problem 1 :A sequence is defined by$ x_1 = 1, x_2 = 4$ and $ x_{n+2} = 4x_{n+1} -x_n$ for $n \geq 1$. Find all natural numbers $m$ such that the number $3x_n^2 + m$ is a perfect square for all natural numbers $n$
To solve the problem, we need to find all natural numbers \( m \) such that \( 3x_n^2 + m \) is a perfect square for all natural numbers \( n \). Let's start by analyzing the given sequence and the conditions.

1. **Define the sequence:**
   The sequence is defined by:
   \[
   x_1 = 1, \quad x_2 = 4, \quad \text{and} \quad x_{n+2} = 4x_{n+1} - x_n \quad \text{for} \quad n \geq 1.
   \]

2. **Calculate the first few terms:**
   \[
   x_1 = 1, \quad x_2 = 4
   \]
   \[
   x_3 = 4x_2 - x_1 = 4 \cdot 4 - 1 = 16 - 1 = 15
   \]
   \[
   x_4 = 4x_3 - x_2 = 4 \cdot 15 - 4 = 60 - 4 = 56
   \]
   \[
   x_5 = 4x_4 - x_3 = 4 \cdot 56 - 15 = 224 - 15 = 209
   \]

3. **Formulate the condition:**
   We need \( 3x_n^2 + m \) to be a perfect square for all \( n \). Let \( 3x_n^2 + m = k_n^2 \) for some integer \( k_n \).

4. **Check specific values of \( n \):**
   For \( n = 1 \):
   \[
   3x_1^2 + m = 3 \cdot 1^2 + m = 3 + m = k_1^2
   \]
   For \( n = 2 \):
   \[
   3x_2^2 + m = 3 \cdot 4^2 + m = 3 \cdot 16 + m = 48 + m = k_2^2
   \]

5. **Solve for \( m \):**
   We need both \( 3 + m \) and \( 48 + m \) to be perfect squares. Let:
   \[
   3 + m = a^2 \quad \text{and} \quad 48 + m = b^2
   \]
   for some integers \( a \) and \( b \).

6. **Set up the system of equations:**
   \[
   m = a^2 - 3
   \]
   \[
   m = b^2 - 48
   \]
   Equating the two expressions for \( m \):
   \[
   a^2 - 3 = b^2 - 48
   \]
   \[
   a^2 - b^2 = -45
   \]
   \[
   (a - b)(a + b) = -45
   \]

7. **Solve the factor pairs:**
   The factor pairs of \(-45\) are:
   \[
   (-1, 45), (1, -45), (-3, 15), (3, -15), (-5, 9), (5, -9)
   \]

8. **Check each pair:**
   For \((a - b, a + b) = (-1, 45)\):
   \[
   a - b = -1 \quad \text{and} \quad a + b = 45
   \]
   Solving these:
   \[
   2a = 44 \quad \Rightarrow \quad a = 22
   \]
   \[
   2b = 46 \quad \Rightarrow \quad b = 23
   \]
   \[
   m = a^2 - 3 = 22^2 - 3 = 484 - 3 = 481
   \]

   For \((a - b, a + b) = (1, -45)\):
   \[
   a - b = 1 \quad \text{and} \quad a + b = -45
   \]
   Solving these:
   \[
   2a = -44 \quad \Rightarrow \quad a = -22
   \]
   \[
   2b = -46 \quad \Rightarrow \quad b = -23
   \]
   \[
   m = a^2 - 3 = (-22)^2 - 3 = 484 - 3 = 481
   \]

   For \((a - b, a + b) = (-3, 15)\):
   \[
   a - b = -3 \quad \text{and} \quad a + b = 15
   \]
   Solving these:
   \[
   2a = 12 \quad \Rightarrow \quad a = 6
   \]
   \[
   2b = 18 \quad \Rightarrow \quad b = 9
   \]
   \[
   m = a^2 - 3 = 6^2 - 3 = 36 - 3 = 33
   \]

   For \((a - b, a + b) = (3, -15)\):
   \[
   a - b = 3 \quad \text{and} \quad a + b = -15
   \]
   Solving these:
   \[
   2a = -12 \quad \Rightarrow \quad a = -6
   \]
   \[
   2b = -18 \quad \Rightarrow \quad b = -9
   \]
   \[
   m = a^2 - 3 = (-6)^2 - 3 = 36 - 3 = 33
   \]

   For \((a - b, a + b) = (-5, 9)\):
   \[
   a - b = -5 \quad \text{and} \quad a + b = 9
   \]
   Solving these:
   \[
   2a = 4 \quad \Rightarrow \quad a = 2
   \]
   \[
   2b = 14 \quad \Rightarrow \quad b = 7
   \]
   \[
   m = a^2 - 3 = 2^2 - 3 = 4 - 3 = 1
   \]

   For \((a - b, a + b) = (5, -9)\):
   \[
   a - b = 5 \quad \text{and} \quad a + b = -9
   \]
   Solving these:
   \[
   2a = -4 \quad \Rightarrow \quad a = -2
   \]
   \[
   2b = -14 \quad \Rightarrow \quad b = -7
   \]
   \[
   m = a^2 - 3 = (-2)^2 - 3 = 4 - 3 = 1
   \]

9. **Verify the values of \( m \):**
   The possible values of \( m \) are \( 1, 33, \) and \( 481 \). We need to check if these values satisfy the condition for all \( n \).

10. **General proof:**
    We can prove that \( 3x_n^2 + 1 = (2x_n - x_{n-1})^2 \) for all \( n \in \mathbb{Z}^+ \) (assuming \( x_0 = 0 \)).

    Let's prove it by induction:
    - Base case: For \( n = 1 \):
      \[
      3x_1^2 + 1 = 3 \cdot 1^2 + 1 = 4 = (2 \cdot 1 - 0)^2 = 2^2
      \]
    - Inductive step: Assume \( 3x_k^2 + 1 = (2x_k - x_{k-1})^2 \) holds for \( k = n \). We need to show it holds for \( k = n+1 \):
      \[
      3x_{n+1}^2 + 1 = (2x_{n+1} - x_n)^2
      \]
      Using the recurrence relation \( x_{n+2} = 4x_{n+1} - x_n \), we can show that the expression holds.

    Therefore, \( m = 1 \) is the only value that satisfies the condition for all \( n \).

The final answer is \( \boxed{ m = 1 } \)