Let 

$$A=\left( \begin{array}{cc}

4 & -\sqrt{5} \\
2\sqrt{5} & -3

\end{array} \right) $$

Find all pairs of integers \(m,n\) with \(n \geq 1\) and \(|m| \leq n\) such as all entries of \(A^n-(m+n^2)A\) are integer.
To solve the problem, we need to find all pairs of integers \((m, n)\) with \(n \geq 1\) and \(|m| \leq n\) such that all entries of \(A^n - (m + n^2)A\) are integers. We start by diagonalizing the matrix \(A\).

1. **Diagonalize the matrix \(A\):**
   The matrix \(A\) is given by:
   \[
   A = \begin{pmatrix}
   4 & -\sqrt{5} \\
   2\sqrt{5} & -3
   \end{pmatrix}
   \]
   To diagonalize \(A\), we need to find its eigenvalues and eigenvectors.

2. **Find the eigenvalues of \(A\):**
   The characteristic polynomial of \(A\) is given by:
   \[
   \det(A - \lambda I) = \det \begin{pmatrix}
   4 - \lambda & -\sqrt{5} \\
   2\sqrt{5} & -3 - \lambda
   \end{pmatrix} = (4 - \lambda)(-3 - \lambda) - (-\sqrt{5})(2\sqrt{5})
   \]
   Simplifying this, we get:
   \[
   (4 - \lambda)(-3 - \lambda) + 10 = \lambda^2 - \lambda - 2
   \]
   Solving the quadratic equation \(\lambda^2 - \lambda - 2 = 0\), we find the eigenvalues:
   \[
   \lambda = 2 \quad \text{and} \quad \lambda = -1
   \]

3. **Find the eigenvectors of \(A\):**
   For \(\lambda = 2\):
   \[
   (A - 2I) \mathbf{v} = 0 \implies \begin{pmatrix}
   2 & -\sqrt{5} \\
   2\sqrt{5} & -5
   \end{pmatrix} \begin{pmatrix}
   v_1 \\
   v_2
   \end{pmatrix} = 0
   \]
   Solving this system, we get the eigenvector \(\mathbf{v}_1 = \begin{pmatrix}
   \sqrt{5} \\
   2
   \end{pmatrix}\).

   For \(\lambda = -1\):
   \[
   (A + I) \mathbf{v} = 0 \implies \begin{pmatrix}
   5 & -\sqrt{5} \\
   2\sqrt{5} & -2
   \end{pmatrix} \begin{pmatrix}
   v_1 \\
   v_2
   \end{pmatrix} = 0
   \]
   Solving this system, we get the eigenvector \(\mathbf{v}_2 = \begin{pmatrix}
   \sqrt{5} \\
   5
   \end{pmatrix}\).

4. **Form the diagonal matrix \(D\) and the matrix \(P\):**
   \[
   D = \begin{pmatrix}
   2 & 0 \\
   0 & -1
   \end{pmatrix}, \quad P = \begin{pmatrix}
   \sqrt{5} & \sqrt{5} \\
   2 & 5
   \end{pmatrix}
   \]
   Then, \(A = PDP^{-1}\).

5. **Compute \(A^n\):**
   \[
   A^n = (PDP^{-1})^n = PD^nP^{-1}
   \]
   Since \(D\) is diagonal, \(D^n\) is easy to compute:
   \[
   D^n = \begin{pmatrix}
   2^n & 0 \\
   0 & (-1)^n
   \end{pmatrix}
   \]

6. **Compute \(A^n - (m + n^2)A\):**
   \[
   A^n - (m + n^2)A = PD^nP^{-1} - (m + n^2)PDP^{-1}
   \]
   Simplifying, we get:
   \[
   A^n - (m + n^2)A = P(D^n - (m + n^2)D)P^{-1}
   \]
   \[
   D^n - (m + n^2)D = \begin{pmatrix}
   2^n - 2(m + n^2) & 0 \\
   0 & (-1)^n + (m + n^2)
   \end{pmatrix}
   \]

7. **Ensure all entries are integers:**
   For the entries to be integers, we need:
   \[
   2^n - 2(m + n^2) \quad \text{and} \quad (-1)^n + (m + n^2)
   \]
   to be integers. This gives us the conditions:
   \[
   \frac{2^n - (-1)^n}{3} = m + n^2
   \]

8. **Check for \(n \geq 8\):**
   For \(n \geq 8\), an induction shows:
   \[
   \frac{2^n - (-1)^n}{3} > n^2 + n
   \]
   This contradicts \(|m| \leq n\). So, we must have \(1 \leq n \leq 7\).

9. **Check for \(1 \leq n \leq 7\):**
   By checking these values, we find the solutions:
   \[
   (m, n) = (0, 1) \quad \text{and} \quad (-6, 7)
   \]

The final answer is \( \boxed{ (m, n) = (0, 1) \quad \text{and} \quad (-6, 7) } \)