Today, to the stack community, I wish to share gist of observations I took in a past few days in specific problems from elementary number theory. These observations might be trivial to prove, and it is likely that I might be missing something to prove/disprove them. The observations are as follows.
Let $p$ be a prime greater than 3. Assume that $p+1$ is square-free throughout the post. The post is divided in two observations, the first is about the form of prime $p$ such that $p+1$ is square-free and the second one is about the dependency of form of prime $p$ on prime factors of $p+1$ square-free number.
Observation 1: If $p$ is a prime such that $p+1$ is square-free, $p$ is surely a number of $4k+1$ form, that is to say, $p \equiv 1\, mod(4)$. To provide an example to this, say $p = 73$, such that $p+1 = 74 = 2*37$ is square-free. From the observation made here, for the given example, it holds true that, $p = 73 = 4k+1.$
Observation 2: If $p$ is a prime such that $p+1$ is square-free, and if $p+1 \not \equiv 0 \,mod (6)$, then, $p\equiv 1 \, mod \, (6)$, or other way around $p$ is of $6t +1$ form. Additionally, the product of, prime factors ($p+1$) modulo 6 except 2, is +1. To illustrate this, let $p = 9337$, $p+1 = 9338 = 2*7*23*29$. Here, except $2$, $7 \equiv 1 \, mod (6) = a_1$, $23 \equiv -1 \, mod (6) = a_2$, and $29 \equiv -1 \, mod (6) = a_3$. Then, $\prod_{i = 1}^{i=3} a_i = 1$ which I believe, reflects the $6k +1$ form of the $p$.
I am not able to draw much to prove the above observations, however, to observe it further, I am providing link of my colab repository from where I programmed these observations for bigger values of prime, $p$.
Attempt on proving Observation 1: Let, $p+1$ be not square-free and it is a number of form $p \equiv 1 \, mod(4)$. That is, there exists a prime $q>2$ such that $p+1 = 2 \times q^2 \times r$ . If $p > 3$, it is guaranteed that $p + 1$ is even, and the other assumption as a means of first case of this proof is, I'm assuming that, it has no powers of 2 in its prime factorization.
In this case, since $q$ has to be a prime of form $6k\pm 1$ and $r$ is an odd number of form $4t \pm 1$, with assumption that $gcd(q^2,r) = 1$. Then we can divide the problem in two cases as follows.
Case 1:
If, $p + 1 = 2 \times (36k^2 \pm 12k + 1) \times (4t + 1)$. If we operate modulo 4 on both the sides, we end up in $p \equiv 1 \, mod(4)$.
Case 2:
If, $p + 1 = 2 \times (36k^2 \pm 12k + 1) \times (4t + 3)$. If we operate modulo 4 on both the sides, we end up in $p \equiv 1 \, mod(4)$. Here, a point to be noted is, I've used $4t + 3$ instead of $4t -1$, however both of these are equivalent in modular arithmetic, hence can be manipulated.
Therefore, in either of these cases, no trivial contradiction is observed. Looking at the case when, $4|p+1$ which is mentioned in comments as well, leads to $p \equiv 3 \, mod(4)$, which is a counter example. This completes the proof of observation 1.
For the proof on observation 2, I'm unable to write along the lines of the proof.
I will be thankful to the stack community, if any guidance on the proof and validity of observations can be provided.
