$. You can simplify that. – Empy2 Apr 22 '17 at 15:21
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    One way to approach 2 is to write: $$P(x) = (x-x_1)\cdots(x-x_n) = (x_1x_2\cdots x_n)\left(\frac{x}{x_1}-1\right)\cdots \left(\frac{x}{x_n}-1\right)$$ and then consider the coefficient of the term $x$ when the right hand side is multiplied out. Do this for the given formula for $P(x)$ and compare. – Winther Apr 22 '17 at 15:32

2 Answers2

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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.

Batominovski
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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 $a_1=n(-1)^{n-1}-1$.

Angina Seng
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