# 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.

Note by Alisa Meier
3 years ago

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## Comments

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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.

Staff - 3 years ago

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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..

- 3 years ago

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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 :)

Staff - 3 years ago

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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).

Staff - 3 years ago

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Yeah, that edit was necessary.

- 3 years ago

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