# Complete residue system for all $p$ implies infinite prime numbers

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?

• Can I just take $\mathcal{P} = \{ n\in \mathbb{N} \,|\, n \text{ is not prime}\}$? – Malcolm Dec 5 '17 at 21:39
• 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. – Jack D'Aurizio Dec 5 '17 at 21:43
• Or you may take $\mathcal{P}$ as the set of "prime numbers plus seven or thirteen". – Jack D'Aurizio Dec 5 '17 at 21:48

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 $p$.