In this post we denote the sum of divisors function as $$\sigma(n)=\sum_{1\leq d\mid n}d,$$ then an even perfect number is a positive integer $n\equiv 0\text{ mod }2$ for which $\sigma(n)=2n.$ As reference it's know for instance the Wikipedia *Perfect number*. In particular it is known the theorems due to Euclides and Euler.

The Euler's totient $\varphi(x)$ function is also a multiplicative function. Using the characterization for even perfect numbers due to Euler and Euclides

$$n=2^{p-1}(2^p-1)$$ where $2^p-1$ is its associated Mersenne prime one has the folowing claim.

**Claim.** *Let* $\lambda\geq 1$ *and* $\mu\geq 1$ *be fixed integers. Define, being* $2^{p}-1$ *a Mersenne prime, the relationship* $$m+1:=2^{p-1}.$$
*Then the identity*
$$\varphi((m+1)^\lambda(2m+1)^\mu)=m(m+1)^\lambda (2m+1)^{\mu-1}$$
*holds.*

I would like to know if it is possible to prove the following conjecture (I've tested it for some segments of integers, and I tried to get the proof for a case).

Question.Prove or refute the following conjecture:

For any choice of$\lambda\geq 1$and$\mu\geq 1$integers, it holds that if an integer$m\geq 1$satisfies$$\varphi((m+1)^\lambda(2m+1)^\mu)=m(m+1)^\lambda (2m+1)^{\mu-1}$$then$$(m+1)(2m+1)$$is an even perfect number.

Many thanks.

Thus in my view it should be a similar theorem/characterization for even perfect numbers by using the Euler's totient function instead of the sum of divisors function.

I hope that my question has a good mathematical content and that there aren't mistakes. Feel free to ask about the check that I did using a Pari/GP program, or criticize if this version of Euclides-Euler theorem is potentitally interesting.

Claimand related conjecture are potentially interesting. I am waiting for your feedback, many thanks. $\endgroup$Question, then (in particular) the following characterization for even perfect numbers will be feasible:An integer$m\geq 1$satisfies the identity$$\varphi((m+1)(2m+1))=m(m+1)$$if and only if$(m+1)(2m+1)$is an even perfect number (being$2m+1=2^p-1$its associated Mersenne prime).$\endgroup$An integer$r\geq 1$satisfies$$\psi\left(\frac{(3r-1)(3r-2)}{2}\right)=\frac{3}{4}(3r-1)^2$$if and only if$\frac{3r-1}{2}(3r-2)$is an even perfect number greater than$6$(being$3r-2=2^p-1$its associated Mersenne prime).As reference the WikipediaPerfect numberthat refers the relationship between even perfect numbers and centered nonagonal numbers. $\endgroup$..since$M$is perfect..., instead of the right claim..since$\frac{M+1}{2} M$is perfect...$\endgroup$