( Soft question ) P vs NP - is such a situation possible?

Question

Currently P vs NP is the holy grail of theoretical computer science. And the nature of the problem is as such that if actually P = NP is proved then most of the proofs for mathematical statements would be trivial to find, and on the exact other end if P != NP is proved then one would appreciate the hardship of finding its own proof finally....

People working in this field are so tired and exhausted of trying to find the solution that many have accepted that they might not even be alive by the time its solved, and many believe we would have to make some other groundbreaking discoveries in math to even be able to think of a rigorous solution.

In fact almost every proof method frequently used to separate complexity classes - like in halting problem's proof - have been proven "insufficient" to solve the problem...

Regardless of the answer being positive or negative, is it possible that if a proof is announced and its actually correct - ONLY for us to realise how trivial it actually was...? like we actually never needed any new concepts... the whole problem was basically hiding right in front of our sight. And maybe or maybe not we require other monumental maths to recognise the trivialness of the problem, but the proof itself, a rigorous and correct one, only needs concepts we have today.....

I know such extreme cases are rare or maybe haven't even occurred yet in maths, but the shear simplicity of the problem - in the sense that a 7th standard student can grasp the essence of the problem efficiently and effectively, just makes me wonder about this situation many a times....

Answer 1

Yes of course it is possible. Everything is possible. Such situations have been rare in mathematics, so I would not bet on it, but we have no way to rule out such a possibility. We do know that most of the standard proof techniques we use every day in complexity theory are unlikely to work, which might be taken as some kind of evidence that it's less likely that the proof will turn out to be simple, but you never know.

In any case, it doesn't really have any actionable implications that I can see, so it's probably not worth worrying about too much.

Answer 2

Consider what happens if I find a polynomial time solution for an NP complete problem. Say a solution taking $n^{100}$ nanoseconds on my computer.

For n = 2 that’s $2^{100} \approx 10^{30}$ nanoseconds. Or $10^{21}$ seconds. Or 30 trillion years. So all but the smallest instance are not trivial, but practically unsolvable.

Now here is a problem that was believed to be very hard and is now known to be trivial: I’ll give you one integer n >= 0 and m >= 1, and you tell me if there are a, b, c >= 1 with at most $10^m$ digits so that $a^n + b^n = c^n$. Finding the answer for n = 3 or n = 4 alone was very, very hard. Today I know that the answer is “Yes” if n=1 or n=2 and “No” otherwise.