Let $S$ be the smallest set of positive integers such that

a) $2$ is in $S,$
b) $n$ is in $S$ whenever $n^2$ is in $S,$ and
c) $(n+5)^2$ is in $S$ whenever $n$ is in $S.$

 Which positive integers are not in $S?$

 (The set $S$ is ``smallest" in the sense that $S$ is contained in any other such set.)
To determine which positive integers are not in the set \( S \), we need to analyze the conditions given:

1. \( 2 \) is in \( S \).
2. If \( n^2 \) is in \( S \), then \( n \) is in \( S \).
3. If \( n \) is in \( S \), then \( (n+5)^2 \) is in \( S \).

We need to find the smallest set \( S \) that satisfies these conditions.

### Step-by-Step Analysis:

1. **Initial Element:**
   - By condition (a), \( 2 \in S \).

2. **Generating New Elements:**
   - By condition (c), since \( 2 \in S \), we have:
     \[
     (2+5)^2 = 7^2 = 49 \in S
     \]
   - By condition (b), since \( 49 \in S \), we have:
     \[
     7 \in S
     \]
   - By condition (c), since \( 7 \in S \), we have:
     \[
     (7+5)^2 = 12^2 = 144 \in S
     \]
   - By condition (b), since \( 144 \in S \), we have:
     \[
     12 \in S
     \]
   - By condition (c), since \( 12 \in S \), we have:
     \[
     (12+5)^2 = 17^2 = 289 \in S
     \]
   - By condition (b), since \( 289 \in S \), we have:
     \[
     17 \in S
     \]

3. **Pattern Recognition:**
   - We observe that starting from \( 2 \), we can generate elements of the form \( 2 + 5k \) for \( k \geq 0 \). This means all numbers of the form \( 2 \pmod{5} \) are in \( S \).

4. **Further Analysis:**
   - We need to check other residues modulo 5:
     - For \( n \equiv 3 \pmod{5} \):
       \[
       2 \to 49 \to 54 \to 59 \to 64 \to 8 \in S
       \]
       This implies all numbers \( \equiv 3 \pmod{5} \) greater than 3 are in \( S \).
     - For \( n \equiv 4 \pmod{5} \):
       \[
       2 \to 49 \to 54^2 = 2916 \to \cdots \to 3136 = 56^2 \to 56 \to 61 \to 66 \to \cdots \to 81 \to 9 \in S
       \]
       This implies all numbers \( \equiv 4 \pmod{5} \) greater than or equal to 9 are in \( S \).
     - For \( n \equiv 1 \pmod{5} \):
       \[
       81 \to 86 \to \cdots \to 121 \to 11 \in S
       \]
       This implies all numbers \( \equiv 1 \pmod{5} \) greater than or equal to 11 are in \( S \).

5. **Remaining Elements:**
   - Since \( 9 \in S \), by condition (b), \( 3 \in S \).
   - To show \( 4 \in S \):
     \[
     11 \to 16 \to 4 \in S
     \]
   - To show \( 6 \in S \):
     \[
     36 \in S \implies 6 \in S
     \]

### Conclusion:
The only positive integers that are not in \( S \) are those that are either \( 1 \) or multiples of \( 5 \).

The final answer is \(\boxed{n=1 \text{ or } 5 \mid n}\).