Cperciva is right, but I think you're mistaken when you say "many problems harder than NP are NP complete", NP complete problems are both NP and NP hard. NP hard is a class of problems such that every problem in NP can be reduced to any problem in NP hard. Meaning if P=NP then every problem in NP(including the NP complete problems) can be solved in polynomial time; all it takes is one algorithm that solves an NP complete problem in polynomial time (note that this does not show that P=NP hard, a much stronger claim).

You're right. I've got the brain hiccups.

> Also, there are problems in NP that are neither in P, nor are they NP- or co-NP-complete.

Err, it seems to me you've got a very serious statement there. I think you mean “that are not known to be …”.

No, there actually are problems in "NP-intermediate" class (if P!=NP) although they are artificial. :) And, of course you are right, for many other problems researchers suspect they might be in NPI. See http://cstheory.stackexchange.com/questions/79/problems-betw...

It's been proven that if P != NP, then there are problems in NP, which are neither in P nor NP-complete. We don't know what those problems are, because identifying such a problem would prove P != NP. There are candidates, such as factoring, for which there are neither polynomial algorithms nor proofs of NP-completeness, but definitively identifying them is at least as hard as proving P != NP, and possibly harder.

Yep. I should just stop posting today. :)

The point I was trying to make (badly) was that the actual claim doesn't imply P = NP, only that P = NP-complete. Given that factoring isn't known to be NP-complete (or co-NP-complete) this paper doesn't imply that factoring is in P.

If a single NP-Complete problem is in P, then P=NP. This is what NP-Complete means. In particular, factoring would be in P. Conversely, factoring is not NP-complete, so if factoring were in P, we could not draw any conclusions about P=?NP.