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$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ is the $n^\text{th}$ prime number and $\sigma(n)$ is the sum of the divisors of $n$. Consider the expression:

$$F(n)=\varphi(|P_{n+2}-\sigma(n)|)+1$$

Conjecture: when $F(n) \equiv 3 \pmod {20}$ then this number is a prime or not. When the number is not a prime it can be a power of prime by calculating $|P_{n+2}-\sigma(n)|=p^k$ ($p$ prime, $k$ a natural number${}> 1$). Vice-versa we introduce the following expression:

$$G(n)=|P_{n+2}-\sigma(n)|$$

When $G(n) \equiv 3 \mod 20$ then this number is a prime or not. When the number is not a prime it can be a power of prime by calculating $\varphi(|P_{n+2}-\sigma(n)|)+1$

Examples:

$n=10270001113$, we have:

$$F(10270001113) =\varphi(|P_{10270001115}-\sigma(10270001113)|)+1 =\varphi(259189944599-10468624896)+1=248721319703$$ which is prime because it ends by $03.$

A counterexample is found with $n=680$:

$$F(680)=\varphi(|P_{682}-\sigma(680)|)+1 = \varphi(5101-1620)+1=3423$$ which is not prime but we have $P_{n+2}-\sigma(n)=p^2$, more precisely it is the square of $59.$

Interestingly for $n \leq 526 388 126$ (calculations with PARI/GP) all counterexamples are the power of prime.

Another example is found for $k=6$, this is $n=526388126$. In this case, we have:

$$F(n)=10549870323$$

which is not prime and $|P_{n+2}-\sigma(n)|=47^6$ (here $k=6$).

Another example is with $n=154$ we have:

$G(154)=|P_{156}-\sigma(154)|=623=7\cdot89$ which is not a prime and not a power of prime. So we calculate $\varphi(|P_{156}-\sigma(154)|)+1=529=23^2$ which is a prime power

The question is: "Are there only these two solutions?

  1. A power of prime if the result is not a prime
  2. Or the result is prime

$F(n) \equiv 3 \mod (20)$ and $P_{n+2}$ is odd. Moreover $F(n)$ is odd if and only if $P_{n+2}-\sigma(n)$ is positive and odd. Consequently $F(n)$ is prime or the power of a prime.

Stefan Kohl
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    Where does this conjecture come from? – gmvh Jan 09 '24 at 16:14
  • This is my conjecture firstly published on math stack exchange but there is no satisfying answer (hint to prove the conjecture for example). – Rédoane Daoudi Jan 09 '24 at 16:19
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    I think the condition $F(n) \equiv 3 \mod 20$ can be changed to $F(n) \equiv 3 \mod 4$. Also then all cases I tested (up to $n=5000000$) fulfill the OPs conjecture. – Fred Hucht Jan 09 '24 at 17:29
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    If $p, q \mid |P_{n+2}-\sigma(n)| = x$ for odd primes $p, q$ then $4 \mid (p - 1)(q - 1) \mid \varphi(x)$ and $F(n) \equiv 1 \pmod{4}$. – Denis Shatrov Jan 09 '24 at 17:47
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    Cross-posted from MSE https://math.stackexchange.com/questions/4836471 – Fred Hucht Jan 09 '24 at 17:53
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    @Denis Shatrov: is it a proof? – Rédoane Daoudi Jan 09 '24 at 17:55
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    @RédoaneDaoudi No. But this shows that $F(n) \equiv 3 \pmod{4}$ is only possible if $x = p^k$ or $x = 2p^k$ or $x = 4$. – Denis Shatrov Jan 09 '24 at 18:00
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    This is hard to test for large numbers, because it requires computing the n-th prime. – joro Jan 10 '24 at 06:28
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    @RédoaneDaoudi But what makes you think this conjecture is true? How did you come up with it? To me, $|\sigma(n)-P_{n+2}|$ does not look like a very natural expression. There must be a reason to consider it. – gmvh Jan 10 '24 at 10:30
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    @gmvh : I have worked about a possible link between prime numbers, Euler's totient function and sum of the divisors since 2019. I have a good intuition when the subject is number theory. – Rédoane Daoudi Jan 10 '24 at 10:55
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    Moreover the calculations with PARI/GP suggest that the conjecture may be true. – Rédoane Daoudi Jan 10 '24 at 11:02
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    For me the real question about naturality is why $P_{n+2}$ should be special. Why not $P_{n+j}$ for any $j > 0$? – Peter Taylor Jan 11 '24 at 20:16
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    @Peter Taylor: I found several counterexamples when $j \ne 2$. And this is a property of prime numbers because when I computed the program with any natural numbers I found several counterexamples. – Rédoane Daoudi Jan 11 '24 at 21:35
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    It is pretty rare for $\sigma(n)$ to not be even so, as Denis Shatrov points out the condition implies that $P_{n+2}-\sigma(n)$ is a power of an odd prime. If it is a cube of a prime, then $p^3-p^2+1$ is not going to be $3\mod 5$, so it has to be at least the fourth power of a prime. This is a rare event, which makes it hard to find by random search. Still, my guess is that the statement is unlikely to be true, as fourth powers of primes do occur among $P_{n+2}-\sigma(n)$. – Lev Borisov Jan 13 '24 at 17:08
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    @LevBorisov : thank you for your comment. You claim that the conjecture is unlikely to be true. So I try to prove that the conjecture is false or to find a counterexample. Have you a proof? – Rédoane Daoudi Jan 13 '24 at 17:56
  • n = 21144097 seems to provide a counterexample – Lev Borisov Jan 13 '24 at 18:30
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    @LevBorisov : no. with n=21144097 we have $\varphi(|P_{21144099}-\sigma(21144097)|)+1 = 370615423$ which is prime. – Rédoane Daoudi Jan 13 '24 at 18:39
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    I see, that's just my bad luck. I get $P_{21144099} −\sigma(21144097)=139^4$, but as luck would have it, $139^4-139^3+1$ is a prime :) – Lev Borisov Jan 13 '24 at 19:13
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    I guess that the $\varphi$ function is necessary for the validity of the conjecture. If $n=154$ we have $P_{156}-\sigma(154)=623=7*89$ which is not a prime and not a power of prime. But when we introduce the $\varphi$ function whe have: $\varphi(|P_{156}-\sigma(154)|)+1=529=23^2$ which is a prime power. – Rédoane Daoudi Jan 14 '24 at 13:53
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    How did you make this conjecture? You couldn't just "cook" a recipe of random arithmetic functions until you reach a primality criterion. I mean, what is the story behind this conjecture? – Konstantinos Gaitanas Jan 16 '24 at 09:48
  • @KonstantinosGaitanas : I worked hard to obtain this conjecture (since 2019). I have a good intuition in number theory. The mixture between intuition and serendipity made me develop this conjecture. I work with wolframalpha to test several formulas and with http://calculis.net/premier to check if a number is prime or not. For large numbers I use PARI/GP. – Rédoane Daoudi Jan 16 '24 at 11:22
  • For the sake of transparency Eric Snyder formulated a first conjecture inspired by my topic: https://math.stackexchange.com/questions/4836413/prime-numbers-which-end-with-03-23-43-63-or-83 then Peter helped me for the calculations with PARI/GP: https://math.stackexchange.com/questions/4836471/square-of-prime-numbers Finally I generalized the conjecture here (on mathoverflow) with my intuition (here I introduce the $F$ and $G$ expressions) and with the help of Peter and Eric Snyder. – Rédoane Daoudi Jan 16 '24 at 13:03

1 Answers1

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$n=165$ gives a counterexample to the second part of the conjecture. Here we have $G(n)=703\equiv 3\pmod{20}$ and it's not prime (since $703=19\cdot 37$), but $$F(n)=\varphi(703)+1=649=11\cdot 59$$ is not a prime power.

Max Alekseyev
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    The OPs conjecture is only made for these $n$ where $F(n)\equiv 3 \pmod{20}$. $F(165)=649 \equiv 9 \pmod{20}$. However, I must admit that the conjecture is not very well formulated... – Fred Hucht Jan 21 '24 at 19:52
  • @FredHucht: $F(n)\equiv 3\pmod{20}$ is stated only in the first part of the conjecture. The second part starts with $G(n)\equiv 3\pmod{20}$ and there is no restriction for $F(n)$.

    Also, the given example for $n=154$ has $F(n)=529\not\equiv 3\pmod{20}$.

    – Max Alekseyev Jan 21 '24 at 19:54
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    This is true. The OP edited the question on Jan 14 and added $G(n)$ as well as the 154 example. This example seems to be incorrect, as it violates the condition. Maybe the OP can clarify. – Fred Hucht Jan 21 '24 at 19:59
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    In fact, thank you. I edited. We focus now on the first part of conjecture – Rédoane Daoudi Jan 22 '24 at 13:17
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    @RédoaneDaoudi: It's a bad idea to delete parts of the question since this makes my answer and some of the comments irrelevant. Please restore your questions and add changes in the form of an update. – Max Alekseyev Jan 23 '24 at 16:22
  • @MaxAlekseyev : I don't know how to restore the question. You can edit it to restore the previous version, thank you. – Rédoane Daoudi Jan 23 '24 at 16:27
  • @RédoaneDaoudi: Check your revisions at https://mathoverflow.net/posts/461849/revisions They have "Edit" and "Rollback" links. – Max Alekseyev Jan 23 '24 at 16:33
  • @MaxAlekseyev : I'm confused with the several versions. Please edit to replace the previous version. – Rédoane Daoudi Jan 23 '24 at 16:43
  • @RédoaneDaoudi: I've rolled back your previous revision. Now you can update it, just please don't delete parts that existing comments and/or answers refer to. – Max Alekseyev Jan 23 '24 at 16:48
  • @MaxAlekseyev : thank you for your help. – Rédoane Daoudi Jan 23 '24 at 16:50