Motivation. If we presume that there exists an odd perfect number $n$, then since the Euler's totient function satisfies $$\varphi(n)=n\cdot\prod_{p\mid n}\left(1-\frac{1}{p}\right),$$ then denoting with $\operatorname{rad}(m)=\prod_{p\mid m}p$ the radical of an integer $m\geq 1$ with $\operatorname{rad}(1)=1$, and $\sigma(m)=\sum_{d\mid m}d$ then sum of divisors function we get that our odd perfect number satisifies $$2\varphi(n)\operatorname{rad}(n)=\sigma(n)\varphi(\operatorname{rad}(n)).$$ That is, our odd perfect number satisifies $$\varphi(n)\operatorname{rad}(n)=n\varphi(\operatorname{rad}(n))\tag{1}$$ and we conclude that our odd perfect number $n$ satisfies by application of the Fermat's little theorem $$2^{n\varphi(\operatorname{rad}(n))}\equiv 1\text{ mod }n,\tag{2}$$ and $$2^{\operatorname{rad}(n)\varphi(n)}\equiv 1\text{ mod }\operatorname{rad}(n).\tag{3}$$

Question 1. Is true or false the following conjecture:

If $n>1$ is an odd integer that satisfies $$m\varphi(n)=n\varphi(m)\tag{4}$$ where $m\mid n$ and $m<n$ then $n$ is an odd perfect number with $\operatorname{rad}(n)=m$.

Thus I am asking if you can provide me a proof of the statement or well a counterexample $(n,m)$ of odd integers such that satisfy $(4)$ and $m\mid n$, but or well $n$ is not an odd perfect nunber or well $m$ is such that $m\neq\operatorname{rad}(n)$.

Question 2. Do you know odd integers $n,m\geq 1$ such that $$2^{n\varphi(m)}\equiv 1\text{ mod }n,$$ and $$2^{m\varphi(n)}\equiv 1\text{ mod }m?$$ (If there are many examples and you want help me, I am especially interested in examples of odd integers $n$ of the form $n\equiv 1\text{ mod }12$ or well of the form $n\equiv 9\text{ mod }36$, and being $m\mid n$ with $m$ without repeated prime factors.) Thanks in advance.


(1) is true for every $n$, and (2) and (3) are true for every odd $n$, so they have nothing to do with being perfect.

Question 1: false, take $m=3$, $n=9$.

Question 2: $7,9$ works (since $\varphi(7)=\varphi(9)$).

  • $\begingroup$ Many thanks then was a bad question. $\endgroup$ – user243301 Apr 19 '17 at 13:10

Question $1$: No, take $(m,n)=(3,9)$. Then $m\mid n$, and $n>1$ is odd, and $$ m\phi(n)=3\phi(9)=3\cdot 6=18=9\cdot 2=n\phi(m). $$ However, $n=9$ is not perfect.

  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 Apr 19 '17 at 13:11

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