# Need help on divisors of the form $4n+1$.

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::

1. 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.

2. 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.

3. 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.

4. I also read elsewhere, that the number of divisors is a function of the power of $5$. Why?

• If you multiply a number of the form $4n+1$ with $5$, you again get a number of the form $4n+1$. So, if you have the number of divisors of the form $4n+1$ which are not divisible by $5$, you have to multiply it with $e+1$ (where $e$ is the exponent in $5^e$) to get the total number of divisors of the form $4n+1$ – Peter Feb 4 '18 at 16:18
• @Peter Please elaborate it to make an answer. Seems interesting and useful. – jitender Feb 4 '18 at 16:20

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.
• So, you mean that (i) only $5$ matters?, why not an even power of $2$, it also leads to the needed form (ii) how is the impact of an even sum of the exponents of $3,7$. Also, please vet: For $210=2\cdot3\cdot7\cdot5$, there need be 2 divisors with a power of $5 = \{5, 105\}$. For $1050= 2\cdot3\cdot7\cdot5^2$, there need be 3 divisors (of stated form) with a power of $5 = \{5, 105, 525\}$. – jitender Feb 4 '18 at 16:34
• An even number cannot be of the form $4k+1$, hence we only need to consider the odd divisors. – Peter Feb 4 '18 at 16:36
• $3^a\cdot 7^b$ is of the form $4k+1$ if and only if $a+b$ is even. This follows from $$3^a\cdot 7^b\equiv (-1)^a\cdot (-1)^b=(-1)^{a+b}\mod 4$$ – Peter Feb 4 '18 at 16:37
• The exponent of $2$ does not matter, as stated. Take the exponents of $3$ and $7$ (suppose, they are $a$ and $b$). List all pairs $(u/v)$ of non-negative integers such that $u+v$ is even and $u\le a$ and $v\le b$ holds. This gives the number of divisors of $3^a\cdot 7^b$ of the form $4k+1$. – Peter Feb 4 '18 at 16:43