4
$\begingroup$

In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function, and [1]. On the other hand I add the reference that Wikipedia has the article Mersenne prime, and that I was inspired in the formula that defines the sequence A072868 from the On-Line Encyclopedia of Integer Sequences.

The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$ with the definition $\psi(1)=1$.

Claim. If we take $n=2^p$ with $2^p-1$ a Mersenne prime, then the equation $$\psi(2(\psi(n)-n)-1)=n\tag{1}$$ holds.

Sketch of proof. Just direct computation using the mentioned representation for the Dedekind psi function.$\square$

I don't know if previous equation is in the literature, one can to state a similar equation than $(1)$ involving the sum of divisors function instead of the Dedekind psi function.

Question. I would like to know if it is possible to prove of refute that if an integer $n\geq 2$ satisfies $(1)$ then $n-1$ is a Mersenne prime. Many thanks.

With a Pari/GP script and for small segments of integers I have not found counterexamples. I'm asking what work can be done for previous question proving the conjecture, or if you can to find a counterexample, before I'm accepting an available answer.

References:

[1] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).

$\endgroup$
  • $\begingroup$ Many thanks for the persons upvoting the post. $\endgroup$ – user759001 Jun 21 at 8:30
  • 2
    $\begingroup$ No counterexample upto $n=10^8$ $\endgroup$ – Peter Jun 21 at 9:14
  • 1
    $\begingroup$ Many thanks for your work @Peter , as usual is excellent. $\endgroup$ – user759001 Jun 21 at 9:45
  • 1
    $\begingroup$ Thank you. I think it is hard to prove this conjecture because we have to apply the psi-function twice. I wonder whether the strategy to prove that the EVEN perfect numbers are equivalent to the Mersenne primes can somehow be used here. Whether odd perfect numbers exist is still not known. Maybe, similar difficulties appear here. $\endgroup$ – Peter Jun 21 at 9:49
1
+50
$\begingroup$

This is a partial answer.

This answer proves the following claims :

Claim 1 : There is no prime $p$ such that $p^3\mid 2\psi(n)-2n-1$.

Claim 2 : If $2\psi(n)-2n-1$ is a prime, then $n-1$ is a Mersenne prime.

Claim 3 : $n$ is not a prime.

Claim 4 : $n+2\le \psi(n)\lt\dfrac{3n+1}{2}$

Claim 5 : If $n$ is even, then $n-1$ is a Mersenne prime.

Claim 6 : $15\not \mid n$

Claim 7 : $21\not \mid n$

Claim 8 : If $n=3^aq^bc$ where $a,b$ are positive integers, $c$ is odd and $q\ge 11$ is a prime such that $\gcd(q,3)=\gcd(c,3q)=1$, then $c\gt \dfrac{16q+15}{q-8}$.


Claim 1 : There is no prime $p$ such that $p^3\mid 2\psi(n)-2n-1$.

Proof : Suppose that there is a prime $p$ such that $$p^3\mid 2\psi(n)-2n-1\tag2$$ then it follows from $(1)$ that $p^2\mid n$ and $p\mid\psi(n)$ from which we have $p\not\mid 2\psi(n)-2n-1$ which contradicts $(2)$. $\quad\blacksquare$


Claim 2 : If $2\psi(n)-2n-1$ is a prime, then $n-1$ is a Mersenne prime.

Proof : If $2\psi(n)-2n-1$ is a prime, then letting $n=\displaystyle\prod_{i=1}^{d}p_i^{e_i}$ where $p_1\lt p_2\lt\cdots\lt p_d$ are primes, we have $$(1)\implies 2\psi(n)-2n-1+1=n\implies 2\prod_{i=1}^{d}(p_i+1)=3\prod_{i=1}^{d}p_i$$

If $d\ge 2$, then we get $p_1=2$ and $\displaystyle\prod_{i=2}^{d}(p_i+1)=\displaystyle\prod_{i=2}^{d}p_i$ which is impossible.

If $d=1$, then $2(p_1+1)=3p_1$ implying $p_1=2$. So, $n$ is of the form $2^a$ where $a\ge 1$, and $n-1$ is a Mersenne prime. $\quad\blacksquare$


Claim 3 : $n$ is not a prime.

Proof : Suppose that $n$ is a prime. Then, it follows from $(1)$ that $n=1$ which contradicts that $n$ is a prime. $\quad\blacksquare$


Claim 4 : $n+2\le \psi(n)\lt\dfrac{3n+1}{2}$

Proof : Suppose that $2\psi(n)-2n-1=1$. Then, it follows from $(1)$ that $n=1$ for which $2\psi(n)-2n-1=1$ does not hold. So, we have $2\psi(n)-2n-1\ge 2$. Since $\psi(m)\gt m$ for $m\ge 2$, we have $n=\psi(2\psi(n)-2n-1)\gt 2\psi(n)-2n-1$, i.e. $\psi(n)\lt\dfrac{3n+1}{2}$. Also, since $m$ is a prime iff $\sigma(m)\le m+1$, we have $n+2\le \psi(n)$. $\quad\blacksquare$


Claim 5 : If $n$ is even, then $n-1$ is a Mersenne prime.

Proof : Suppose that there is an odd integer $q\gt 1$ such that $n=2^aq$ where $a\ge 1$. Then, $$\psi(n)\lt\frac{3n+1}{2}\implies q+2\le \psi(q)\lt q+\frac{1}{6}$$ which is impossible. So, we have $n=2^a$. Then, $(1)\implies \psi(2^{a}-1)=2^a$ which implies that $2^a-1$ is a prime. $\quad\blacksquare$


Claim 6 : $15\not \mid n$

Proof : Suppose that $n=3^a5^bc$ where $a,b$ are positive integers, and $c$ is odd such that $\gcd(c,15)=1$, then $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{45c+1}{48}\implies c\lt -\frac{95}{3}$$ which contradicts that $c$ is positive. $\quad\blacksquare$


Claim 7 : $21\not \mid n$

Proof : Suppose that $n=3^a7^bc$ where $a,b$ are positive integers, and $c$ is odd such that $\gcd(c,21)=1$, then $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{63c+1}{64}\implies c\lt -127$$which contradicts that $c$ is positive. $\quad\blacksquare$


Claim 8 : If $n=3^aq^bc$ where $a,b$ are positive integers, $c$ is odd and $q\ge 11$ is a prime such that $\gcd(q,3)=\gcd(c,3q)=1$, then $c\gt \dfrac{16q+15}{q-8}$.

Proof : We have $$\psi(n)\lt\frac{3n+1}{2}\implies c+2\le\psi(c)\lt \frac{9qc+1}{8(q+1)}\implies c\gt \frac{16q+15}{q-8}\quad\blacksquare$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Many thanks for your detailed propositions, I'm going to study these. $\endgroup$ – user759001 Aug 1 at 16:38

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.