Find all perfect squares $n$ such that if the positive integer $a\ge 15$ is some divisor $n$ then $a+15$ is a prime power.

[i]Proposed by Saudi Arabia[/i]
To find all perfect squares \( n \) such that if the positive integer \( a \geq 15 \) is some divisor of \( n \), then \( a + 15 \) is a prime power, we proceed as follows:

1. **Initial Perfect Squares**:
   From the definition, the numbers \( n = 1, 4, 9 \) satisfy the required property because they do not have any divisors \( \geq 15 \).

2. **Assume \( n = t^2 \geq 15 \)**:
   Suppose \( n = t^2 \geq 15 \) also satisfies the required property. We need to check the cases where \( n \) is a power of 2 or has an odd prime divisor.

3. **Case 1: \( n \) is a power of 2**:
   - If \( n = 2^k \), then \( a = 2^m \) for some \( m \geq 4 \) (since \( 2^4 = 16 \geq 15 \)).
   - We check the values:
     \[
     2^4 + 15 = 31 \quad (\text{prime})
     \]
     \[
     2^5 + 15 = 47 \quad (\text{prime})
     \]
     \[
     2^6 + 15 = 79 \quad (\text{prime})
     \]
     \[
     2^7 + 15 = 143 = 11 \cdot 13 \quad (\text{not a prime power})
     \]
   - Therefore, \( n \) can only be \( 2^4 = 16 \) or \( 2^6 = 64 \).

4. **Case 2: \( n \) has an odd prime divisor \( p \)**:
   - If \( p > 3 \), then \( p^2 \mid n \) and \( p^2 > 15 \), implying \( p^2 + 15 \) must be a prime power.
   - Since \( p \) is odd, \( p^2 + 15 \) is even, so \( p^2 + 15 = 2^m \).
   - Considering modulo 3:
     \[
     p^2 + 15 \equiv 0 \text{ or } 1 \pmod{3}
     \]
     \[
     2^m \equiv 1 \pmod{3} \implies m \text{ is even}
     \]
   - Let \( m = 2k \):
     \[
     (2^k - p)(2^k + p) = 15
     \]
     - The factors of 15 are \( (1, 15) \) and \( (3, 5) \). Since \( p \geq 10 \):
       \[
       2^k - p = 1 \quad \text{and} \quad 2^k + p = 15 \implies p = 7, k = 3
       \]
     - Thus, \( p = 7 \) and \( 2^3 = 8 \):
       \[
       n = 7^2 = 49
       \]

5. **Combining Results**:
   - We have \( n = 4^x \cdot 49^z \) for some non-negative integers \( x, z \).
   - Since \( 27 \) is not nice, \( y \leq 1 \).
   - Numbers \( 18 \) and \( 21 \) are not nice, so \( x, y \) and \( y, z \) cannot both be positive.
   - Therefore, we consider \( n = 4^x \cdot 49^z \) with \( z \geq 1 \).

6. **Checking \( z = 1 \)**:
   - \( 7^2 + 15 = 2^6 \)
   - \( 2 \cdot 7^2 + 15 = 113 \)
   - \( 4 \cdot 7^2 + 15 = 211 \)
   - \( 8 \cdot 7^2 \) is not nice, so only \( n = 49, 196 \) satisfy the required property.

Therefore, the perfect squares \( n \) that satisfy the given property are:
\[
n = 1, 4, 9, 16, 49, 64, 196
\]

The final answer is \(\boxed{1, 4, 9, 16, 49, 64, 196}\)