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Let be $\mathcal{P} \subseteq \mathbb{N} $ with the following property:

For all $p \ge 2$, $p$ prime number, $ \exists \ \{a_1, a_2, \ldots , a_p \} \subseteq \mathcal{P}$ which is a complete residue system modulo $p.$

Does this implies that $\mathcal{P}$ has infinite prime numbers?

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  • $\begingroup$ Can I just take $\mathcal{P} = \{ n\in \mathbb{N} \,|\, n \text{ is not prime}\}$? $\endgroup$ – Malcolm Dec 5 '17 at 21:39
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    $\begingroup$ No. You may take $\mathcal{P}$ as the set of the non-prime sums of two squares, for instance. This particular $\mathcal{P}$ has density zero and does not contain any prime. $\endgroup$ – Jack D'Aurizio Dec 5 '17 at 21:43
  • $\begingroup$ Or you may take $\mathcal{P}$ as the set of "prime numbers plus seven or thirteen". $\endgroup$ – Jack D'Aurizio Dec 5 '17 at 21:48
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No, it does not. $$ \mathcal{P} = \{ n\in \mathbb{N} \,|\, n \text{ is not prime}\} $$ contains no primes and contains a complete residue system mod $p$ for all primes $p$.

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  • $\begingroup$ Thanks, that is a good counterexample. $\endgroup$ – 674123173797 - 4 Dec 10 '17 at 2:51

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