This conjecture is inspired by the comment of @Eric Snyder: Prime numbers which end with 03, 23, 43, 63 or 83
$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ is the $n^{th}$ prime number and $\sigma(n)$ is the sum of the divisors of n. Consider the expression:
$$F(n)=\varphi(|\sigma(n)-P_{n+2}|)+1$$
Conjecture: when $F(n) \equiv 3 \mod 20$ then this number is a prime or $P_{n+2}-\sigma(n)=p^2$ (p prime) when the number is not a prime.
Examples:
$n=10270001113$, we have:
$$F(10270001113)=\varphi(\sigma(10270001113)-P_{10270001115})+1=\varphi(10468624896-259189944599)+1=248721319703$$ which is prime because it ends by 03.
A counterexample is found with n=680:
$$F(680)=\varphi(\sigma(680)-P_{680})+1=\varphi(1620-5101)+1=3423$$ which is not prime but we have $P_{n+2}-\sigma(n)=p^2$, more precisely it is a square of 59.
Interestingly for $n \leq 30 000 000$ all counterexamples have $P_{n+2}-\sigma(n)=p^2$
- Prove that there are infinity $P_{n+2}-\sigma(n)=p^2$?
- Prove that there are only two possibilities: the result ends with 03, 23, 43, 63 or 83 and it is a prime or the result ends with 03, 23, 43, 63 or 83 and it is not a prime. In this case $P_{n+2}-\sigma(n)=p^2$
Let $n$ be a natural number $\geq 1$, $\varphi(n)$ is the Euler's totient function, $\sigma(n)$ is the sum of divisors of $n$.
The important result is that if n is odd and $\varphi(n)=4k+2$ then $n$ has only one prime divisor. If we focus on numbers $a+b$ with $a$ even ($a=\sigma(k)$) and $b$ odd ($b=P_{k+2})$ the conjecture is true but we can generalize this result for all odd numbers (so this is the reason for which the formula is true with prime numbers).
Here I claim that only exceptions are $p^2$ but in reality the only exceptions are $p^k$.