If any nonsquare number has four different rectangular arrays, how many factors does it have?
If a number has four different rectangular arrays, it means that it can be written as the product of two different numbers in four different ways.
Let's consider a number $N$ and its prime factorization: $N=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}\cdots$.
If $N$ has four different rectangular arrays, it means that $N$ can be expressed as $N=(p_1^{b_1} \cdot p_2^{b_2})\cdot (p_3^{c_1} \cdot p_4^{c_2})$ and $(p_1^{b_1} \cdot p_3^{c_1})\cdot (p_2^{b_2} \cdot p_4^{c_2})$ and so on, where $b_1$, $b_2$, $c_1$, $c_2$ are positive integers such that $b_1+b_2=a_1$ and $c_1+c_2=a_2$.
The number of factors of $N$ is given by $(a_1+1)(a_2+1)(a_3+1)\cdots$.
In this case, we can rewrite $(a_1+1)(a_2+1)$ as $(b_1+b_2+1)(c_1+c_2+1)$, where $(b_1+b_2+1)$ represents the number of different rectangular arrays with $p_1^{b_1} \cdot p_2^{b_2}$ and $(c_1+c_2+1)$ represents the number of different rectangular arrays with $p_3^{c_1} \cdot p_4^{c_2}$.
Therefore, the number of factors of $N$ is $(b_1+b_2+1)(c_1+c_2+1)(a_3+1)(a_4+1)\cdots$.
Since each factor increases the total number of factors, the number of factors of $N$ is greater than or equal to $(a_1+1)(a_2+1)$.
Hence, any nonsquare number that has four different rectangular arrays has at least $(a_1+1)(a_2+1)$ factors.