# On odd perfect numbers and $x\varphi(y)=y\varphi(x)$, where $\varphi(n)$ is the Euler's totient function

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 $$m then $$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 $\varphi(7)=\varphi(9)$).

• Many thanks then was a bad question.
– 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.