For the case of a number of the form $2^a\cdot 3^b\cdot5^c\cdot7^d,$ need to find the divisors of the form $4n+1$. Please vet my knowledge of the same::
Only the number of terms of $3,7$ matter when the sum of their powers is even, as the product of the residue classes of both leads to the needed residue class, i.e.: $(4n-1)(4n'-1) = 4n''+1$.
Have a confusion here, as it is stated elsewhere that $2$ as a multiplier would not affect the roots. I am confused about it, particularly the residue class of $2$ is $4n+2$, and of $2^2$ is $4n+1$. So, is it the odd power of $2$ that does not affect, or is it the even power too.For a small number as $2^2\cdot3\cdot7\cdot5 = 420$, it leads to divisors of the stated form as: $5, 21, 105$. However, the divisors possible are having the exponents values for sum of $a,b$ as even, i.e. both odd, or both even:
(i) $a=1, b=1$ (ii) $a=0, b=0.$
For the latter case, the divisor can be $5$, as it is not divisible by $3,7$. While for divisor $21$, it does not matter to have $5$ as factor. For $105$, all three matter.
However, $2$ as occuring in pair is not visible in any of the divisors.To check for an odd power of $2$, let us take another number $2\cdot3\cdot7\cdot5 = 210$, it leads to the same divisors as earlier, i.e.: $5, 21, 105$. How to explain this fact.
I also read elsewhere, that the number of divisors is a function of the power of $5$. Why?
Answer to the last part : Suppose, you have the number of divisors of $$2^a\cdot 3^b\cdot 7^d$$ which have the form $4k+1$ , denote if with $n$. The the number of divisors of $$2^a\cdot 3^b\cdot 5^c\cdot 7^d$$ of the form $4k+1$ is $(c+1)\cdot n$ because for every divisor $q$ of the form $4k+1$, the numbers $5q,25q,\cdots 5^c\cdot q$ are also divisors of the form $4k+1$, hence for each $q$ we have $c+1$ such divisors.