1
$\begingroup$

I would like to know a full characterization of the solutions of next equation involving the sum of divisors function $\sigma(m)=\sum_{d\mid m}d$ and the Euler's totient function denoted as $\varphi(m)$

$$\varphi\left(n^n\sigma(n)\right)=\varphi(n^{n+1}).\tag{1}$$ The equation arises from my experiments with odd perfect numbers, since one has the following claim.

Claim. If there exists an odd perfect number then satisfies $(1)$ (and as a consequence we can get also an statement invoking Euler-Fermat $2^{\varphi(n^n\sigma(n))}\equiv 1\text{ mod }n^{n+1}$, I know the first few solutions of this congruence).

Just a comment is that I was playing with more reasonable powers before this $n^n$.

Proof. The proof is easy from the fact that the Euler's totient function is multiplicative, $\varphi\left(n^n\sigma(n)\right)=\varphi(2n^{n+1})=\varphi(n^{n+1}).$$\square$

Now, it was fun when I ran a program to know what integers satisfy our equation $(1)$. The sequence starts as $1, 2, 8, 128$, that are the terms that I can compute with my program. But due my belief I suspect that if fact every term from the OEIS A058891 satisfy $(1)$.

Question. I would like to know a full characterization (I am saying what reasonable work can be done) of solutions of previous equation $(1)$. I know the Claim and I suspect that the integers of the form $$2^{2^{\lambda-1}-1}\tag{2}$$ satisfy our equation, where $\lambda$ runs over integers $\geq 1$. What can you prove about it? Can you prove that $(2)$ satisfies our equation (I know that it should be easy but is required patience)? Are there more solutions? Many thanks.

Please if you know the equation $(1)$ from the literature please answer this as a reference request and I try to find and read those facts from the literature.

$\endgroup$
  • $\begingroup$ If you run a program in your computer and you find a sufficiently large odd solution of $$2^{\varphi(n^n\sigma(n))}\equiv 1\text{ mod }n^{n+1},$$ please add it as a comment. I am saying an odd integer say us $>100$. Many thanks. $\endgroup$ – user243301 Mar 7 '18 at 23:17
  • $\begingroup$ Just a comment if some user was interested in this kind of equations is that it is possible to prove (easily by cases) that each integer $m\geq 1$ satisfies $\varphi(4m\sigma(m))=2\varphi(2m\sigma(m))$. And also the fact that if $n$ is an odd perfect number and $a\geq 2$ an integer then $\varphi(2^a n\sigma(n))=2^{a-1}\varphi(n^2)$. $\endgroup$ – user243301 Mar 8 '18 at 11:59
0
$\begingroup$

About $2^{2^{\lambda-1}-1}$:

Assume $2^a$ satisfies the equation. Then, since $\sigma(2^a)=2^{a+1}-1$, we have that

$$\varphi\left(2^{a2^a}\left(2^{a+1}-1\right)\right)=\varphi\left(2^{a2^a+a}\right).$$

Since $\varphi(mn)=\varphi(m)\varphi(n)$ when $\gcd(m,n)=1$ and $\phi(2^a)=2^{a-1},$ we have that then

$$2^{a2^a-1}\varphi\left(2^{a+1}-1\right) = 2^{a2^a+a-1}.$$

$$\varphi\left(2^{a+1}-1\right) = 2^a.$$

You wish to show that

$$\varphi\left(2^{2^k}-1\right) = 2^{2^k-1}.$$

Let

$$F_n=2^{2^n}+1.$$

Then

$$2^{2^n}-1=\left(2^{2^{n-1}}-1\right)\left(2^{2^{n-1}}-1\right),$$

so we have inductively that

$$2^{2^n}-1 = F_0\cdots F_{n-1}.$$

Since all Fermat numbers are coprime, this holds iff we have that

$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) = \prod_{k=0}^{n-1} \left(2^{2^k}\right).$$

Indeed, since $\varphi(x)\leq x-1$, with equality holding iff $x$ is prime, we have that

$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) \leq \prod_{k=0}^{n-1} \left(2^{2^k}\right),$$

with equality holding iff each of the first $n-1$ Fermat numbers are prime. Thus, your conjecture is equivalent to that made by Fermat that they are all prime, and thus fails first at $n=6$, and similarly everywhere thereafter (as $F_5$ is not prime).

$\endgroup$
  • $\begingroup$ I am going to study your answer. I can not believe that the same mistake was made again (the same false conjecture that did Fermat). Many thanks. $\endgroup$ – user243301 Mar 9 '18 at 8:18
  • $\begingroup$ I'm sure that your answer is right 100% but I have not studied it yet, thus I'm sorry for not accepting it yet. $\endgroup$ – user243301 Mar 17 '18 at 16:31
  • 1
    $\begingroup$ @user243301 That's fine - and although it's highly unlikely that a complete characterization of the solutions to this can be provided, this answer doesn't fully respond to all you asked, so it's reasonable to want to leave the question open in case anyone else has anything to provide on it. $\endgroup$ – Carl Schildkraut Mar 17 '18 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy