Find the magnitude of the product of all complex numbers $c$ such $x_{n+1} = (c^2 - 2c)^2 x_n x_{n-1} + 2x_n - x_{n-1}$ Find the magnitude of the product of all complex numbers $c$ such that the recurrence defined by $x_1 = 1$, $x_2 = c^2 - 4c + 7$, and $x_{n+1} = (c^2 - 2c)^2 x_n x_{n-1} + 2x_n - x_{n-1}$ also satisfies $x_{1006} = 2011$.
My try:
Lemma 1: $x_n$ is a polynomial in $c$, where the first coefficient is 1.
Proof: Induction.
Lemma 2:$$P_{n}(0)=6n-5$$
Since $P_{n+1}(c) = (c^2 - 2c)^2 P_n(c) P_{n-1}(c) + 2P_n(c) - P_{n-1}(c),$ we can plug in zero to get the recursion $P_{n+1}(0) = 2P_n(0) - P_{n-1}(0).$ The starting values are $P_1(0) = 1, P_2(0) = 7,$ and it is easy to prove by induction that $P_n (0) = 6n - 5.$
then I can't find this problem ,because if $P_{n}(x)=2n-1$ have no  double roots.I can do it,But I can't solve this equation have double roots case.so How do it? Thanks
 A: Adopting the notation of function sug,  $x_n$ is a monic polynomial $P_n(c)$ with variable $c$ and integer coefficients.
Consider $Q_n(c) = P_n(c)-P_n(2)$. By induction, we can prove that $Q_n(c)$ has $(c-2)^2$ as a factor (for the factorization over $\mathbb{Z}[c]$).
We have $P_n(0)=6n-5$, $P_n(2)=2n-1$ also by induction. Thus,
$Q_n(0)= 4n-4$. For $n=1006$, $Q_n(0)= 4\cdot 1005 = 4020$. Then we have
$$
Q_{1006}(0)=4\cdot 3 \cdot 5\cdot 67.
$$
The numbers $c$ we are looking for are the roots of the equation
$$
Q_{1006}(c)=0.
$$
We factor the polynomial $Q_{1006}(c)$ into a product of monic irreducible factors over $\mathbb{Q}[c]$ (coefficients are integers by Gauss lemma). Write
$$
Q_{1006}(c)=(c-2)^2 q_1(c)^{e_1} \cdots q_r(c)^{e_r}, 
$$
where $q_1, \ldots, q_r$ are distinct irreducible factors and $e_i\geq 1$.
Evaluating at $c=0$, we obtain that
$$
4 q_1(0)^{e_1} \cdots q_r(0)^{e_r} = 4 \cdot 3\cdot 5\cdot 67.
$$
As $3,5,67$ are primes,  there is a nonempty set $B$ of indices up to $r$, such that
$$
i\in B \Longleftrightarrow |q_i(0)|\in \{3,5,67, 3\cdot 5, 5\cdot 67, 3\cdot 67, 3\cdot 5\cdot 67\},
$$
and for any other $j\in \{1,\ldots,r\}-B$, we have
$$
|q_j(0)|=1.
$$
Comparing the prime factorization, we must have for the index $i\in B$, $e_i=1$.
Therefore the roots from $q_i(c)$, $i\in B$, are simple.
On the other hand, for $j\notin B$, the product of absolute values of the roots from $q_j$ must be $1$. Regardless of counting with or without multiplicity, the product contributes to $1$.
Hence, the product of absolute values of distinct roots of $Q_{1006}(c)=0$ is
$$
2 \cdot 3\cdot 5\cdot 67 = 2010.
$$
