What is the maximum cardinality of a set $Z \subset \mathbb{N}$ such that for any distinct $a_1, a_2, a_3 \in Z$, their sum $a_1 + a_2 + a_3$ is prime?

This means: $Z = \{a_1, a_2, a_3, \cdots, a_n\}$, $a_i \in N^{*}$, $a_1 < a_2 < \cdots < a_n$, then $a_i + a_j + a_k$ ($i\neq j\neq k$) is prime. Find $|Z|_{max}$.

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    $\begingroup$ This makes no sense. If English is a real problem, perhaps you could put in a number of examples to illustrate what you are talking about. $\endgroup$ – Will Jagy Mar 16 '13 at 4:29
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    $\begingroup$ It is not clear to me what the question is asking. Almost everything is the sum of primes. For example $9=2+2+2+3$. Do you mean $2$ primes? Or $2$ distinct primes? By "any of the three" do you mean the sum of any three"? $\endgroup$ – André Nicolas Mar 16 '13 at 4:30
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    $\begingroup$ This is very unclear. If English isn't your native language, then add the original text to this post, and use the (translation-request) tag. Hopefully, someone can help you out. $\endgroup$ – Cameron Buie Mar 16 '13 at 4:35
  • $\begingroup$ I've edited the question to how Zander and I interpreted the question. If this is incorrect, please let us know. $\endgroup$ – TMM Mar 29 '13 at 14:11

If I have interpreted the question correctly, you want the size of the largest set $Z$ of positive integers such that the sum of any three elements of $Z$ is prime.

$Z$ cannot have 5 elements, otherwise by the pigeonhole principle there would exist $a,b,c\in Z$ satisfying either $$ a\equiv b\equiv c \pmod{3} $$ or $$ a \equiv b+1 \equiv c+2 \pmod{3} $$ In either case $a+b+c=3k, k>1$ is composite.

However $Z=\{1,5,7,11\}$ satisfies the condition so $|Z|_{max}=4$.

  • $\begingroup$ Please forgive me if I answered the wrong question. $\endgroup$ – Zander Mar 29 '13 at 13:56
  • $\begingroup$ This is also how I understand the question, and your answer seems correct, so (+1) from me! $\endgroup$ – TMM Mar 29 '13 at 14:04

I am still confused with your question. I hope this theorem is of relative importance to your question.

Theorem: (Vinogradov)

Any sufficiently large odd integer n can be written as a sum of three prime numbers.

By "sufficiently large" we mean $n > 10^{43000}$.

  • $\begingroup$ I don't know what Chen and Wang proved, but the fact that every large enough odd number is the sum of three primes is due to Vinogradov. $\endgroup$ – Chris Eagle Mar 16 '13 at 6:40
  • $\begingroup$ You're right, Chen and Wang just improved on the bound. $\endgroup$ – Rustyn Mar 16 '13 at 6:41
  • $\begingroup$ this mean : $Z=\{a_{1},a_{2},a_{3},\cdots,a_{n}\},a_{i}\in N^{*}$,then $a_{i}+a_{j}+a_{k},(i\neq j\neq l)$ was primes,find $|Z|_{max}$ $\endgroup$ – math110 Mar 16 '13 at 6:45
  • $\begingroup$ yes,Thank you my frend $\endgroup$ – math110 Mar 16 '13 at 6:49

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