<s>[INST] Number of divisors of the form $(4n+1)$ 
Find the number of divisors of $$2^2\cdot3^3\cdot5^3\cdot7^5$$ which are of the form $(4n+1)$  

I know how to find the total number of divisors. But, to find the number of divisors of the form $(4n+1)$, I'm thinking of listing down the divisors and then finding, but that'd be very tedious. Is there any elegant way to do this?  
Any help will be appreciated.
Thanks.
 [/INST] Number of divisors of $$N= 2^2\times3^3\times5^3\times7^5,$$which are of the form $4n+1$ exculuding 
$$\begin{align}1 &=\{ \text{number of terms in product}\}\\
&=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+\{\text{number of terms in product}\}\\
&= ( 3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1\\
&= 2\times4\times3+2\times3\times4-1\\
&= 47
\end{align}$$
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