How many ways are there to divide a $2\times n$ rectangle into rectangles having integral sides, where $n$ is a positive integer?
To solve the problem of finding the number of ways to divide a \(2 \times n\) rectangle into rectangles with integral sides, we can use the recurrence relation provided. The recurrence relation is given by:

\[ r_n = 6r_{n-1} - 7r_{n-2} \]

with initial conditions:

\[ r_1 = 2 \]
\[ r_2 = 8 \]

We will solve this recurrence relation using the characteristic equation method.

1. **Form the characteristic equation:**
   The characteristic equation for the recurrence relation \( r_n = 6r_{n-1} - 7r_{n-2} \) is obtained by assuming a solution of the form \( r_n = \lambda^n \). Substituting this into the recurrence relation, we get:

   \[
   \lambda^n = 6\lambda^{n-1} - 7\lambda^{n-2}
   \]

   Dividing through by \( \lambda^{n-2} \), we obtain the characteristic equation:

   \[
   \lambda^2 = 6\lambda - 7
   \]

   \[
   \lambda^2 - 6\lambda + 7 = 0
   \]

2. **Solve the characteristic equation:**
   We solve the quadratic equation \( \lambda^2 - 6\lambda + 7 = 0 \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

   \[
   \lambda = \frac{6 \pm \sqrt{36 - 28}}{2}
   \]

   \[
   \lambda = \frac{6 \pm \sqrt{8}}{2}
   \]

   \[
   \lambda = \frac{6 \pm 2\sqrt{2}}{2}
   \]

   \[
   \lambda = 3 \pm \sqrt{2}
   \]

   So, the roots of the characteristic equation are \( \lambda_1 = 3 + \sqrt{2} \) and \( \lambda_2 = 3 - \sqrt{2} \).

3. **Form the general solution:**
   The general solution to the recurrence relation is a linear combination of the solutions corresponding to the roots of the characteristic equation:

   \[
   r_n = A(3 + \sqrt{2})^n + B(3 - \sqrt{2})^n
   \]

4. **Determine the constants \( A \) and \( B \):**
   We use the initial conditions to find the constants \( A \) and \( B \).

   For \( r_1 = 2 \):

   \[
   2 = A(3 + \sqrt{2}) + B(3 - \sqrt{2})
   \]

   For \( r_2 = 8 \):

   \[
   8 = A(3 + \sqrt{2})^2 + B(3 - \sqrt{2})^2
   \]

   Calculate \( (3 + \sqrt{2})^2 \) and \( (3 - \sqrt{2})^2 \):

   \[
   (3 + \sqrt{2})^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}
   \]

   \[
   (3 - \sqrt{2})^2 = 9 - 6\sqrt{2} + 2 = 11 - 6\sqrt{2}
   \]

   Substituting these into the second initial condition:

   \[
   8 = A(11 + 6\sqrt{2}) + B(11 - 6\sqrt{2})
   \]

   We now have a system of linear equations:

   \[
   2 = A(3 + \sqrt{2}) + B(3 - \sqrt{2})
   \]

   \[
   8 = A(11 + 6\sqrt{2}) + B(11 - 6\sqrt{2})
   \]

   Solving this system, we find:

   \[
   A = \frac{3 - \sqrt{2}}{4}
   \]

   \[
   B = \frac{3 + \sqrt{2}}{4}
   \]

5. **Write the final solution:**
   Substituting \( A \) and \( B \) back into the general solution, we get:

   \[
   r_n = \frac{3 - \sqrt{2}}{4}(3 + \sqrt{2})^n + \frac{3 + \sqrt{2}}{4}(3 - \sqrt{2})^n
   \]

The final answer is \( \boxed{ r_n = \frac{3 - \sqrt{2}}{4}(3 + \sqrt{2})^n + \frac{3 + \sqrt{2}}{4}(3 - \sqrt{2})^n } \).