[bawolff]

I was thinking, if you could definitively prove these assumptions are hard, that would prove P != NP, because if P=NP that would imply there would be an algorithm to solve these types of problems, since they are the type of thing that can be solved in polynomial time with the key, but cannot without a key. (I'm a bit out of my depth here)

[adastra22]

For the stuff underlying asymmetric keys, yes. The hash function stuff doesn’t have backdoors.

[gpm]

Hash functions too. If P=NP then you can reverse a hash in polynomial time.

NP is the set of all functions that you can verify a solution to in polyomial time, and the solution of the inverse-hash function is just a plaintext that hashes to the right value, and obviously you can check if a plaintext is right in polynomial time by just hashing it and comparing the hashes. Thus reversing a hash function is in NP, so if P=NP it's in P.

There's some subtlety here in that "reversing" a hash function really just means coming up with a plaintext that generates the right hash, not the original one, but you can put any polynomial-time set of constraints on the plaintext and finding a plaintext that satisfies those constraints (and hashes to the right value) is still in NP, so the subtlety really doesn't save you much.

Edit:

Side point, but since we're talking quantum, we should really be saying BQP=NP not P=NP, BQP being the problems solvable in polynomial time on a quantum computer, it's a superset of P and a subset of NP, but we don't know if it's equal to either or both. I.e. P=NP implies BQP=NP, BQP != NP implies P != NP, BQP != P implies P != NP, but the reverse of all of those statements isn't known to be true.