Find all natural integers $n$ such that $(n^3 + 39n - 2)n! + 17\cdot 21^n + 5$ is a square. 
To find all natural integers \( n \) such that \((n^3 + 39n - 2)n! + 17 \cdot 21^n + 5\) is a square, we can proceed as follows:

1. **Consider the expression modulo 3:**
   \[
   (n^3 + 39n - 2)n! + 17 \cdot 21^n + 5 \pmod{3}
   \]
   We need to analyze each term modulo 3.

2. **Simplify each term modulo 3:**
   - \( n^3 \pmod{3} \) can be 0, 1, or 2 depending on \( n \).
   - \( 39n \equiv 0 \pmod{3} \) since 39 is divisible by 3.
   - \( -2 \equiv 1 \pmod{3} \).
   - \( n! \pmod{3} \):
     - For \( n \geq 3 \), \( n! \) contains the factor 3, so \( n! \equiv 0 \pmod{3} \).
     - For \( n = 1 \) or \( n = 2 \), \( n! \) is 1 and 2 respectively.
   - \( 21^n \equiv 0 \pmod{3} \) since 21 is divisible by 3.
   - \( 17 \cdot 21^n \equiv 0 \pmod{3} \).
   - \( 5 \equiv 2 \pmod{3} \).

3. **Combine the terms for \( n \geq 3 \):**
   \[
   (n^3 + 39n - 2)n! + 17 \cdot 21^n + 5 \equiv (n^3 + 1) \cdot 0 + 0 + 2 \equiv 2 \pmod{3}
   \]
   However, a perfect square modulo 3 can only be 0 or 1. Therefore, for \( n \geq 3 \), the expression cannot be a perfect square.

4. **Check \( n = 1 \) and \( n = 2 \):**
   - For \( n = 1 \):
     \[
     (1^3 + 39 \cdot 1 - 2) \cdot 1! + 17 \cdot 21^1 + 5 = (1 + 39 - 2) \cdot 1 + 17 \cdot 21 + 5 = 38 + 357 + 5 = 400
     \]
     \( 400 = 20^2 \), which is a perfect square.
   - For \( n = 2 \):
     \[
     (2^3 + 39 \cdot 2 - 2) \cdot 2! + 17 \cdot 21^2 + 5 = (8 + 78 - 2) \cdot 2 + 17 \cdot 441 + 5 = 84 \cdot 2 + 7497 + 5 = 168 + 7497 + 5 = 7670
     \]
     \( 7670 \) is not a perfect square.

5. **Conclusion:**
   The only natural integer \( n \) such that \((n^3 + 39n - 2)n! + 17 \cdot 21^n + 5\) is a square is \( n = 1 \).

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