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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

math110
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  • I hope the trick to this problem can be revealed. The recursion is not nice at all. – Display name Dec 25 '20 at 03:15
  • I think that a first step to a solution is to rewrite the recurrence relation as $$ \frac{(x_{n+1}-x_{n})-(x_{n}-x_{n-1})}{x_{n}x_{n-1}} = c^2(c-2)^2 $$ – md2perpe Dec 25 '20 at 11:07
  • For $c=2$ we have $x_n=2n-1$ so $x_{1006}=2011$. Therefore $c=2$ is among the factors. Is it perhaps the only one? – md2perpe Dec 25 '20 at 11:57
  • I have run some tests using Python code. It seems like $x_n$ grows very fast (exponentially or even faster) if $c \not\in { 0, 2 }.$ – md2perpe Dec 26 '20 at 17:03

1 Answers1

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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 $$ 2 \cdot 3\cdot 5\cdot 67 = 2010. $$

Sungjin Kim
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