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For integers $A,B\geq 1$ we define the difference $\sigma(A)\sigma(B)-\sigma(AB)$, denoting it as $[A,B]$, where $\sigma(n)=\sum_{1\leq d\mid n}d$ denotes the sum of divisors function. It is possible to get in closed-form the parity of the arithmetical function $[A,B]$, I mean $[A,B]\text{ mod }2$. I was studying propositions about this when I wondered about the veracity of some conjectures.

Question. I would like to know if it you can to find a counterexample or well a (partial) proof for some of the following conjecture (please see bellow and comments). Many thanks.

Conjecture 2. Let $x\geq 1$ be an integer that satisfies $[x,x]=x$, then $x$ is a prime number.

Please if these conjectures are in the literature answer the question as a reference request, or add a comment.

I relegate two conjectures to an Appendix, y deleted the old Conjecture 1 since this is false (my problem was an implementation with three integer values in my program, instead two, the I knew the counterexample, I wrote it in my notebook, but I relate this exception with an open question; any case Conjecture 1 has a bunch of counterxamples).

Appendix: Adding two conjectures from the genuine version of the post. When I can I'm going to revise the old Conjecture 1.

Conjecture 3. Let $x,y\geq 1$ be positive integers such that $y=\sigma(x)$, $\gcd(x,y)>1$, $xy$ isn't a perfect square, $\sigma(y)\equiv0\text{ mod }2$, the integer $x$ is a triangular number and $[x,y]\equiv 1\text{ mod }2$. Then $x$ is an even perfect number.

Conjecture 4. If there exists an integer $N>1$ such that $[N,\sigma(N)]=3\cdot[N,N]$ holds, then $N$ is an odd perfect number.

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  • $\begingroup$ I've edited in Mathematics Stack Exchange the post with identificator 4423186, on the other hand in an incoming post I would like to ask in one of these Stack Exchange sites (MSE, PSE or MO) about if one can to deduce for the definition $[A,B]$ or $[A,B]\text{ mod 2}$ commutation relations and multiplets (if these notions as companion to the artihmetic function $[A,B]\text{ mod 2}$ do make sense, my belief is that yes, I'm trying understand basics about isospin). If some user or moderator can to add his/her thoughts about it, this might be helpful for the post I plan to ask. Many thanks. $\endgroup$
    – user142929
    Commented Apr 8, 2022 at 17:10
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    $\begingroup$ $\sigma(5^2)=31\equiv3\bmod4$, but $31+1$ is not four times an odd number. $\endgroup$ Commented Apr 9, 2022 at 2:41
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    $\begingroup$ It would be nice to know what, if any, evidence you have for these conjectures. $\endgroup$ Commented Apr 9, 2022 at 2:46
  • $\begingroup$ Many thanks @GerryMyerson maybe there is a typo (when I can I revise your claim and my Conjecture 1). On the other hand I wrote a conjecture for you: If $y=\sigma(\sigma(n))$ is prime with $\sigma(n)$ a perfect square, then $y$ is a Mersenne prime and $n$ itself is a product of distinct Mersenne primes. I don't know if it is in the OEIS or in the literature (if you know it from OEIS or the literature add a comment please). I call it the GM conjecture (isn't the Great Mersenne conjecture, the "GM" is a tribute to your name). My computational evidence for some of my conjectures is humble $\endgroup$
    – user142929
    Commented Apr 9, 2022 at 17:58
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    $\begingroup$ You are right @GerryMyerson I've edited the post with my explanation, and agreement that Conjecture 1 was false since it has a bunch of counterexaples. Many thanks again. $\endgroup$
    – user142929
    Commented Apr 11, 2022 at 16:52

1 Answer 1

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Please restrict to one question per post (standard policy). Here is a proof of Conjecture 2.

For any $s\in\mathbb{C}$, the function $\sigma_s(n):=\sum_{d\mid n} d^s$ satisfies the Hecke multiplicativity relation $$\sigma_s(m)\sigma_s(n)=\sum_{d\mid(m,n)}d^s\sigma_s\left(\frac{mn}{d^2}\right).$$ This is straightforward to prove by restricting $m$ and $n$ to be powers of the same prime, since $\sigma_s$ is a multiplicative function. On a deeper level, $\sigma_s(n)$ is the $n$-th Hecke eigenvalue of a certain Eisenstein series that is a Hecke eigenform.

At any rate, setting $s=1$ and $m=n=x$, we get $$\sigma(x)^2=\sum_{d\mid x}d\sigma\left(\frac{x^2}{d^2}\right).$$ Assume that $x\geq 2$. Then the right-hand side is at least $\sigma(x^2)+x$, because the divisors $d=1$ and $d=x$ are present. If the left-hand side equals this, then there are no further divisors, hence $x$ is a prime number.

P.S. I suggest that you open a new question for Conjecture 1, another one for Conjecture 3, and a third one for Conjecture 4. This question should be closed as I proved Conjecture 2.

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    $\begingroup$ You're right about the policy of ask a question per post, I'm sorry. I'm going to read your answer when I can. I accept your answer now. After I'm going to think what of the other conjectures should I ask in the site. Many thanks. $\endgroup$
    – user142929
    Commented Apr 9, 2022 at 17:52
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    $\begingroup$ @user142929 Thank you. Of course the restriction of "one question per post" is practical. Imagine that you get a response for Conjecture 1, another one for Conjecture 2 etc., then it is unclear which response you should accept. One question makes things fair and simple. $\endgroup$
    – GH from MO
    Commented Apr 9, 2022 at 18:46

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