Problem 575

We begin by determining the prime factorization of $2025$. $$\begin{align*} 2025&= 25\times81\\ &=5\times5\times9\times9\\ &=5\times5\times3\times3\times3\times3\\ &=3^4\times5^2\end{align*}$$ There are exactly $15$ positive factors of $2025$. These are:

$1$, $3$, $3^2$, $3^3$, $3^4$, $5$, $5^2$, $3\times5$, $3^2\times5$, $3^3\times5$, $3^4\times 5$, $3\times5^2$, $3^2\times5^2$, $3^3\times5^2$, and $3^4\times 5^2$

We are asked to express $2025$ as
the product of two positive integers $n$ and $m^2$, where $m^2$ is a perfect square. Of the $15$ positive factors, the following are
perfect squares:

$1$, $3^2$, $3^4$, $5^2$, $3^2\times5^2$, and $3^4\times 5^2$

These are all the possible values of $m^2$, and so the values of $m$ are: $1$, $3$, $5$, $3^2$, $3\times5$, and $3^2\times5$.

Thus, the ordered pairs of positive integers $(m,n)$ for which $m^2\times n=2025$ are:

$(1,3^4\times5^2)$, $(3,3^2\times5^2)$, $(5,3^4)$, $(3^2,5^2)$, $(3\times5,3^2)$, $(3^2\times5,1)$

Therefore, there are $6$ such
ordered pairs.

It is interesting to note that each of the $6$ values of $n$ is also a perfect square. Can you see
why this occurs?