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.

  • $\begingroup$ 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. $\endgroup$
    – user142929
    Oct 28 '19 at 17:44
  • $\begingroup$ 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). $\endgroup$
    – user142929
    Oct 28 '19 at 17:53
  • $\begingroup$ 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. $\endgroup$
    – user142929
    Oct 30 '19 at 6:16
  • $\begingroup$ In my first comment was added that ..since $M$ is perfect..., instead of the right claim ..since $\frac{M+1}{2} M$ is perfect... $\endgroup$
    – 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.

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

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