Problem 1315

1. Parity Consideration:
- We start by noting that if three numbers in the set $A$ have the same parity (all odd or all even), then their pairwise differences would be even. Since the only even prime number is 2, all these differences would have to be 2, which is impossible for three distinct numbers. Therefore, the set $A$ can contain at most two odd and two even numbers.

2. Set Size:
- Given that $n \geq 4$ and the set $A$ must contain exactly four elements (two odd and two even), we can proceed to construct such sets.

3. Odd Elements:
- Since $2015$ is an odd number and must be included in the set, the other odd number must be 2 units away from 2015 to ensure the difference is prime. Thus, the other odd number can be either $2013$ or $2017$.

4. Even Elements:
- The two even numbers must also be 2 units apart to ensure their difference is prime. These even numbers can either be both less than 2013 or both greater than 2015.

5. Case Analysis:
- Case 1: The odd numbers are $2013$ and $2015$.
- The even numbers must be 2 units apart and less than 2013. The only pair that fits this criterion is $2008$ and $2010$.
- Thus, one possible set is $\{2008, 2010, 2013, 2015\}$.

- Case 2: The odd numbers are $2013$ and $2015$.
- The even numbers must be 2 units apart and greater than 2015. The only pair that fits this criterion is $2018$ and $2020$.
- Thus, another possible set is $\{2013, 2015, 2018, 2020\}$.

- Case 3: The odd numbers are $2015$ and $2017$.
- The even numbers must be 2 units apart and less than 2015. The only pair that fits this criterion is $2010$ and $2012$.
- Thus, another possible set is $\{2010, 2012, 2015, 2017\}$.

- Case 4: The odd numbers are $2015$ and $2017$.
- The even numbers must be 2 units apart and greater than 2017. The only pair that fits this criterion is $2020$ and $2022$.
- Thus, another possible set is $\{2015, 2017, 2020, 2022\}$.

6. Verification:
- We verify that in each of these sets, the pairwise differences are prime:
- For $\{2008, 2010, 2013, 2015\}$:
- Differences: $2, 5, 7, 3, 5, 2$ (all prime).
- For $\{2013, 2015, 2018, 2020\}$:
- Differences: $2, 5, 7, 3, 5, 2$ (all prime).
- For $\{2010, 2012, 2015, 2017\}$:
- Differences: $2, 5, 7, 3, 5, 2$ (all prime).
- For $\{2015, 2017, 2020, 2022\}$:
- Differences: $2, 5, 7, 3, 5, 2$ (all prime).

The final answer is $\boxed{ \{2008, 2010, 2013, 2015\}, \{2013, 2015, 2018, 2020\}, \{2010, 2012, 2015, 2017\}, \{2015, 2017, 2020, 2022\} }$.