Uhm $2016$ is so weak? There are exactly $10$ prime factors smaller than $30$. Look at a number $N \in A$ and it's prime factorization. \[ N = 2^{e_1}3^{e_2} \dots 29^{e_{10}}. \]Look at $e_1,e_2,\dots,e_{10}$ in modulo $2$. Then there are $2^{10}$ possibilities. As $2^{10}<2016$, there are two numbers $a,b \in A$, such that all their exponents are equivalent modulo $2$ by the Pigeonhole Principle. Now remove $a,b$ from $A$, we still have $2^{10}<2014$, so repeat the process to pick $c,d \in A$. Well, observe that $ab$ and $cd$ are both squares. That implies that $abcd$ is a perfect square as well. Done. $\hfill \square$

See here and here for similar problems.