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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.

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    $\begingroup$ Ignore that specific answer. It is confusing because it makes wrong use of the equal sign (it appears to claim $1=\ldots = 47$) $\endgroup$ – Hagen von Eitzen Feb 4 '18 at 0:00
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$4n+1$ 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.

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  • $\begingroup$ Have two issues: (i) Cannot understand the logic behind the answer coming in pack of $4$. (ii) I agree that $2$ or $2^2$ form will not occur by being even (rather than odd as needed by the $4n+1$ form), but how to understand it is confusing. As per me, any multiplier will matter, and hence I took the residue class of $2$ and squared it for $2^2$. Also, would like a small example for $84=2^2\cdot3\cdot7$, with $a=1,b=1,a+b$ is even; & the only factor of the $4n+1$ form is: $21$. $\endgroup$ – jitender Feb 4 '18 at 0:24
  • $\begingroup$ I mean in the earlier comment, that by your logic the number of divisors should be given by : The choices are : $0\le a \le 1, 0 \le b \le 1$. So,the possible choices are : $a=0,b=0; a=1,b=1$, with the former case yielding no value as both $a,b$ being null, and the product term ($2^2$) by itself cannot lead to any odd divisor. The left is only one choice for $3.7=21$. Here, I am unable to understand the application of 'divisors coming in packs of $4$'. $\endgroup$ – jitender Feb 4 '18 at 0:46
  • $\begingroup$ I hope you meant for the reason that the exponent for $5$ is $3$, and if the exponent for $5$ is chosen as $1$, then the divisors of the stated form for $84*5=420$ are : $5, 21, 105$ only. This defies the assumption that now the pack should be of two (as power of $5$ is $1$). $\endgroup$ – jitender Feb 4 '18 at 1:04

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