Zero-knowledge proof for $P \ne NP$? In computational complexity, $P \ne NP$ is a widely believed conjecture. Suppose that someone discovered a proof for it. He wants to publish a proof that he correctly proved the conjecture. I am aware that NP-complete problems have Zero-knowledge proofs. Assuming $P \ne NP$, can we have Zero-knowledge proof (or morally equivalent one) for  $P \ne NP$?
 A: If your friend is willing to part with an upper bound $B$ on the length of the proof in some formal system, then you can

*

*Take a computer program which verifies proofs in that formal
system, which (presumably) runs in polynomial time

*Convert it into a circuit which takes $B$ bits as input and outputs True if it is a valid proof of $P \neq NP$ and False otherwise.

*Reduce CircuitSat to the existence of a Hamiltonian cycle using the NP-completeness of Hamiltonian cycles, and find the Hamiltonian cycle corresponding to the proof.

*Use the zero knowledge proof scheme for Hamiltonian cycles.

A: First of all, it does not really make sense to talk about zero-knowledge proofs for a single statement such as $P\ne NP$.  You really should be asking about whether there is a zero-knowledge proof for a computational problem, such as: "Given a statement $S$ in the first-order language of set theory, and a positive integer $n$, does there exist a ZFC-proof of $S$ of length at most $n$?"
With this reformulation, you have already partially answered your own question, by noting that there is a standard theorem about the existence of zero-knowledge proofs for any $NP$-complete language. However, there are two caveats to be aware of in that standard theorem. The first caveat is that it assumes the existence of a one-way function, which is a stronger assumption than $P\ne NP$. So if you only have the hypothesis $P\ne NP$ available to you, then that's not quite good enough.  The second caveat is that the usual proof relies on a commitment scheme, which has two parts to it, the binding part and the hiding part.  Ideally we'd like both parts to be statistically secure and not just computationally secure, but it turns out that it is impossible for both parts to be statistically secure unless the polynomial hierarchy collapses to the second level (Fortnow, "The complexity of perfect zero knowledge").  So if you want the strongest possible form of zero knowledge, then the answer to your question is likely no.
A: This is just a long comment, but it probably depends on the actual structure of the proof.
Imagine some Ramsey-theoretical question, and you have proved that all graphs with property P are Hamiltonian. One zero-knowledge proof would be your friend supplying you with a graph with property P, and you produce a Hamiltonian cycle in it.
But, proving that something is Hamiltonian is not the same as being able to (efficiently) producing the Hamiltonian cycle, so you are stuck.
One can imagine a probabilistic type proof, showing that the expected number of 'obstructions' to $P=NP$ is positive, but this is not at all the same as being able to produce such obstructions concretely (which is perhaps what a zero-knowledge proof would be).
However, there are certainly many ways to give zero-knowledge proofs of $P=NP$,
perhaps the easiest one is to break some encryption scheme and get rich.
Also, this paper seems very much related.
