$q = x_a({p_a}^2)+4$
Let $p_n$ be a sequence of consecutive prime numbers starting from 3. ($p_a$ represents a prime number in the sequence, and $x_a$ is the corresponding $x$ to the prime number.)
We are able to set nonzero natural number $x_a$ freely. How does one find such nonzero natural number $q$ that satisfies the aforementioned given the number of prime numbers in the sequence?
Also, what happens if
1) we allow $x_a$ and $q$ to be nonzero integer, not just natural number?
2) $p_n$ is defined as a sequence of consecutive ${p_a}^{b_a}$ where $p_a$ is defined as aforementioned We allow to set nonzero natural number $b_a$ freely. In this the equation changes to $q = x_a({p_a}^{2b_a})+4$.
3) If we remove "consecutive" from the definition of $p_n$ (either the first definition or the definition in 2).)
4) Combination of 1) and 2) or combination of 1) and 3)
Add: Is there any such $q$ that is smaller than the product of numbers $p_a$ in case of the origianl case? Also is there any such $q$ that is smaller than the product of numbers ${p_a}^{b_a}$ in case of 2)?