Requesting assistance regarding unique factorization representation of numbers by s_p primes

Let $S_p := \{np+1 | n \in \mathbb{N_0} \} = \{1, p+1, 2p+1, \dots \}$

An element $s_p \in S_p$ is called $s_p$ prime, if and only if it's only divisors in S_p are $1$ and $s_p$ .

In Apostol's book "An Introduction to Number Theory" I found an exercises, in which one had to show that every number in $S_4$ is either an $s_4$ -prime or a product of $s_4$ -primes.

A number $p$ that suffices this property be now called $p$ -complete. Respectively such a set \(S_p\ will be called complete.

Now one can ask: Which $p \in \mathbb{N}$ suffice this property?

Well, let $x,y \in S_P$ , then there $\exists$ unique $m,n \in \mathbb{N}$ with $k(np+1) = mp+1$ for a yet unspecified $k \in \mathbb{N}$

$k$ itself has unique representation: $k = p*s+t$ with $s \in \mathbb{N}$ and $0 \le t \le p-1$

Thus one gets the equation:

$(sp+t)(np+1) = mp+1 \Leftrightarrow spnp +sp+np+t = mp +1 \Leftrightarrow p(nsp +sp+np-m) + t = 1$ and can immediately confirm: $S_p$ is complete for any $p \in \mathbb{N}$

Now I am interested in all sets $S_p$ , in which all numbers have a unique prime factorization. I would call such a set $S_p$ perfect. However which sets $S_p$ are perfect?

How do I tackle this problem? What is a good approach? Any constructive help, recommendation of reading material, comment or answer is appreciated. Thanks in advance.

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What have you tried?
Is $S_2$ complete? Why, or why not?
Is $S_3$ complete? Why, or why not?
Is $S_4$ complete? Why, or why not?
Is $S_5$ complete? Why, or why not?

The areas that this involves is Modular Arithmetic and related concepts.

Calvin Lin Staff - 5 years, 8 months ago

I also got a result now: $S_p$ is only perfect for p = 1 or p = 2

The proof of that was also not too difficult. Maybe this can be turned into a nice problem for brilliant..

Alisa Meier - 5 years, 8 months ago

That's great! It's not too hard, once you figure out the slight trick involved. Looking at small cases can help, which is why I asked.

I look forward to seeing the question that you pose. It could be made really interesting :)

You should clarify the definition of " $s_p$ prime". I believe what you mean is that "the only divisors of $s_p$ that are in $S_p$ are 1 and $s _ p$ ".

For example, $9 \in S_ 4$ , and the only divisors are not 1 and 9 (since it has a divisor of 3).

Yeah, that edit was necessary.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$