Let  $p$ a prime number, $p\geq 5$. Find the number of polynomials of the form
\[ x^p + px^k + p x^l + 1, \quad k > l, \quad k, l \in \left\{1,2,\dots,p-1\right\},  \] which are irreducible in $\mathbb{Z}[X]$.

[i]Valentin Vornicu[/i]
1. Consider the polynomial \( P(x) = x^p + px^k + px^l + 1 \) where \( p \) is a prime number, \( p \geq 5 \), and \( k > l \) with \( k, l \in \{1, 2, \ldots, p-1\} \).

2. To determine the irreducibility of \( P(x) \) in \( \mathbb{Z}[x] \), we first apply the Rational Root Theorem. According to this theorem, if \( P(x) \) has a rational root, it must be an integer that divides the constant term, which is 1. Therefore, the possible rational roots are \( x = 1 \) and \( x = -1 \).

3. **Case 1: \( x = 1 \)**
   \[
   P(1) = 1^p + p \cdot 1^k + p \cdot 1^l + 1 = 1 + p + p + 1 = 2 + 2p
   \]
   Since \( p \geq 5 \), \( 2 + 2p \) is clearly not zero. Therefore, \( x = 1 \) is not a root of \( P(x) \).

4. **Case 2: \( x = -1 \)**
   \[
   P(-1) = (-1)^p + p(-1)^k + p(-1)^l + 1
   \]
   Since \( p \) is an odd prime, \( (-1)^p = -1 \). Thus,
   \[
   P(-1) = -1 + p(-1)^k + p(-1)^l + 1 = p(-1)^k + p(-1)^l
   \]
   For \( P(-1) \) to be zero, we need:
   \[
   p(-1)^k + p(-1)^l = 0 \implies (-1)^k + (-1)^l = 0
   \]
   This implies that \( (-1)^k \) and \( (-1)^l \) must be of opposite signs, i.e., \( k \) and \( l \) must have opposite parities (one is even and the other is odd).

5. We need to count the number of pairs \( (k, l) \) such that \( k \) and \( l \) have opposite parities. There are \( \frac{p-1}{2} \) even numbers and \( \frac{p-1}{2} \) odd numbers in the set \( \{1, 2, \ldots, p-1\} \).

6. The number of ways to choose one even number and one odd number is:
   \[
   \left( \frac{p-1}{2} \right) \times \left( \frac{p-1}{2} \right) = \left( \frac{p-1}{2} \right)^2
   \]

7. Since \( k > l \), we need to consider the order of selection. For each pair \( (k, l) \) where \( k \) and \( l \) have opposite parities, there are two possible orders: \( (k, l) \) and \( (l, k) \). Therefore, we divide by 2 to avoid double-counting:
   \[
   \frac{1}{2} \left( \frac{p-1}{2} \right)^2 = \frac{(p-1)^2}{8}
   \]

8. The number of irreducible polynomials is the total number of pairs \( (k, l) \) minus the number of pairs where \( k \) and \( l \) have opposite parities:
   \[
   \binom{p-1}{2} - \frac{(p-1)^2}{8}
   \]

The final answer is \(\boxed{2 \binom{\frac{p-1}{2}}{2}}\).