# Around a characterization for even perfect numbers, similar than Euclides-Euler theorem, in terms of totatives

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.

• All, the comparison with the characterization for even perfect numbers due to Euclides and Euler comes from the case $\lambda=\mu=1$, and from the fact that as a consequence of Euclides-Euler theorem for even perfect numbers $n=2^{p-1}\cdot(2^p-1)$ is a triangular number $n=\frac{M+1}{2}\cdot M$ where $M=2^p-1$ is its associated Mersenne prime. That's $$\sigma\left(\frac{M(M+1)}{2}\right)=M(M+1)$$ since $M$ is perfect. As it was said feel free to provide me the opinion if my Claim and related conjecture are potentially interesting. I am waiting for your feedback, many thanks. – user142929 Oct 28 '19 at 17:44
• Thus if you can to prove the conjecture in my 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). – user142929 Oct 28 '19 at 17:53
• As aside comment I think that a similar claim and conjecture (a third and last characterization similar than Euclides-Euler theorem) are feasible in terms of the Dedekind psi function $\psi(n)$, in particular the conjecture: 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 Wikipedia Perfect number that refers the relationship between even perfect numbers and centered nonagonal numbers. – user142929 Oct 30 '19 at 6:16
• In my first comment was added that ..since $M$ is perfect..., instead of the right claim ..since $\frac{M+1}{2} M$ is perfect... – user142929 Nov 14 '19 at 11:49

This is not a complete answer, by far. Just some observations, and solutions to some cases for $$m$$.

Using the identity $$\phi(n^m)=n^{m-1}\phi(n)$$ and the fact that $$(m+1,2m+1)=1$$, we can say

$$\phi((m+1)^\lambda (2m+1)^\mu) = (m+1)^{\lambda-1} (2m+1)^{\mu-1} \phi(m+1) \phi(2m+1).$$

This should then equal $$m(m+1)^\lambda (2m+1)^{\mu-1}$$. After division, we are left with

$$\phi(m+1)\phi(2m+1) = m(m+1).$$

In particular, this identity is equivalent to the one in your question.

Let us make the substitution $$s = m + 1$$. Then $$\phi(s)\phi(2s-1) = s(s-1)$$

Suppose that $$s=p^\alpha$$. Then $$(p-1) \phi(2p^{\alpha}-1) = p (p^\alpha - 1)$$.

If $$p=2$$, this says that $$\phi(2^{\alpha+1} - 1) = (2^{\alpha+1} - 1) - 1$$. Because the only numbers for which $$\phi(x)=x-1$$ are primes, $$2^{\alpha+1}-1$$ is a prime and $$s(2s-1)$$ is indeed an even perfect number.

Otherwise, let us denote $$2p^\alpha - 1$$ as $$x$$. Thus, $$\phi(x)=\frac p {2(p-1)} (x-1)$$. Suppose $$x=q^\beta r$$, so that it is not squarefree ($$\beta > 1$$). This says that $$2(p-1)q^{\beta-1}(q-1)\phi(r) = p(q^\beta r - 1)$$. Taking this modulo $$q$$, we get a contradiction. Thus, $$x$$ is squarefree.

We can also see that $$x$$ cannot be a prime. If $$x$$ were a prime, we would get $$\phi(x) = x-1 = \frac p {2(p-1)} (x-1)$$, or that $$p=2$$.

However, I don't see how to proceed if $$s$$ is not a prime power, or if $$2s-1$$ is a squarefree composite number.

• Many thanks for your answer. I'm going to study it! I think that your simplifications are elegant and will be useful. – user142929 Oct 30 '19 at 15:55