Number of divisors of the form $(4n+1)$

Any positive divisor of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form $4k+1$ is a number of the form: $$ 3^a\cdot 5^b\cdot 7^c $$ with $0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $a+c$ being even. There are: $$ \frac{4\cdot 4\cdot 6}{2}=\color{red}{48} $$ ways to choose $a,b,c$ that way.

Hint:

$(4n+3)(4k+3)=4(4nk+3(n+k)+2)+1$, applicable to $3, \; 7$; and

$(4n+1)^2=4(4n^2+2n)+1$, applicable to $5$.