Is pigeonhole principle here?

If the product of 2 elements of the set is a perfect square, means that we can express this product as $2^{2a}\times3^{2b}\times5^{2c}\times7^{2d} = (2^{a}\times3^{b}\times5^{c}\times7^{d})^2$ . That implies that all $a_1+a_2, b_1+b_2,c_1+c_2, d_1+d_2$ are even (given the product $(2^{a_1}\times3^{b_1}\times5^{c_1}\times7^{d_1})\times(2^{a_2}\times3^{b_2}\times5^{c_2}\times7^{d_2})$ ). To simplify things, we'll use values for $a,b,c,d$ equal to 0 or 1 . Thus, if the sum of exponents for the same prime is 0 or 2 for all primes, the product will be a perfect square, since 0 and 2 are even.

We can write all combinations of $(a,b,c,d)$ like binary numbers $\overline{abcd}$ , and if a number repeats, the product is a perfect square. Then, between 0 and $(1111)_2=(15)_{10}$ we have 16 distinct numbers. Therefore, using the pigeonhole principle, if we have one more element in the set, there must be at least 2 that their product is a perfect square. So $\boxed{17}$ is the solution.