Sure, here is the translation:

```
II. Let the polynomial sequence $\left\{P_{n}(x)\right\}$ satisfy:
$$
P_{1}(x)=x^{2}-1, P_{2}(x)=2 x\left(x^{2}-1\right) .
$$

and $P_{n+1}(x) P_{n-1}(x)=\left(P_{n}(x)\right)^{2}-\left(x^{2}-1\right)^{2}$, $n=2,3, \cdots$.

Let $s_{n}$ be the sum of the absolute values of the coefficients of $P_{n}(x)$, for any positive integer $n$, find the non-negative integer $k_{n}$, such that $2^{-k} k_{n}$ is an odd number.
(Huang Yumin, problem contributor)
```
From equation (1), we have
\[ P_{n+1}(x) P_{n-1}(x) - P_{n}^2(x) \]
\[ = P_{n}(x) P_{n-2}(x) - P_{n-1}^2(x), \quad n=3,4, \cdots, \]
\[ P_{n-1}(x) \left[ P_{n+1}(x) + P_{n-1}(x) \right] \]
\[ = P_{n}(x) \left[ P_{n}(x) + P_{n-2}(x) \right]. \]
By induction, we get
\[ \frac{P_{n+1}(x) + P_{n-1}(x)}{P_{n}(x)} = \frac{P_{n}(x) + P_{n-2}(x)}{P_{n-1}(x)} \]
\[ = \cdots = \frac{P_{3}(x) + P_{1}(x)}{P_{2}(x)}. \]
(2)
Since \( P_{1}(x) = x^2 - 1 \) and \( P_{2}(x) = 2x(x^2 - 1) \),
\[ P_{3}(x) = \frac{P_{2}^2(x) - (x^2 - 1)^2}{P_{1}(x)} = (x^2 - 1)(4x^2 - 1), \]
\[ \frac{P_{3}(x) + P_{1}(x)}{P_{2}(x)} \]
\[ = \frac{(x^2 - 1)(4x^2 - 1) + (x^2 - 1)}{2x(x^2 - 1)} = 2x. \]

Substituting equation (3) into equation (2), we get
\[ P_{n+1}(x) = 2x P_{n}(x) - P_{n-1}(x), \quad n=2,3, \cdots. \]
Defining \( P_{0}(x) = 0 \), equation (4) also holds for \( n=1 \). The characteristic equation corresponding to equation (4) is
\[ \lambda^2 - 2x \lambda + 1 = 0, \]
with roots \( x \pm \sqrt{x^2 - 1} \) for \( |x| > 1 \). We have
\[ P_{n}(x) = \frac{1}{2} \sqrt{x^2 - 1} \left[ \left( x + \sqrt{x^2 - 1} \right)^n - \left( x - \sqrt{x^2 - 1} \right)^n \right], \quad n=0,1,2, \cdots. \]

Using the binomial theorem, we get
\[ P_{n}(x) = \sum_{m=0}^{s_{n}^{n}} C_{n}^{2m+1} x^{n-2m-1} (x^2 - 1)^{m+1}, \]
where \( \sigma_{n} = \begin{cases} \left[ \frac{n}{2} \right], & \text{if } n \text{ is odd,} \\ \frac{n}{2} - 1, & \text{if } n \text{ is even.} \end{cases} \)
Thus,
\[ P_{n}(x) = (x^2 - 1) \sum_{m=0}^{n} C_{n}^{2m+1} x^{n-2m-1} \]
\[ \cdot \sum_{i=0}^{m} (-1)^t C_{m}^{t} x^{2m-2t} \]
\[ = (x^2 - 1) \sum_{i=0}^{n} (-1)^t x^{n-2t-1} \sum_{m=1}^{0_{n}^{n}} C_{m}^{2_{m}^{m+1} C_{m}^{t}}. \]

Let \( a_{i} = \sum_{m=1}^{n} C_{n}^{2_{m}+1} C_{m}^{l} \), substituting into equation (5), we get
\[ P_{n}(x) = \left[ a_{0} x^{n+1} - a_{1} x^{n-1} + a_{2} x^{n-3} - \cdots \right. \]
\[ + (-1)^{o_{n}} a_{o_{n}} x^{n-2o_{n}+1} \right] \]
\[ - \left[ a_{0} x^{n-1} - a_{1} x^{n-3} + \cdots \right. \]
\[ + (-1)^{o_{n}-1} a_{o_{n}-1} x^{n-2a_{n}+1} \]
\[ + (-1)^{s_{n}} a_{o_{n}} x^{n-2o_{n}-1} \right] \]
\[ = a_{0} x^{n+1} - (a_{1} + a_{0}) x^{n-1} + (a_{2} + a_{1}) x^{n-3} - \cdots \]
\[ + (-1)^{o_{n}} (a_{o_{n}} + a_{o_{n}-1}) x^{n-2s_{n}+1} + (-1)^{o_{n}+1} a_{o_{n}} x^{n-2o_{n}-1}. \]

Thus, we have
\[ s_{n} = a_{0} + (a_{1} + a_{0}) + \cdots + (a_{o_{n}} + a_{o_{n}-1}) + a_{n} \]
\[ = 2 (a_{0} + a_{1} + \cdots + a_{o_{n}}) \]
\[ = 2 \sum_{i=0}^{0} \sum_{m=1}^{o_{n}^{n}} C_{m}^{2m+1} C_{m}^{t} \]
\[ = 2 \sum_{m=0}^{\infty} C_{n}^{2m+1} \sum_{i=0}^{m} C_{m}^{t} = 2 \sum_{m=0}^{\infty} C_{n}^{2m+1} 2^m. \]

Therefore,
\[ s_{n} = \frac{1}{\sqrt{2}} \left[ (1 + \sqrt{2})^n - (1 - \sqrt{2})^n \right], \quad n=0,1,2, \cdots. \]
Let \( t_{n} = (1 + \sqrt{2})^n + (1 - \sqrt{2})^n, \quad n=0,1,2, \cdots \), then
\( s_{n} \) and \( t_{n} \) are non-negative integers, \( s_{0} = 0, s_{1} = 2, t_{0} = t_{1} = 2 \) and
\[ t_{n} + \sqrt{2} s_{n} = 2 (1 + \sqrt{2})^n, \quad n=0,1,2, \cdots. \]
\[ \text{When } n \geqslant 2, \]
\[ t_{n} = 2 \sum_{0 \leqslant 2m \leqslant n}^{2m} 2^{2m} = 2 \left( 1 + \sum_{0 < 2m \leqslant n} C_{m}^{2m} 2^m \right). \]

It is clear that \( \frac{1}{2} t_{n} \) is an odd number when \( n=0,1,2, \cdots \).
Let \( n, m \) be non-negative integers, from equation (7) we get
\[ t_{n+m} + \sqrt{2} s_{n+m} = 2 (1 + \sqrt{2})^{n+m} \]
\[ = \frac{1}{2} (t_{n} + \sqrt{2} s_{n})(t_{m} + \sqrt{2} s_{m}) \]
\[ = \frac{1}{2} t_{n} t_{m} + s_{n} s_{m} + \frac{\sqrt{2}}{2} (t_{n} s_{m} + t_{m} s_{n}). \]
Thus, we have
\[ t_{n+m} = \frac{1}{2} t_{n} t_{m} + s_{n} s_{m}, \quad s_{n+m} = \frac{1}{2} (t_{n} s_{m} + t_{m} s_{n}). \]
Let \( n = m \), we get
\[ s_{2n} = \left( \frac{1}{2} t_{n} \right) \cdot 2 s_{n}, \quad n=0,1,2, \cdots. \]
Since \( s_{1} = 2 \) and \( \frac{1}{2} t_{n} \) is an odd number, by mathematical induction, we know
\[ 2^{-(m+1)} s_{2m} \] is an odd number, \( m=0,1,2, \cdots \).
(2)

That is, \( k_{2m} = m + 1 \).
From equation (8), if \( k_{n} \neq k_{m} \), then
\[ k_{n+m} = \min(k_{n}, k_{m})! \].
For any positive integer, let
\[ n = 2^{m_0} + 2^{m_1} + \cdots + 2^{m_l} \].

where \( m_0, m_1, \cdots, m_l \) are integers, and \( 0 \leqslant m_0 < m_1 < \cdots < \).