It is well knwon that for positive integers such that $gcd(a,n)=1$, Euler-Fermat Theorem states than $a^{\phi(n)}\equiv 1 \mod n$, where $\phi(n)$ is the Euler totient function counting positive integers $1\leq k\leq n$ which are coprime with $n$. Is a multiplicative function (an arithmetic function is called multiplicative if for all integers $r,t$ such that $gcd(r,t)=1$ then $f(r\cdot t)=f(r)\cdot f(t))$. Both divisors functions, $\sigma(n)$, the sum of positive divisors of $n$ and $\sigma_0(n)$ counting the number of these divisors, are multiplicative functions too. Its values in prime powers is easly computed in a closed form, see this Math Stack Exchange or Wikipedia. Thus is easy to prove the following

Proposition. If $n\geq 1$ is an odd integer sufficiently large then $$\begin{align} \sigma(2^{\phi(n)-2}\cdot n)=(2^{\phi(n)-1}-1)\cdot \sigma(n)&\text{} \tag 1 \\ 8\phi(2^{\phi(n)-2}\cdot n)-\phi(n)\equiv 0\mod n\phi(n)&\text{} \tag 2 \\ 2\sigma_0(2^{2^{\phi(n)-1}-1}\cdot n)-\sigma_0(n)\equiv 0\mod n\sigma_0(n)&\text{} \tag 3\\ \end{align}$$

My questions are

Question. a) Can you find any application of previous statements? References and suggestions will well received. b) I am looking to close what odd integers satisfy $\phi(n)\geq 2$, and same question for $\phi(n)\geq 3$, in an easy and rigorous way (optional you can study the case $\phi(n)\geq a$, for odd integers, as you see).

My only goal, in this ocassion, with this question is learning fun and obtain some points of reputation. Thanks in advance.

My attempts to obtain some applications are first, from (1), using Euler-Fermat for odd perfect numbers implies $$\sigma(2^{\phi(n)-2}\cdot n)\equiv -n\mod n^2$$ but this don't say nothing to me. Another attempt was from (1) and (3), but neither I don't extract any information with my computations for the question

Conjecture (Chen and Fang, 2007). There are no odd composite $n$ such that $n$ divides $2\sigma(n)-\sigma_0(n)$.

See [2] as free paper of the online free journal INTEGERS.


[1] Wikipedia or this Mathematics Stack Exchange, sum of divisor function, number of positive divisor function, Euler's totient function, Euler-Fermat Theorem, multiplicative function.

[2] Yong-Gao Chen, Jin-Hui Fang, On $N\mid \varphi(N)D(N)+2$ and $N\mid\varphi(N)\sigma(N)+1$, INTEGERS: THE ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8(1), (2008), article A#07.


The following might be of some use to you, but it could be inapplicable as well. A proof appears in this publication.

If $N := q^k n^2$ is an odd perfect number in Eulerian form, then $$\frac{\sigma(N/q^k)}{q^k}=\frac{2(N/q^k)}{\sigma(q^k)}=\frac{D(N/q^k)}{\sigma(q^{k-1})}=\gcd\left((N/q^k),\sigma(N/q^k)\right),$$ where $D(x) := 2x - \sigma(x)$ is the deficiency of $x \in \mathbb{N}$.

The number $\sigma(N/q^k)/q^k$ is called the index of the Euler prime $q$.

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  • $\begingroup$ Thank you very much you are very kind. $\endgroup$ – user243301 Sep 9 '17 at 19:44

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