[Example 5.5.5] If $M$ is a subset of the set $\{1,2,3, \cdots, 15\}$, and the product of any three distinct elements in $M$ is not a square number, use $|M|$ to denote the number of elements in the set $M$, find the maximum value of $|M|$.
Let $A=\{1,2,3, \cdots, 15\}$, the problem is to find a subset $M$ of $A$ such that the product of any three distinct elements in $M$ is not a perfect square. This is not easy to think through, especially when we need to find the subset $M$ that contains the maximum number of elements that meet the requirement, which is even more challenging.

We can only approach this from the opposite direction, thinking about which three elements in set $A$ have a product that is a perfect square, and then try to remove some numbers from $M$.
Which three numbers in $A$ have a product that is a perfect square?
Since $1 \times 4 \times 9=36=6^{2}, 2 \times 6 \times 12=144=12^{2}, 3 \times 5 \times 15=15^{2}, 7 \times$ $8 \times 14=28^{2}$, we have
$$
\{1,4,9\},\{2,6,12\},\{3,5,15\},\{7,8,14\}
$$

These 4 triplets have a product that is a perfect square.
Since the product of any three distinct elements in $M$ is not a perfect square, none of the above 4 triplets can appear in set $M$. To maximize the number of elements in $M$, a simple idea is to remove one element from each of the 4 triplets, removing a total of 4 elements. This might meet the requirement, so $|M| \leqslant 15-4=11$.
Therefore, our first task is to prove that $|M| \geqslant 12$ is impossible.
If $|M| \geqslant 12$, then there are at most 3 elements in set $A$ that do not belong to $M$. Thus, one of the 4 triplets must have all its elements in set $M$, which contradicts the requirement that the product of any three distinct elements in $M$ is not a perfect square.
Therefore, $|M| \geqslant 12$ is impossible, and $|M| \leqslant 11$.
Thus, we must remove at least one element from each of the 4 triplets in $A$.

We also need to note that none of the 4 triplets include the element 10, and 10 is a special element because $2 \times 5 \times 10=10^{2}, 6 \times 15 \times 10=900=30^{2}$. There are also 2 elements not included in the triplets, which are 11 and 13. Since 11 and 13 are prime numbers, they do not have the same adaptability in forming perfect squares as 10.
Therefore, we need to pay special attention to element 10.
Our second task is to discuss the relationship between element 10 and set $M$.
(1) If $10 \notin M$, this means that we not only remove one element from each of the 4 triplets in $\{1,2,3, \cdots, 15\}$, but also remove 10, so $|M| \leqslant 10$.
(2) If $10 \in M$, we can still prove that $|M| \leqslant 10$.
Due to the role of 10, we can recombine the subsets $\{2,6,12\},\{3,5,15\}$ into $\{2,5\},\{6,15\},\{3,12\}$. The reason for this recombination is that the first two pairs, when combined with 10, form a triplet whose product is a perfect square.
Now let's discuss $\{3,12\}$.
First case: At least one of 3 and 12 is not in $M$, i.e., we remove one of 3 and 12 from $M$, and then remove 3 more elements from the other 12 elements (excluding 3, 12, and 10).

And keep $10 \in M$, this is a subset $M$ composed of 11 elements, i.e., $M$ is composed of 9 elements from $\{1,2,4, 5,6,7,8,9,11,13,14,15\}$, plus 10, and one of 3 or 12.
Since $\{2,5\},\{6,15\},\{1,4,9\},\{7,8,14\}$ are 4 pairwise disjoint subsets (excluding 3 and 12), if $|M|=11$, and since $M$ must include 10 and one of 3 or 12, to ensure that the product of any three elements in $M$ is not a perfect square, $M$ should include 11 and 13. At this point, $M$ already has $10,11,13$ and one of 3 or 12, and $M$ still needs 7 more elements. If we take 7 elements from $\{2,5\},\{6,15\},\{1,4,9\},\{7,8,14\}$, one of the subsets must be in $M$, otherwise, if we take 1 element from the pairs $\{2,5\},\{6,15\}$ and 2 elements from the triplets $\{1,4,9\},\{7,8,14\}$, there are only 6 elements in total, so we need to take 1 more element, which will make one of the 4 subsets completely in $M$. If $\{2,5\}$ or $\{6,15\}$ is in $M$, adding 10 will form a triplet whose product is a perfect square. If $\{1,4,9\}$ or $\{7,8,14\}$ is in $M$, there will be a triplet whose product is a perfect square.
From the above analysis, $|M| \neq 11$, i.e., $M \leqslant 10$.
Second case: Both 3 and 12 are in $M$, since $3 \times 12=36=6^{2}$, at this point $\{1\}$, $\{4\},\{9\},\{2,6\},\{5,15\},\{7,8,14\}$, none of these can be a subset of $M$, otherwise there will be $1 \times 3 \times 12=6^{2}, 3 \times 4 \times 12=12^{2}, 3 \times 9 \times 12=18^{2}, 2 \times 6 \times 12=12^{2}, 3 \times 5 \times 15=15^{2}, 7 \times 8 \times 14=28^{2}$, all of which are perfect squares. To ensure that the product of any three elements in $M$ is not a perfect square, each of the 6 sets $\{1\},\{4\},\{9\},\{2,6\},\{5,15\},\{7,8,14\}$ must have at least one element not in $M$, so $M$ can have at most $15-6=9$ elements, and $|M|$ cannot be maximized.
Therefore, $|M| \leqslant 10$.
Now let's construct a subset $M$ that meets the requirements and has 10 elements.
From $\{1,4,9\}$, remove 9, leaving $\{1,4\}$;
From $\{3,6,12\}$, remove 2, leaving $\{6,12\}$;
From $\{3,5,15\}$, remove 3 and 15, leaving $\{5\}$;
From $\{7,8,14\}$, remove 8, leaving $\{7,14\}$.
Add $10,11,13,12$ to these 4 numbers,
Thus, we form the set $M=\{1,4,5,6,7,10,11,12,13,14\}$, in which
the product of any three numbers is not a perfect square, and this set has 10 elements, i.e., $|M|$
$=10$.