For how many two-element subsets $\{a, b\}$ of the set $\{1, 2, 3, 4, 5, 6, \dots , 36\}$ is the product $ab$ a perfect square? Problem:
For how many two-element subsets $\{a, b\}$ of the set $\{1, 2, 3, 4, 5, 6, \dots , 36\}$ is the product $ab$ a perfect square?
Note:
The numbers $a$ and $b$ are distinct, so the $a$ and $b$ must be different integers from the set of numbers $1 - 36$.
Reiteration:
Using two integers selected from the number group, $1-36$, how many combinations are possible so that the product of the two numbers selected is a perfect square.
Attempt:
First list out the square numbers between 1-36 (inclusive):
1, 4, 9, 16, 25, 36.
NOTE: The square number “1” does not work because a and b need to be distinct.
Rewrite the numbers paired with the number “1” so that it is in the format {a, b}.
(4, 1)
(9, 1)
(16, 1)
(25, 1)
(36, 1)
Next, since the number “16” for example, it can be broken down into 4•4 or 8•2, the subset {8, 2} also works.
4 : 2•2 (Note: 2•2 does not work because a and b need to be distinct)
16 : 4•4 or 8•2 (Note: 4•4 does not work because a and b need to be distinct)
36 : 2•18 or 3•12 or 4•9
Currently, we have found 9 solutions.
Issue:
I currently found 9 out of the total number of solutions that satisfy this problem. But, I have ran into an issue : What is the most efficient way to organize your solutions and find other combinations?
Attempt Continued:
I then continued to list out the combinations but started with the sets with “1” in the front and then 2, then 3, then 4, etc.
Subsets which start with “1”:
{1, 4}
{1, 9}
{1, 16}
{1, 25}
{1, 36}
Subsets which start with “2”:
{2, 8}
{2, 18}
{2, 32}
Subsets which start with “3”:
{3, 12}
{3, 27}
Subsets which start with “4”:
{4, 9}
{4, 16}
{4, 25}
{4, 36}
Subsets which start with “5”:
{5, 20}
Subsets which start with “6”:
{6, 24}
Subsets which start with “7”:
{7, 28}
Subsets which start with “8”:
{8, 18}
{8, 16}
Conclusion:
The list keeps going on. The correct answer for this question is 27. If you carefully calculate and list out the subsets, you can get the answer.

The challenge here is to think of some efficient way to "see" the solution. Here's one idea, based on the OP's first step, namely isolating the squares:
Partition the integers from $1$ to $36$ into square multiples of squarefree numbers:
$$\begin{align}
&\{1,4,9,16,25,36\}\\
&\{2,8,18,32\}\\
&\{3,12,27\}\\
&\{5,20\}\\
&\{6,24\}\\
&\{7,28\}\\
\end{align}$$
where we don't need to go any further because every other number belongs to a singleton. Now pairs $a$ and $b$ that multiply to a square must both come from the same set. We thus have
$${6\choose2}+{4\choose2}+{3\choose2}+{2\choose2}+{2\choose2}+{2\choose2}=15+6+3+1+1+1=27$$
pairs in all.
