Requesting assistance regarding unique factorization representation of numbers by s_p primes

Let Sp:={np+1nN0}={1,p+1,2p+1,} S_p := \{np+1 | n \in \mathbb{N_0} \} = \{1, p+1, 2p+1, \dots \}

An element spSp s_p \in S_p is called sps_p prime, if and only if it's only divisors in S_p are 11 and sps_p .

In Apostol's book "An Introduction to Number Theory" I found an exercises, in which one had to show that every number in S4S_4 is either an s4s_4-prime or a product of s4s_4-primes.

A number pp that suffices this property be now called pp-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,ySP x,y \in S_P , then there \exists unique m,nN m,n \in \mathbb{N} with k(np+1)=mp+1 k(np+1) = mp+1 for a yet unspecified kN k \in \mathbb{N}

kk itself has unique representation: k=ps+tk = p*s+t with sN s \in \mathbb{N} and 0tp10 \le t \le p-1

Thus one gets the equation:

(sp+t)(np+1)=mp+1spnp+sp+np+t=mp+1p(nsp+sp+npm)+t=1(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: SpS_p is complete for any pN p \in \mathbb{N}

Now I am interested in all sets SpS_p, in which all numbers have a unique prime factorization. I would call such a set SpS_p perfect. However which sets SpS_pare 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
4 years ago

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What have you tried?
Is S2 S_2 complete? Why, or why not?
Is S3 S_3 complete? Why, or why not?
Is S4 S_4 complete? Why, or why not?
Is S5 S_5 complete? Why, or why not?

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

Calvin Lin Staff - 4 years ago

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I also got a result now: SpS_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 - 4 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 :)

Calvin Lin Staff - 4 years ago

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

For example, 9S4 9 \in S_ 4 , and the only divisors are not 1 and 9 (since it has a divisor of 3).

Calvin Lin Staff - 4 years ago

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Yeah, that edit was necessary.

Alisa Meier - 4 years ago

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