Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers. Find the total number of ordered pairs $(m,n)$ such that $m^2n=20^{20}$ where $m,n$ are positive integers. 
My Approach: Simplifying $$m^2n=20^{20}=(2^2 \times 5)^{20}=2^{40} \times 5^{20}$$
Hence one way is $n=5^{20}, m=2^{20}$ or vice-versa or $m=2^{10},n=10^{20}$ and $m=5^{10},n=4^{20}$. These are the only ways I could think of initially but there are several combinations possible for values of $m,n$. If anyone can help spot a pattern or what are the possible values of $m,n$, I 'll try again to proceed further. I believe that there are many integers possible but I am not able to apply the correct logic.
Thank You.

Let $m=2^a5^b$ and $n=2^c5^d$.
So,
\begin{align*}
m^2n&=2^{2a}5^{2b}2^c5^d\\
&=2^{2a+c}5^{2b+d}=2^{40}5^{20}
\end{align*}
which implies
$$2a+c=40\\2b+d=20$$
Since $2a+c=40$, $c$ has to even. So, we can put $c=2,4,6\dots$ and find the corresponding values of $a$. We can do the same for $b$ and $d$.
Note that $40\leq c\leq 0$. So, there are $21$ solutions for the first equation, and similarly $11$ solutions for the second equation. Since each such $(a,c)$ and each $(b,d)$ gives you a solution, the total number of solutions is $21\times 11=231$.
This completes the answer.
