The remainder on dividing the polynomial $p(x)$ by $x^2 - (a+b)x + ab$ (where $a \not = b$) is $mx + n$. Find the coefficients $m, n$ in terms of $a, b$. Find $m, n$ for the case $p(x) = x^{200}$ divided by $x^2 - x - 2$ and show that they are integral.
1. **Understanding the problem**: We need to find the remainder when the polynomial \( p(x) \) is divided by \( x^2 - (a+b)x + ab \). Given that the remainder is \( mx + n \), we need to express \( m \) and \( n \) in terms of \( a \) and \( b \). Specifically, we need to find \( m \) and \( n \) for \( p(x) = x^{200} \) when divided by \( x^2 - x - 2 \).

2. **Using the Remainder Theorem**: The polynomial \( x^2 - (a+b)x + ab \) has roots \( a \) and \( b \). Therefore, \( p(a) = ma + n \) and \( p(b) = mb + n \).

3. **Setting up the system of equations**:
   \[
   \begin{cases}
   p(a) = ma + n \\
   p(b) = mb + n
   \end{cases}
   \]
   For \( p(x) = x^{200} \), we have:
   \[
   \begin{cases}
   a^{200} = ma + n \\
   b^{200} = mb + n
   \end{cases}
   \]

4. **Solving for \( m \) and \( n \)**:
   Subtract the two equations:
   \[
   a^{200} - b^{200} = m(a - b)
   \]
   Since \( a \neq b \), we can divide by \( a - b \):
   \[
   m = \frac{a^{200} - b^{200}}{a - b}
   \]

5. **Finding \( n \)**:
   Substitute \( m \) back into one of the original equations:
   \[
   n = a^{200} - ma
   \]

6. **Special case \( p(x) = x^{200} \) and \( x^2 - x - 2 \)**:
   Here, \( x^2 - x - 2 = (x-2)(x+1) \), so \( a = 2 \) and \( b = -1 \).

7. **Calculating \( m \) and \( n \)**:
   \[
   m = \frac{2^{200} - (-1)^{200}}{2 - (-1)} = \frac{2^{200} - 1}{3}
   \]
   \[
   n = 2^{200} - m \cdot 2 = 2^{200} - 2 \cdot \frac{2^{200} - 1}{3} = \frac{2^{200} + 2}{3}
   \]

8. **Verification**:
   \[
   m = \frac{2^{200} - 1}{3}
   \]
   \[
   n = \frac{2^{200} + 2}{3}
   \]

9. **Checking integrality**:
   Since \( 2^{200} \equiv 1 \pmod{3} \), both \( m \) and \( n \) are integers.

The final answer is \( \boxed{ m = \frac{2^{200} - 1}{3} } \) and \( n = \frac{2^{200} + 2}{3} \).