$ M$ is a subset of $ \{1, 2, 3, \ldots, 15\}$ such that the product of any three distinct elements of $ M$ is not a square. Determine the maximum number of elements in $ M.$
To solve the problem, we need to determine the maximum number of elements in a subset \( M \) of \(\{1, 2, 3, \ldots, 15\}\) such that the product of any three distinct elements of \( M \) is not a perfect square.

1. **Initial Consideration**:
   - Let \( |M| \) denote the number of elements in the set \( M \).
   - We need to show that if \( |M| = 11 \) does not work, then \( |M| = 12, 13, 14, 15 \) also do not work.

2. **Claim 1**:
   - If \( |M| = 11 \) does not work, then \( |M| = 12, 13, 14, 15 \) also do not work.
   - **Proof**: Consider any 11 elements from \( M \geq 12 \). We must find 3 elements \( a, b, c \) from \( M \) such that \( abc = x^2 \). As we assumed \( |M| = 11 \) does not work, for every \( |M| = 11 \), we will get three elements whose product is a perfect square. \(\blacksquare\)

3. **Claim 2**:
   - \( |M| = 11 \) does not work.
   - **Proof**:
     - Consider the following sets of triples:
       \[
       \begin{aligned}
       &A_1 = (2, 1, 8), \\
       &A_2 = (2, 3, 6), \\
       &A_3 = (2, 4, 8), \\
       &A_4 = (2, 5, 10), \\
       &A_5 = (2, 6, 12), \\
       &A_6 = (2, 7, 14), \\
       &A_7 = (2, 8, 9).
       \end{aligned}
       \]
     - Let \( 2 \in M \). Then we have \( 6, 8 \notin M \). Also, one of \( (5, 10) \) and one of \( (7, 14) \) cannot be in \( M \).
     - Consider \( T = \{1, 4, 9\} \). All of these must be in \( M \), but \( 1 \times 4 \times 9 = 36 = 6^2 \), which is a contradiction.
     - Therefore, \( 2 \notin M \).

4. **Further Consideration**:
   - Consider the following sets of triples:
     \[
     \begin{aligned}
     &B_1 = (3, 1, 12), \\
     &B_2 = (3, 2, 6), \\
     &B_3 = (3, 4, 12), \\
     &B_4 = (3, 5, 15), \\
     &B_5 = (3, 6, 8), \\
     &B_6 = (3, 9, 12).
     \end{aligned}
     \]
   - Let \( 3 \in M \). Then we have \( 6, 12 \notin M \). Also, one of \( (5, 15) \) cannot be in \( M \).
   - Again, consider \( T = \{1, 4, 9\} \). All of these must be in \( M \), which is a contradiction.
   - Therefore, \( 3 \notin M \).

5. **Final Consideration**:
   - We need to find two more elements which cannot be in \( M \). Consider:
     \[
     \begin{aligned}
     &T = \{1, 4, 9\}, \\
     &N = \{5, 8, 10\}, \\
     &K = \{5, 12, 15\}, \\
     &L = \{7, 8, 14\}.
     \end{aligned}
     \]
   - Let \( 5 \in M \). Then one of \( (8, 10) \) and one of \( (12, 15) \) cannot be in \( M \). But then again consider \( T = \{1, 4, 9\} \), which is a contradiction.
   - Therefore, \( 5 \notin M \).

6. **Conclusion**:
   - From these considerations, we can construct \( |M| = 10 \) elements by excluding \( 1, 2, 3, 5, 7 \):
     \[
     M = \{4, 6, 8, 9, 10, 11, 12, 13, 15\}.
     \]
   - This set satisfies the condition that the product of any three distinct elements is not a perfect square.

The final answer is \( \boxed{10} \).