I want to add another important aspect, the aspect of NP completeness. There is a bunch of problems considered "NP complete" and they are all related such that it is easy to translate one problem into another (easy as in "quickly").

This means, first: If P == NP, then all of these problems become easy, and second: if P == NP and we find an algorithm that solves only one of the NP complete problems quickly, then this algorithm can solve all algorithms quickly.

Reversely, if now this paper's proof is correct so P != NP, then there is no algorithm that solves any of these problems quickly.