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?

  • 1
    $\begingroup$ 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 $4n+1$ $\endgroup$ – Peter Feb 4 '18 at 16:18
  • $\begingroup$ @Peter Please elaborate it to make an answer. Seems interesting and useful. $\endgroup$ – 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.

| cite | improve this answer | |
  • $\begingroup$ 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 $5 = \{5, 105, 525\}$. $\endgroup$ – jitender Feb 4 '18 at 16:34
  • 1
    $\begingroup$ An even number cannot be of the form $4k+1$, hence we only need to consider the odd divisors. $\endgroup$ – Peter Feb 4 '18 at 16:36
  • $\begingroup$ $3^a\cdot 7^b$ is of the form $4k+1$ if and only if $a+b$ is even. This follows from $$3^a\cdot 7^b\equiv (-1)^a\cdot (-1)^b=(-1)^{a+b}\mod 4$$ $\endgroup$ – Peter Feb 4 '18 at 16:37
  • 1
    $\begingroup$ 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 $4k+1$. $\endgroup$ – Peter Feb 4 '18 at 16:43
  • 1
    $\begingroup$ @jitender There is no "counting issue". You can just multiply. Duplicates cannot occur. $\endgroup$ – Peter Feb 4 '18 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.