A finite sequence of integers $ a_0, a_1, \ldots, a_n$ is called quadratic if for each $ i$ in the set $ \{1,2 \ldots, n\}$ we have the equality $ |a_i \minus{} a_{i\minus{}1}| \equal{} i^2.$

a.) Prove that any two integers $ b$ and $ c,$ there exists a natural number $ n$ and a quadratic sequence with $ a_0 \equal{} b$ and $ a_n \equal{} c.$

b.) Find the smallest natural number $ n$ for which there exists a quadratic sequence with  $ a_0 \equal{} 0$ and $ a_n \equal{} 1996.$
### Part (a)

1. **Understanding the Problem:**
   We need to show that for any two integers \( b \) and \( c \), there exists a natural number \( n \) and a quadratic sequence \( a_0, a_1, \ldots, a_n \) such that \( a_0 = b \) and \( a_n = c \).

2. **Rewriting the Condition:**
   The condition \( |a_i - a_{i-1}| = i^2 \) implies that \( a_i - a_{i-1} = \epsilon_i \cdot i^2 \) where \( \epsilon_i = \pm 1 \). Therefore, we need to find \( \epsilon_i \) such that:
   \[
   a_n - a_0 = \sum_{i=1}^n \epsilon_i \cdot i^2
   \]
   This simplifies to:
   \[
   c - b = \sum_{i=1}^n \epsilon_i \cdot i^2
   \]

3. **Constructing the Sequence:**
   We need to show that any integer \( c - b \) can be expressed as \( \sum_{i=1}^n \epsilon_i \cdot i^2 \). We will consider the remainder of \( c - b \) modulo 4.

4. **Case Analysis Based on Remainder:**
   - **Case 1: \( r = 0 \)**
     \[
     c - b = 4k \quad \text{for some integer } k
     \]
     We can write:
     \[
     4 = 1^2 - 2^2 - 3^2 + 4^2
     \]
     Repeating this pattern \( k \) times gives:
     \[
     c - b = \sum_{i=1}^{4k} \epsilon_i \cdot i^2
     \]

   - **Case 2: \( r = 1 \)**
     \[
     c - b = 4k + 1
     \]
     We can write:
     \[
     1 = 1^2
     \]
     and then:
     \[
     4 = 2^2 - 3^2 - 4^2 + 5^2
     \]
     Repeating the pattern for \( k \) times gives:
     \[
     c - b = 1^2 + \sum_{i=2}^{4k+1} \epsilon_i \cdot i^2
     \]

   - **Case 3: \( r = 2 \)**
     \[
     c - b = 4k + 2
     \]
     We can write:
     \[
     2 = -1^2 - 2^2 + 4^2
     \]
     and then:
     \[
     4 = 5^2 - 6^2 - 7^2 + 8^2
     \]
     Repeating the pattern for \( k \) times gives:
     \[
     c - b = -1^2 - 2^2 + 4^2 + \sum_{i=5}^{4k+2} \epsilon_i \cdot i^2
     \]

   - **Case 4: \( r = 3 \)**
     \[
     c - b = 4k + 3
     \]
     We can write:
     \[
     3 = -1^2 + 2^2
     \]
     and then:
     \[
     4 = 3^2 - 4^2 - 5^2 + 6^2
     \]
     Repeating the pattern for \( k \) times gives:
     \[
     c - b = -1^2 + 2^2 + \sum_{i=3}^{4k+3} \epsilon_i \cdot i^2
     \]

5. **Conclusion:**
   Since we can always find such \( \epsilon_i \) for any \( c - b \), we have shown that for any two integers \( b \) and \( c \), there exists a natural number \( n \) and a quadratic sequence with \( a_0 = b \) and \( a_n = c \).

\(\blacksquare\)

### Part (b)

1. **Understanding the Problem:**
   We need to find the smallest natural number \( n \) such that there exists a quadratic sequence with \( a_0 = 0 \) and \( a_n = 1996 \).

2. **Sum of Squares:**
   The sum of the first \( k \) squares is given by:
   \[
   S_k = \sum_{i=1}^k i^2 = \frac{k(k+1)(2k+1)}{6}
   \]

3. **Finding the Upper Bound:**
   We need \( S_n \geq 1996 \). Calculate \( S_{17} \) and \( S_{18} \):
   \[
   S_{17} = \frac{17 \cdot 18 \cdot 35}{6} = 1785
   \]
   \[
   S_{18} = \frac{18 \cdot 19 \cdot 37}{6} = 2109
   \]
   Since \( 1785 < 1996 < 2109 \), we need \( n \geq 18 \).

4. **Parity Check:**
   The sum of squares modulo 2:
   \[
   S_{18} \equiv 2109 \pmod{2} \equiv 1 \pmod{2}
   \]
   Since 1996 is even, \( n \neq 18 \). Thus, \( n \geq 19 \).

5. **Constructing the Sequence for \( n = 19 \):**
   \[
   S_{19} = \frac{19 \cdot 20 \cdot 39}{6} = 2470
   \]
   We need to express:
   \[
   2470 - 1996 = 474
   \]
   as a sum of distinct squares. We have:
   \[
   474 = 14^2 + 5^2 + 4^2
   \]
   Therefore:
   \[
   1996 = 1^2 + 2^2 + 3^2 - 4^2 - 5^2 + 6^2 + 7^2 + \cdots + 13^2 - 14^2 + 15^2 + \cdots + 19^2
   \]

6. **Conclusion:**
   The smallest \( n \) for which there exists a quadratic sequence with \( a_0 = 0 \) and \( a_n = 1996 \) is \( n = 19 \).

\(\blacksquare\)

The final answer is \( \boxed{ n = 19 } \)