Prove that for any number $m$ there is always $n$, such that $m - n$ and $m + n$ are prime numbers

Suppose $m = 0$. Then $n$ can just be any prime, since $-n$ would then also be a prime number, right? I think so, but you might not, in which case a few small $m$ might be the only counterexamples.

Then, without loss of generality, we can limit the inquiry to positive $m$, since the negative $m$ will have the same solutions except for opposite signs.

For $m = 1$, let's choose $n = 4$. Then $m + n = 5$ and $m - n = -3$. That checks out.

For $m = 2$, $n = 5$ might be the only choice.

For $m = 3$, $n = 10$ is one out of many possible choices.

This is Goldbach's conjecture and is still not proved.