The action of the unitary divisors group on the set of divisors and odd perfect numbers

Let n be a natural number. Let Un={dNdn and gcd(d,n/d)=1} be the set of unitary divisors, Dn be the set of divisors and Sn={dNd2n} be the set of square divisors of n.

The set Un is a group with ab:=abgcd(a,b)2. It operates on Dn via:


The orbits of this operation “seem” to be

Und=dUnd2 for each dSn

From this conjecture it follows (also one can prove this directly since both sides are multiplicative and equal on prime powers):


where σ denotes the sum of unitary divisors.

Since σ(k) is divisible by 2ω(k) if k is odd, where ω= counts the number of distinct prime divisors of k, for an odd perfect number n we get (Let now n be an odd perfect number):


where kd=σ(n/d2)2ω(n/d2) are natural numbers.
Let ˆd be the largest square divisor of n. Then:

Hence we get:

for some natural numbers ld.

If the prime 2 divides not the prime power 2ω(n/ˆd2), we must have ω(n/ˆd2)=0 hence n=ˆd2 is a square number, which is in contradiction to Eulers theorem on odd perfect numbers.

So the prime 2 must divide the prime power 2ω(n/ˆd2) and we get:


with ld=σ(n/d2)2ω(n/d2). Hence the odd perfect number, satisifies:


Hence an odd perfect number satisifies:


So my idea was to study the function a(n), which is multiplicative on odd numbers, on the right hand side and what properties it has to maybe derive insights into odd perfect numbers.

The question is if it ever can happen that an odd number n satisfies: n=a(n)? (checked for n=2k+1 and 1k107)

Conjecture: For all odd n3 we have a(n)<n. This would prove that there exists no odd perfect number.

This conjecture could be proved as follows:
Since a(n) is multiplicative, it is enough to show that for an odd prime power pk we have


The values of a at prime powers are not difficult to compute and they are:




However, I am not very good at proving inequalities, so:

If someone has an idea how to prove the following inequalities for odd primes p that would be very nice:

p2k+1>p2(k+1)12(p1), for all k0


p2k>p2k+1+pk+1pk12(p1), for all k1

Thanks for your help!

The inequalities have been proved here:


Here are some general comments:

  1. You don't need to bring these actions of abelian groups on various sets of divisors. The identity
    is easy to check directly, without appeal to anything fancy.

  2. Let's call α(n) the number of prime divisors of n which appear with an odd exponent in the factorization of n. This is what you call ω(n/ˆd2). You are right in observing that 2α(n) divides σ(n). This is where Euler's result comes from: If n is an odd perfect number then α(n)=1.

  3. It seems you want to define a new function a(n)=σ(n)2α(n), and you conjecture that a(n)<n
    for all odd numbers n. If true this conjecture would imply that there are no odd perfect numbers. Unfortunately it is false. For example the inequality is reversed at n=335272.

Source : Link , Question Author : Community , Answer Author : Gjergji Zaimi

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