Is every integer $\geq5$ the sum of two primes and a power of a prime (where $1$ is included in the prime powers)?
I don't really expect someone to prove this here, but I wonder if the question has been studied.
Goldbach's conjecture has been tested very far, and no exceptions are known. It says every even number greater than or equal to $4$ is the sum of two primes. If true, it immediately follows that given any positive even and odd numbers $a$ and $b$, as long as $2n-a$ and $2n-1-b$ are greater than or equal to $4$, we can form representations of $2n$ and $2n-1$ as $a$ plus the Goldbach representation of $2n-a$ and $b$ plus the Goldbach representation of $2n-1-b$.
In particular, we could take $a=2$ and $b=3$, extending Vinogradov's theorem to even numbers. Those choices would also work for this problem, but we could also take $b=1$. The point is to illustrate that this problem is implied by Goldbach's conjecture, so is likely true and easier to prove.
We already know it is true for odd numbers, so how much easier is it for even numbers? There are really only three cases, since a sum must contain an even number of odd number to be even (there are two non-trivial ways to choose two odds and one way to choose none):
(wherever written, $p,q,r$ are odd primes)
$2n=2^a+q+r$
$2n=2^a+4$
$2n=p^a+2+r$
For the first two, $a$ must be positive, but $a=0$ is also okay for the last one. The first and second combined imply that all we need is for $2n-2^a$ to be a "Goldbach number" for some positive integer $a$, and the last says that $2n-2$ having a representation as the sum of a prime and a prime power is sufficient. It is known that infinitely many positive integers do not have a representation as the sum of a prime and a prime power, but of course none of the ones known are even as this one should be.
Edit:
I'm not confident that much is known on this. In particular, I'm not sure it's known even for three prime powers in the even case.