Do any decision problems exist outside NP and NP-Hard?

Is it possible that P could equal NP, but that it's impossible to provide explicit constructions of such algorithms that could solve our "hard" problems in polynomial time?

• In the same sense that the Banach-Tarski Paradox doesn't provide an explicit construction, despite showing that that a sphere can be decomposed into a finite number of pieces and reassembled without stretching/funny business. I realize that this question is somewhat ill-posed since what "hard" would mean may not be clear if P=NP, but I'd appreciate an answer that gets to the heart of the question rather than a snarky answer that merely points out this ill-posedness.

• Answer:

  There is a possibility that even if P=NP we cannot prove it so.  There's even the possibility that we might be able to prove that P vs NP is independent of ZFC. At first this seems strange, how can you prove something is unresolvable if a single example could resolve it.  Doesn't that imply that no example can exist which for this problem would mean P!=NP?  The issue is that there may be polynomial time algorithms for known hard problems that we cannot prove run in polynomial time (or alternatively, always produce a correct answers).  An independence result would mean that an efficient algorithm might exist and we might be able to stumble upon it, but we'll never be able to prove it is efficient/correct. The existence of a universal polynomial time NP solver if P=NP implies that it is not actually possible for us to be able to prove P=NP yet still be unable to find an efficient algorithm to known hard problems, as points out.  I've edited my initial answer with this in mind.