Determine all natural numbers n for which the number $A = n^4 + 4n^3 +5n^2 + 6n$ is a perfect square of a natural number.
To determine all natural numbers \( n \) for which the number \( A = n^4 + 4n^3 + 5n^2 + 6n \) is a perfect square, we start by factoring \( A \).

1. **Factorization of \( A \):**
   \[
   A = n^4 + 4n^3 + 5n^2 + 6n = n(n+3)(n^2+n+2)
   \]

2. **Greatest Common Divisors (GCD) Analysis:**
   - \( \gcd(n, n+3) \) is either 1 or 3.
   - \( \gcd(n, n^2+n+2) \) is either 1 or 2.
   - \( \gcd(n+3, n^2+n+2) \) is either 1, 2, 4, or 8. This is because:
     \[
     n^2 + n + 2 = (n-2)(n+3) + 8
     \]
     If \( p \) divides both \( n+3 \) and \( n^2+n+2 \), then \( p \) must divide 8.

3. **Case Analysis:**
   - **Case 1: All \( n, n+3, n^2+n+2 \) are square numbers.**
     - If \( n \) and \( n+3 \) are square numbers, then \( n = 1 \) (since 1 is the only natural number such that both \( n \) and \( n+3 \) are squares).
     - For \( n = 1 \):
       \[
       n^2 + n + 2 = 1^2 + 1 + 2 = 4
       \]
       Hence, \( n = 1 \) is a solution.

   - **Case 2: \( n = 3k^2 \Rightarrow n+3 = 3(k^2+1) \).**
     - \( n+3 \) is never three times a square number.

   - **Case 3: \( n = 2k^2 \Rightarrow n+3 = 2k^2 + 3 \).**
     - \( n+3 \) is never a square number because if \( 2k^2 + 3 = r^2 \) for integer \( r \), then:
       \[
       2k^2 \equiv r^2 \pmod{3}
       \]
       This implies \( k \equiv 0 \pmod{3} \) and \( r \equiv 0 \pmod{3} \), leading to a contradiction.

   - **Case 4: \( n+3 = 2k^2 \Rightarrow n = 2k^2 - 3 \).**
     - \( n \) is never a square number (same argument as in Case 3).

   - **Case 5: \( n+3 = 4k^2 \Rightarrow n = (2k)^2 - 3 \).**
     - For \( n \) to be a square number, \( k = 1 \):
       \[
       n = 1
       \]
       We have already seen that \( n = 1 \) is a solution.

   - **Case 6: \( n+3 = 8k^2 = 2(2k)^2 \).**
     - This is the same as Case 4.

4. **Conclusion:**
   - From the above cases, \( A \) will be a perfect square if and only if \( n = 1 \).

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