I came across this answer to the question of finding the number of divisors of $2^2\cdot3^3\cdot5^3\cdot7^5$ of the form $4n+1$ on MSE, and am unable to understand as :
(i) The answer (like others on the question) ignores the factor $2^2$ completely. My reasoning for the same is that $2$ will give a factor of the form of $\exists n \in \mathbb{Z}, 4n+2.$ So, an even power of $2$ will give $4n+1$ form. Had there been given an odd power to $2$, then the same residue class, i.e. $4n+2$ would have occurred. I want to know that what should have been the answer had the question been : $2^3\cdot3^3\cdot5^3\cdot7^5$.
(ii) The answer given to the post is confusing, and is not clear as to what is meant.
(iii) Why the answer be $47$, and not $48$, as given in one of earlier answers.
$4n+1$ is odd so it cannot be a multiple of $2$ and even less a multiple of $2^2$; therefore, we can ignore the $2^2$ part completely.
All in all, the number of such divisors is certainly a multiple of $4$ because they always come in packs of $4$ (namely $u, 5u, 25u, 125u$ where $u$ is not a multiple of $5$) - and this fact immediately rules out $47$ as an answer.
Now how many different $u$ are possible? $u$ must be of the form $3^a7^b$ with $0\le a\le 3$ and $0\le b\le 5$ and $a+b$ even; so whether $a\in\{0,2\}$ and $b\in\{0,2,4\}$ (leading to $2\cdot 3=6$ choices) or $a\in\{1,3\}$ and $b\in\{1,3,5\}$ (leading to another $2\cdot 3=6$ choices). So we have $4\cdot (6+6)=48$ such divisors.