Given the polynomial sequence $(p_{n}(x))$ satisfying $p_{1}(x)=x^{2}-1$, $p_{2}(x)=2x(x^{2}-1)$, and $p_{n+1}(x)p_{n-1}(x)=(p_{n}(x)^{2}-(x^{2}-1)^{2}$, for $n\geq 2$, let $s_{n}$ be the sum of the absolute values of the coefficients of $p_{n}(x)$. For each $n$, find a non-negative integer $k_{n}$ such that $2^{-k_{n}}s_n$ is odd.
1. **Claim 1: \( P_{n+1}(x) = 2xP_n(x) - P_{n-1}(x) \)**

   We will prove this by induction.

   - **Base Case: \( n = 2 \)**

     \[
     P_3(x)P_1(x) = (2xP_2(x) - P_1(x))P_1(x) = P_2(x)^2 - P_1(x)^2
     \]

     This holds true as given in the problem.

   - **Inductive Step:**

     Assume the statement is true for all values less than \( n \geq 3 \). Then,

     \[
     P_{n+1}(x)P_{n-1}(x) = P_n(x)^2 - P_1(x)^2
     \]

     Using the inductive hypothesis:

     \[
     P_n(x) = 2xP_{n-1}(x) - P_{n-2}(x)
     \]

     Substituting this into the equation:

     \[
     P_{n+1}(x)P_{n-1}(x) = (2xP_{n-1}(x) - P_{n-2}(x))^2 - (x^2 - 1)^2
     \]

     Simplifying, we get:

     \[
     P_{n+1}(x) = 2xP_n(x) - P_{n-1}(x)
     \]

     Hence, the claim is proven by induction.

2. **Claim 2: Each term of \( 2xP_n(x) \) and \( -P_{n-1}(x) \) has the same sign**

   - Note that \( P_n(x) \) is a polynomial of degree \( n+1 \). If \( n \) is even, it only has terms with odd degrees, and if \( n \) is odd, it only has terms with even degrees.
   - The coefficient of the highest degree term is positive, and the signs alternate thereafter.

   For example:
   \[
   P_1(x) = x^2 - 1, \quad P_2(x) = 2x^3 - 2x
   \]

   From our first claim, we can see that:

   \[
   P_{n+1}(x) = 2xP_n(x) - P_{n-1}(x)
   \]

   The degree of \( P_{n+1}(x) \) is \( n+2 \), and the terms appear in only the same parity of degree (i.e., \( n+2, n, n-2, \ldots \)). The signs alternate as well.

3. **Claim 3: \( S_1 = 2, S_2 = 4, S_{n+1} = 2S_n + S_{n-1} \)**

   This follows immediately from our second claim. Solving the recursion, we get:

   \[
   S_n = \frac{1}{\sqrt{2}} \left[ (1+\sqrt{2})^n - (1-\sqrt{2})^n \right]
   \]

4. **Claim 4: \( k_n = 2L(n) \), where \( L(n) = v_2(n) \)**

   Keep in mind that \( k_n = L(S_n) \). Using the Binomial Theorem:

   \[
   S_n = 2^{-\frac{1}{2}} \left[ \sum_{i=0}^n \binom{n}{i} 2^{\frac{i}{2}} - \sum_{i=0}^n \binom{n}{i} (-1)^i 2^{\frac{i}{2}} \right]
   \]

   Simplifying, we get:

   \[
   S_n = \sum_{0 \leq 2 \nmid i \leq n} \binom{n}{i} 2^{\frac{i+1}{2}}
   \]

   Let \( i = 2r+1 \):

   \[
   \binom{n}{2r+1} = \frac{n}{2r+1} \binom{n-1}{2r}
   \]

   Therefore:

   \[
   S_n = \sum_{0 \leq r \leq \frac{n-1}{2}} \binom{n}{2r+1} 2^{r+1} = 2n + \sum_{1 \leq r \leq \frac{n-1}{2}} \binom{n-1}{2r} \cdot \frac{2^r \cdot 2n}{2r+1}
   \]

   If \( 1 \leq r \leq \frac{n-1}{2} \), then \( 2^{L(n)+1} \mid 2^r \cdot 2n \), and:

   \[
   2^{L(2n)+1} \mid \binom{n-1}{2r} \cdot 2^r \cdot 2n
   \]

   Since \( 2r+1 \) is odd:

   \[
   2^{L(n)+1} \mid \binom{n-1}{2r} \cdot \frac{2^r \cdot 2n}{2r+1}
   \]

   Therefore:

   \[
   2^{L(2n)+1} \mid \sum_{1 \leq r \leq \frac{n-1}{2}} \binom{n-1}{2r} \cdot \frac{2^r \cdot 2n}{2r+1}
   \]

   We can write \( 2n \) as \( 2n = 2^{L(2n)}(2t+1) \) for some \( t \in \mathbb{N} \), and:

   \[
   \sum_{1 \leq r \leq \frac{n-1}{2}} \binom{n-1}{2r} \cdot \frac{2^r \cdot 2n}{2r+1} = 2^{L(n)+1}s
   \]

   for some \( s \in \mathbb{N} \). Therefore:

   \[
   S_n = 2^{L(n)}(2t+1) + 2^{L(2n)+1}s = 2^{L(2n)}(2(t+s)+1)
   \]

   Hence, our proof is complete. Therefore, \( k_n = 2v_2(n) \).

\(\blacksquare\)

The final answer is \( \boxed{ k_n = 2v_2(n) } \)