No. of subsets whose element multiply to give a square number
Factor each element $x_i$ to produce a vector $v_i$ of the powers of all primes which are at most 70. The product $\prod_{i \in S} x_i$ is a perfect square if every prime appears in an even power, that is, if $\sum_{i \in S} v_i = \mathbf0 \pmod{2}$. This suggests considering $v_i$ as a vector over $\mathbb{Z}_2$ (the integers mod 2). You are now given a matrix, and wish to know how many subsets of rows sum to the zero vector. This is a problem in linear algebra, and the answer depends on the rank of the matrix: if the matrix has $n$ rows, $m$ columns, and rank $r$, then the answer is $2^{n-r}$.

Here is a worked out example for $1,2,4,5,8$. In this case there are two relevant primes: $2,5$. The full matrix (before reducing modulo 2) is
$$
\begin{bmatrix}
0 &amp; 0 \\ 1 &amp; 0 \\ 2 &amp; 0 \\ 0 &amp; 1 \\ 3 &amp; 0
\end{bmatrix}
$$
After reducing modulo 2, the matrix becomes
$$
\begin{bmatrix}
0 &amp; 0 \\ 1 &amp; 0 \\ 0 &amp; 0 \\ 0 &amp; 1 \\ 1 &amp; 0
\end{bmatrix}
$$
This matrix has full rank (rank 2), and so the answer is $2^{5-2} = 8$.

Here are the 8 subsets:
$$
\emptyset \\
\{1\} \\
\{4\} \\
\{1,4\} \\
\{2,8\} \\
\{1,2,8\} \\
\{2,4,8\} \\
\{1,2,4,8\}
$$

As noted in the comments, you might want to disallow the empty set (which multiplies to 1). In this case, the formula becomes $2^{n-r}-1$.