TITLE: About interesting numbers
QUESTION [9 upvotes]: We call a natural number interesting if it can be decomposed into natural factors, each of which is less than $30$. Prove that out of $10000$ interesting numbers, we can always choose two, whose product is a complete square.
My attempt. Order a set of $10000$ thousand interesting numbers in ascending order
$$
a_1 < a_2 < \ldots < a_{10000}.
$$
Suppose, that $a_1 < 10000$. We note that the estimate $10000 = 100^2 < a_{10000}$. Then it is clear that under these restrictions there are a number $b < 100$, that $b^2 = a_n$, for some $n \in \{1,
\ldots,10000\}$.
But how to prove if $a_1 \ge 10000$?

REPLY [29 votes]: Interesting numbers can only have prime factors $<30$. There are $10$ such primes. For the purposes of this problem for each interesting number $n$ and each prime $<30$ it is only important whether this prime occurs with an odd or an even exponent in $n$. In other words, the interesting numbers considered here fall into $2^{10}=1024$ classes ${}^*)$, whereby the product of any two numbers in the same class is a square. It follows that  among $\geq1025$ interesting numbers there is a pair of the desired kind.
${}^*)$ Each interesting $n$ is of the form $n=2^{x_1}\cdot 3^{x_2}\cdot\ldots\cdot 29^{x_{10}}$. The parities (even or odd) of the $x_k$  can be captured in a binary word $(b_1,\ldots, b_{10})\in\{0,1\}^{10}$. There are $2^{10}$ such words.