Problem 712

From the observation, if $(a, b)$ is a pair of numbers among the 48 given numbers, for example, $a=2^{7} \cdot 3^{8} \cdot 5^{9} \cdot 11, b=2^{8} \cdot 3^{7} \cdot 5^{l} \cdot 11 \cdot 13$, then the product $a b=\left(2^{4} \cdot 3^{7} \cdot 5^{*} \cdot 11\right)^{2} \cdot 3 \cdot 13$. Since $\left(2^{8} \cdot 3^{7} \cdot 5^{9} \cdot 11\right)^{2}$ is already a perfect square, we only need to consider the remaining prime numbers 3 and 13 in the product $a b$.

Let $(a, b)$ be any two numbers among the 18 given numbers, and express the product $a b$ as a perfect square and a set of prime numbers that are not squared. For example, the pair $(a, b)$ corresponds to the set $\{3,13\}$.

Since there are $C_{48}^{2}=1128$ different pairs of numbers chosen from 48 numbers, and there are $2^{n}=1024$ subsets of a set of 10 different prime numbers, and since 1128 > 1024, there must be two different pairs $(a, b)$ and $(c, d)$ that correspond to the same subset of prime numbers $\left\{p_{1}, p_{2}, \cdots, p_{k}\right\}$, where $0 \leqslant k \leqslant 10$. Therefore,
$$\begin{array}{l} a b=m^{2} p_{1} p_{2} \cdots p_{k}, c d=n^{2} p_{1} p_{2} \cdots p_{k}, \\ a b c d=\left(m n p_{1} p_{2} \cdots p_{\mathrm{k}}\right)^{2} \end{array}$$
is a perfect square.
If the two pairs $(a, b)$ and $(c, d)$ have no common elements, then $a, b, c, d$ are the desired numbers.
If these two pairs have a common element, let's say $b=d$, then $a c$ must be a perfect square.
In this case, consider the remaining 46 numbers. Since the product of these 46 numbers still has no more than 10 different prime factors, and there are $C_{48}^{2}=1035>1024=2^{n}$, we can certainly find two different pairs $(x, y)$ and $(u, v)$ such that $x y u v$ is a perfect square. If these two pairs have no common elements, then $x$, $y, u, v$ are the desired numbers; if they have a common element, say $x=v$, then $y u$ is a perfect square. In this way, $a, c, y, u$ are the desired numbers.