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Definitions:

Here I present a novel conjecture using basic mathematical tools like the sum of the divisors of an integer $n$ called $\sigma(n)$, the sum of the squares of the positive divisors of n called $\sigma_2(n)$ I also use the prime-counting function which is the function counting the number of prime numbers less than or equal to some real number n. The prime-counting function is called $\pi(n)$

Conjecture:

We introduce the following expression called $A$:

$$A = \sigma_2(\pi(n) − \sigma(n + 2))$$

We focus on numbers ends with 2. I calculate $A − 1$ and so the new number ends with 1. Then I calculate the square root of this number ends with 1. When the number is an integer, it is always prime.

Example:

Let $n = 100547$, we have $A = \sigma_2(9639 − \sigma(100549)) = 8264809922$ We have $A − 1 = 8264809921$ and we calculate the square root of 8264809921 and we have $\sqrt{A−1} = \sqrt{8264809921} = 90911$ and 90911 is prime

Computing

The conjecture has been checked up to $n=2,000,000$ with the following Python program:

https://onlinegdb.com/0C8cDRpu6

The conjecture is true 74408 times.

Questions

  1. Is this conjecture interesting?
  2. Is it possible to prove this conjecture or to find counter-example?
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    $\begingroup$ The conjecture looks very ad hoc to me, so I am pretty certain that it is false. How many positive examples did you find up to $n=2000000$? $\endgroup$
    – GH from MO
    Commented Aug 5 at 23:45
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    $\begingroup$ Why do you expect that I (or anyone here) know how to run a C++ program? We are mathematicians, not programmers. It's your program, so please use it to answer my question. $\endgroup$
    – GH from MO
    Commented Aug 6 at 0:12
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    $\begingroup$ I think what you observe is this. If $m$ is prime, then $\sigma_2(m)-1$ is a square, namely $m^2$. If $m$ is composite, then $\sigma_2(m)-1$ is rarely a square, and it happens even more rarely that such an $m$ is of the form $\sigma(n+2)-\pi(n)$. $\endgroup$
    – GH from MO
    Commented Aug 6 at 1:13
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    $\begingroup$ My point is that all the positive examples (in the given range), where you end up with a prime, are such that $\sigma(n+2)-\pi(n)$ is itself a prime, the same prime. For example, in the case of $n=100547$, the final prime $90911$, which is just $\sigma(n+2)-\pi(n)$. Your question is not really about $\sigma(n+2)-\pi(n)$, but about the integers $m$ for which $\sigma_2(m)-1$ is a square number. Most of these $m$'s are primes: it is virtually impossible to find a composite $m$ with this property as such $m$'s are so rare. $\endgroup$
    – GH from MO
    Commented Aug 6 at 2:07
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    $\begingroup$ @WillJagy Yes, I was sloppy when I wrote that comment. The set of non-prime $m$'s up to $10^7$ for which $\sigma_2(m)-1$ is a square equals $\{1,6,40,136,2696,3352,46976,223736,5509736\}$. Probably there are infinitely many such $m$'s even when you add the condition that $\sigma_2(m)\equiv 2\pmod{10}$ and $m$ lies in the range of $\sigma(n+2)-\pi(n)$. And such an $n$ would falsify the OP's conjecture. Probably the OP's conjecture is the result of an evolution where artificial conditions were gradually added to make the counterexamples disappear. $\endgroup$
    – GH from MO
    Commented Aug 6 at 3:41

2 Answers 2

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I am now inclined to believe that the OP's conjecture is true, and I present some ideas to support this. If $n$ is a counterexample, then $m:=\sigma(n+2)-\pi(n)$ is a composite number such that $\sigma_2(m)-1$ is a square number ending in $1$. The goal would be to prove that there is no such $m$ (regardless of $n$).

Let us focus on the composite numbers $m$ such that $\sigma_2(m)-1$ is a square number. These numbers are listed at OEIS, and it seems that they are all even. Let us assume this, and let $k>0$ be the exponent of $2$ in $m$. I claim that $k$ is odd. To see this, observe that every divisor of $\sigma_2(2^k)$ is congruent to $1$ modulo $4$, because it is an odd number dividing $\sigma_2(m)$, which is a square number plus $1$. However, if $k$ is even, then $$\sigma_2(2^k)=\frac{4^{k+1}-1}{3}=\frac{(2^{k+1}+1)(2^{k+1}-1)}{3}$$ is divisible by $2^{k+1}-1$, which is congruent to $3$ modulo $4$. So $k$ is odd (under the standing assumption that $m$ is even). But this implies that $\sigma_2(2^k)$ is divisible by $5$, because in the previous display $4^{k+1}=16^{(k+1)/2}$ is congruent to $1$ modulo $5$. Hence $\sigma_2(m)$ is divisible by $5$, and therefore $\sigma_2(m)-1$ does not end in $1$.

To summarize, the OP's conjecture follows from the more natural conjecture that the linked OEIS sequence consists of even numbers.

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  • $\begingroup$ Thank you for your answer. Is it a proof? $\endgroup$
    – Sulfura
    Commented Aug 7 at 2:06
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    $\begingroup$ @Sulfura It is not a complete proof. What I show is that as long as $m=\sigma(n+2)-\pi(n)$ is even, your conjecture is true. In addition, it is plausible that the remaining case, i.e. when $m=\sigma(n+2)-\pi(n)$ is odd, produces no counterexample. $\endgroup$
    – GH from MO
    Commented Aug 7 at 2:50
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    $\begingroup$ I'm not so sure about absence of odd terms in OEIS A318169. Here is a near-miss: were $p:=411706627786612628571$ a prime, $11^4\cdot p$ would be such a term as $\sigma_2(11^4)(1+p^2)-1$ is a square. $\endgroup$
    – Max Alekseyev
    Commented Aug 7 at 20:17
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    $\begingroup$ @MaxAlekseyev Yes, it is hard to make a guess. The question has a similar flavor (at least to me) as the existence of odd perfect numbers. I agree that a more extensive search is necessary here (perhaps a clever search, not a bruce force search). Please keep us updated! $\endgroup$
    – GH from MO
    Commented Aug 7 at 20:48
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This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ where } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ is also a prime number.

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    $\begingroup$ Yes, I emphasized this in my response: "The goal would be to prove that there is no such $m$ (regardless of $n$)." $\endgroup$
    – GH from MO
    Commented Aug 7 at 1:22
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    $\begingroup$ It was proven with a simple code that there is no counterexample for$ \ \ n<100e6 \ \ $(just adjust the Python code in the main post). $\endgroup$
    – Aline Houfty
    Commented Aug 7 at 17:45
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    $\begingroup$ My response and the linked OEIS page make it clear that there is no counterexample up to $2\cdot 10^{12}$. Also, it suffices to check odd numbers only (cf. my response). $\endgroup$
    – GH from MO
    Commented Aug 7 at 18:42

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