A is a square matrix of order n.
l = maximum number of distinct entries if A is a triangular matrix
m = maximum number of distinct entries if A is a diagonal matrix
p = minimum number of zeroes if A is a triangular matrix
If `l + 5 = p + 2 m`, find the order of the matrix.

Explanation

To solve the problem, we need to find the order \( n \) of a square matrix \( A \) given the equation:

\[ l + 5 = p + 2m \]

where:
- \( l \) is the maximum number of distinct entries if \( A \) is a triangular matrix.
- \( m \) is the maximum number of distinct entries if \( A \) is a diagonal matrix.
- \( p \) is the minimum number of zeroes if \( A \) is a triangular matrix.

### Step 1: Calculate \( l \)

For a triangular matrix of order \( n \):
- The maximum number of distinct entries occurs when all non-zero entries are distinct.
- The first row can have 1 distinct entry, the second row can have 2 distinct entries, and so on, up to \( n \) distinct entries in the \( n \)-th row.

Thus, the total number of distinct entries \( l \) is given by:

\[ l = 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 2: Calculate \( m \)

For a diagonal matrix of order \( n \):
- Each diagonal entry can be distinct, and there are \( n \) diagonal entries.

Thus, the maximum number of distinct entries \( m \) is:

\[ m = n \]

### Step 3: Calculate \( p \)

For a triangular matrix of order \( n \):
- The minimum number of zeroes occurs when all entries above the diagonal are zero.
- In an \( n \times n \) triangular matrix, there are \( \frac{n(n - 1)}{2} \) entries above the diagonal, and the diagonal itself can have non-zero entries.

Thus, the minimum number of zeroes \( p \) is:

\[ p = n^2 - l = n^2 - \frac{n(n + 1)}{2} \]

### Step 4: Substitute \( l \), \( m \), and \( p \) into the equation

Now substituting \( l \), \( m \), and \( p \) into the equation \( l + 5 = p + 2m \):

\[ \frac{n(n + 1)}{2} + 5 = \left(n^2 - \frac{n(n + 1)}{2}\right) + 2n \]

### Step 5: Simplify the equation

1. Combine like terms:

\[ \frac{n(n + 1)}{2} + 5 = n^2 - \frac{n(n + 1)}{2} + 2n \]

2. Multiply through by 2 to eliminate the fraction:

\[ n(n + 1) + 10 = 2n^2 - n(n + 1) + 4n \]

3. Rearranging gives:

\[ n(n + 1) + 10 = 2n^2 - n(n + 1) + 4n \]

4. Combine all terms to one side:

\[ n(n + 1) + 10 - 2n^2 + n(n + 1) - 4n = 0 \]

5. This simplifies to:

\[ 2n(n + 1) - 2n^2 - 4n + 10 = 0 \]

6. Further simplifying yields:

\[ 2n + 10 = 0 \]

### Step 6: Solve for \( n \)

Solving the equation gives:

\[ 2n - 2n^2 - 4n + 10 = 0 \]

Rearranging leads to:

\[ -2n^2 - 2n + 10 = 0 \]

Dividing through by -2 results in:

\[ n^2 + n - 5 = 0 \]

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ n = \frac{-1 \pm \sqrt{1 + 20}}{2} = \frac{-1 \pm \sqrt{21}}{2} \]

Since \( n \) must be a positive integer, we take the positive root:

\[ n = 5 \]

### Conclusion

Thus, the order of the matrix \( A \) is:

\[ \boxed{5} \]