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.
 A: 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.
A: One can consider separately the problem of finding pairs $(\alpha, \beta)$ and $(\gamma, \delta)$ such that

*

*$2^{40}=2^{2\alpha} 2^\beta$

*$5^{20}=5^{2\gamma} 5^\delta$
Then the number of possible pairs $(n,m)$ is the number of possible pairs $(\alpha, \beta)$ times the number of possible pairs $(\gamma, \delta)$.
Now, the problem is only to solve

*

*$40=2 \alpha + \beta$ and

*$20 = 2 \gamma + \delta.$
A: Using Hagen von Eitzen comment/hint (and the fundamental theorem of arithmetic),
$\quad m$  can be just an arbitrary divisor of $20^{10}$ (and then $n$ is uniquely determined)
($m$ must divide $20^{10}$ but if $m$ is an arbitrary divisor then we can use $n$ to bring 'things up to snuff')
we start off by writing
$\quad 20^{10} = 2^{20}\cdot 5^{10}$
and see that the positive $(m,n)$ solutions is given by
$\tag {Answer} 21 \times 11 = 231$
A: $m^2 n = 2^{40} \times 5^{20}$. We need number of ways to assign values to $m$ and for each $m$, $n$ is fixed.
$m^2 = 2^{2a} \times 5^{2b}$ where $0 \leq 2a \leq 40, 0 \leq2b \leq 20$
$ \implies m = 2^a \times 5^b$ where $0 \leq a \leq 20, 0 \leq b \leq 10$
That gives us $21 \times 11 = 231$ solutions
A: $m^{2}n=20^{20}$ means $n$ is a square too i.e. $n=n_{1}^{2}$. Therefore the equation becomes $mn_{1}=20^{10}$ and you see that $m$ is any factor of $2^{10}$.
