Find all integers $a$ such that there is a prime number of $p\ge  5$ that divides
 ${p-1 \choose 2}$ $+ {p-1 \choose 3} a$ $+{p-1 \choose 4} a^2$+ ...+$ {p-1 \choose p-3} a^{p-5} .$
1. **Define the problem and the sum \( S \):**
   We need to find all integers \( a \) such that there exists a prime number \( p \geq 5 \) that divides the sum:
   \[
   S = \binom{p-1}{2} + \binom{p-1}{3} a + \binom{p-1}{4} a^2 + \cdots + \binom{p-1}{p-3} a^{p-5}.
   \]

2. **Assume \( a \) is good:**
   Assume \( a \) is good, meaning there exists a prime \( p \) such that \( p \mid S \). We denote the given sum by \( S \).

3. **Check if \( p \mid a \):**
   Note that \( p \nmid a \). If \( p \mid a \), then \( p \mid \binom{p-1}{2} \), which is not possible because:
   \[
   \binom{p-1}{2} = \frac{(p-1)(p-2)}{2}
   \]
   and \( p \) does not divide \( (p-1)(p-2) \).

4. **Use the binomial theorem:**
   Using the binomial theorem, we can express \( S \) as:
   \[
   S = \frac{(1+a)^{p-1} - a^{p-1} - (p-1)a^{p-2} - (p-1)a - 1}{a^2}.
   \]
   Therefore, if \( p \mid S \), then:
   \[
   p \mid (1+a)^{p-1} - a^{p-1} - (p-1)a^{p-2} - (p-1)a - 1.
   \]

5. **Simplify using Fermat's Little Theorem:**
   If \( a \equiv -1 \pmod{p} \), then:
   \[
   (1+a)^{p-1} \equiv 0 \pmod{p}.
   \]
   This implies \( p = 2 \), which is not possible since \( p \geq 5 \). Hence, \( a \not\equiv -1 \pmod{p} \).

6. **Apply Fermat's Little Theorem:**
   Using Fermat's Little Theorem, \( (1+a)^{p-1} \equiv 1 \pmod{p} \) and \( a^{p-1} \equiv 1 \pmod{p} \). Therefore:
   \[
   (1+a)^{p-1} \equiv a^{p-1} \pmod{p}.
   \]
   This implies:
   \[
   p \mid a^{p-2} + a + 1.
   \]

7. **Check the polynomial \( a^2 + a + 1 \):**
   We need \( p \mid a^2 + a + 1 \). Note that \( a^2 + a + 1 \) is always odd, so such a \( p \) always exists unless \( a^2 + a + 1 \) is a power of three.

8. **Check if \( a^2 + a + 1 = 3^k \):**
   Since \( a^2 + a + 1 \) is not divisible by 9 (otherwise \( 9 \mid (2a+1)^2 + 3 \) and \(-3\) is not a quadratic residue modulo 9), \( a \) is good unless \( a^2 + a + 1 = 3 \).

9. **Solve \( a^2 + a + 1 = 3 \):**
   Solving \( a^2 + a + 1 = 3 \) gives:
   \[
   a^2 + a - 2 = 0.
   \]
   Factoring, we get:
   \[
   (a+2)(a-1) = 0.
   \]
   Therefore, \( a = -2 \) or \( a = 1 \).

Conclusion:
Thus, \( a \) is good unless \( a \in \{-2, 1\} \).

The final answer is \( \boxed{ a \in \mathbb{Z} \setminus \{-2, 1\} } \).