Are all even numbers the difference of prime powers

Question

Does there exist an even positive integer greater than $100$ (to eliminate trivial cases) that cannot be expressed in the form:

• $p^2-q$
• $p-q^2$
• $p^2-q^2$
• $p^3-q^3$

where $p$ and $q$ are primes.

Answer 1

For sure, there are infinitely many even integers that cannot be written as $p^2-q^2$ where $p,q$ are primes. You only have to look at integers of the form $4k+2$ (or $8k+4$), where $k\in\mathbb{Z}$.

Also, there are infinitely many even integers that cannot be written as $p^3-q^3$. Just look at integers of the form $2r$, where $r$ is a positive prime such that $r \equiv -1\pmod{3}$.

If $n$ is an even number such that $n \equiv -1\pmod{3}$ and $n+9$ is not prime, then $n$ cannot be written as $p-q^2$, where $p,q$ are primes. For example, you can take $n=30k-4$ for any positive integer $k$.

If $n$ is an even number such that $n \equiv +1\pmod{3}$ and $n-9$ is not prime, then $n$ cannot be written as $p^2-q$, where $p,q$ are primes. For example, you can take $n=30k+4$ for any positive integer $k$.

Three conditions can simultaneously be satisfied. For example, in the case of $226$ (or any number of the form $2(210k+113)$ such that $k$ is a nonnegative integer and $210k+113$ is prime), the $p^2-q$, $p^2-q^2$, and $p^3-q^3$ conditions hold. In the case of $716$ (or any number of the form $4(210k+179)$ such that $k$ is a nonnegative integer and $210k+179$ is prime), the $p-q^2$, $p^2-q^2$, and $p^3-q^3$ conditions hold. I don't know if there is any even number that satisfies all these four conditions, but my gut feeling tells me that it is impossible to find one. In fact, I don't think that there is any number satisfying both the $p-q^2$ and the $p^2-q$ conditions simultaneously.