Square roots and prime numbers

Question

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.

Questions

1. Is this conjecture interesting?
2. Is it possible to prove this conjecture or to find counter-example?

Generalization of the conjecture

If $\sqrt{A-1}$ is an integer so $\sqrt{A-1}$ is always prime.

Answer 1

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.

Answer 2

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.

Answer 3

GH from MO's answer suggests to study (odd) composite numbers $m$ such that $\sigma_2(m)−1$ is a square. Therefore $\sigma_2(m)$ is not divisible by $4$ and has no (prime) divisors equal $3$ modulo $4$. Let $m=p_1^{a_1}\dots p_k^{a_k}$ be the prime decomposition of $m$. Then

$$\sigma_2(m)=(1+p_1^2+\dots+p_1^{2a_1})\dots (1+p_k^2+\dots+p_k^{2a_k})=$$ $$\frac{p_1^{2a_1+2}-1}{p_1^2-1}\cdots \frac{p_k^{2a_k+2}-1}{p_k^2-1}.$$

In particular, if $m$ is odd then no $a_i$ equals $2$ or $3$ modulo $4$ and at most one $a_i$ is odd (and in this case it equals $1$ modulo $8$). Moreover, for each natural $i\le k$ the greatest common divisor of $p_i^2-1$ and $a_i+1$ has no divisors $q$ equal $3$ modulo $4$, because otherwise $1+p_i^2+\dots+p_i^{2a_i}$ divides $q$.

Moreover, suppose that $\sigma_2(m)\equiv 2\pmod {10}$, as required. Then exactly one $a_i$, say, $a_1$ equals $1$ modulo $8$ and each other $a_i$ is divisible by $4$.

I am trying to advance this study farther.

Answer 4

My conjecture is false.

With $n=31880$ we have $A=\sigma_2(|\pi(31880)-\sigma(31882)|)=\sigma_2(|3424-50400|)=\sigma_2(46976)=2942303050$

We have $A-1=2942303049$ and $\sqrt{2942303049}=54243$

Here we have $\sqrt{A-1}=54243$ that is an integer but $54243$ is not prime ($54243=3^3*7^2*41$)