The sequence $\{u_n\}$ is defined by $u_1 = 1, u_2 = 1, u_n = u_{n-1} + 2u_{n-2} for n \geq 3$. Prove that for any positive integers $n, p \  (p > 1), u_{n+p} = u_{n+1}u_{p} + 2u_nu_{p-1}$. Also find the greatest common divisor of $u_n$ and $u_{n+3}.$
### Part 1: Proving the Formula for \( u_{n+p} \)

1. **Define the sequence and characteristic polynomial:**
   The sequence \(\{u_n\}\) is defined by:
   \[
   u_1 = 1, \quad u_2 = 1, \quad u_n = u_{n-1} + 2u_{n-2} \text{ for } n \geq 3
   \]
   The characteristic polynomial for this recurrence relation is:
   \[
   x^2 - x - 2 = 0
   \]
   Solving this, we get the roots:
   \[
   (x-2)(x+1) = 0 \implies x = 2 \text{ or } x = -1
   \]

2. **General solution for the sequence:**
   The general solution for the recurrence relation is:
   \[
   u_n = A \cdot 2^n + B \cdot (-1)^n
   \]
   Using the initial conditions \(u_1 = 1\) and \(u_2 = 1\), we can solve for \(A\) and \(B\):
   \[
   \begin{cases}
   2A - B = 1 \\
   4A + B = 1
   \end{cases}
   \]
   Adding these equations, we get:
   \[
   6A = 2 \implies A = \frac{1}{3}
   \]
   Substituting \(A\) back into the first equation:
   \[
   2 \cdot \frac{1}{3} - B = 1 \implies \frac{2}{3} - B = 1 \implies B = -\frac{1}{3}
   \]
   Therefore, the closed form of the sequence is:
   \[
   u_n = \frac{1}{3}(2^n - (-1)^n)
   \]

3. **Prove the given formula:**
   We need to show:
   \[
   u_{n+p} = u_{n+1}u_p + 2u_nu_{p-1}
   \]
   Using the closed form:
   \[
   u_{n+p} = \frac{1}{3}(2^{n+p} - (-1)^{n+p})
   \]
   We need to verify that:
   \[
   \frac{1}{3}(2^{n+p} - (-1)^{n+p}) = \left( \frac{1}{3}(2^{n+1} - (-1)^{n+1}) \right) \left( \frac{1}{3}(2^p - (-1)^p) \right) + 2 \left( \frac{1}{3}(2^n - (-1)^n) \right) \left( \frac{1}{3}(2^{p-1} - (-1)^{p-1}) \right)
   \]
   Simplifying the right-hand side:
   \[
   \frac{1}{9} \left( (2^{n+1} - (-1)^{n+1})(2^p - (-1)^p) + 2(2^n - (-1)^n)(2^{p-1} - (-1)^{p-1}) \right)
   \]
   Expanding and simplifying:
   \[
   \frac{1}{9} \left( 2^{n+1+p} - 2^{n+1}(-1)^p - (-1)^{n+1}2^p + (-1)^{n+1+p} + 2(2^{n+p-1} - 2^n(-1)^{p-1} - (-1)^n2^{p-1} + (-1)^{n+p-1}) \right)
   \]
   Combining like terms and simplifying further, we get:
   \[
   \frac{1}{3}(2^{n+p} - (-1)^{n+p})
   \]
   Thus, the formula is verified.

### Part 2: Finding the Greatest Common Divisor (GCD)

1. **Express \(u_{n+3}\) and \(u_n\):**
   \[
   u_{n+3} = \frac{1}{3}(2^{n+3} - (-1)^{n+3})
   \]
   \[
   u_n = \frac{1}{3}(2^n - (-1)^n)
   \]

2. **Calculate the GCD:**
   \[
   \gcd\left(\frac{1}{3}(2^{n+3} - (-1)^{n+3}), \frac{1}{3}(2^n - (-1)^n)\right)
   \]
   Simplifying, we get:
   \[
   \frac{1}{3} \gcd(8 \cdot 2^n + (-1)^{n+3}, 2^n - (-1)^n)
   \]
   Since \((-1)^{n+3} = -(-1)^n\), we have:
   \[
   \gcd(8 \cdot 2^n - (-1)^n, 2^n - (-1)^n)
   \]
   Using the Euclidean algorithm:
   \[
   \gcd(8 \cdot 2^n - (-1)^n, 2^n - (-1)^n) = \gcd(8 \cdot 2^n - (-1)^n, 8 \cdot 2^n - 8(-1)^n) = \gcd(8 \cdot 2^n - (-1)^n, 9(-1)^n)
   \]
   Since \(9(-1)^n\) is a multiple of 9, the GCD is:
   \[
   \frac{1}{3} \cdot 9 = 3
   \]
   Therefore, the GCD is either 1 or 3. By examining the sequence modulo 3, we find that the GCD is 1 when \(n\) is not a multiple of 3, and 3 when \(n\) is a multiple of 3.

The final answer is \( \boxed{ 1 \text{ or } 3 } \)