One of the reasons this would be a huge result is that even a very expensive polynomial-time solution for any NP-complete problem (say, one which is O(n^2000) and would take several billion years to run against a 256-bit key) is only going to get faster over time. Also, the experience of cryptography is that once cracks start appearing in an algorithm the mathematicians start trying to wedge their chisels in and hammer those cracks into fractures, into holes, and eventually into whatever you call something which no longer resembles a crack so much it resembles absence-of-wall.

There are specific mechanisms by which this would happen, too. A near miss on P=NP would mean funding for professors looking into refining it, and professional interest generally, would suddenly explode. Nobody wants to fund or work on duds, but big contributions to this problem would make careers.

Not incidentally, while the government funds quite a bit of academic research, this particular branch of mathematics is of special interest to one type of government customer which has ways of getting funding for projects expedited if they are in the national interest... and it is virtually inconceivable that they would not find very, very strong national interest in the vicinity of this result.

Edit : For some reason I can't reply to posts so I'll do it here.

RiderOfGiraffes> It would will leave in question those problems in NP, but not NPC. Currently it is generally believed, but not known, that factoring is such a problem.

This is exactly my point. P = NPC doesn't imply anything about factorisation. Since factorisation is in NP we haven't shown that P = NP

kd0amg> NP-complete problems ... are, in essence, the hardest problems in class NP.

From what I understand they a 'thought to be' the toughest but this isn't proven. If this article turned out to be true then they are reduced to being as hard as P. So who is still standing as a contender for the hardest problem in NP? - factorization of course as it has never been demonstrated to be polynomially reducible to NPC.

Proving that any single NPC problem is in P will be enough to prove that every NP problem is in P, and not just the NPC ones. Suppose A is is NP, B is in NPC, and further suppose that solving B is polynomial. Reduce A to B (a polynomial operation because B is in NPC), solve B (a polynomial operation by assumption), and convert the solution back to a solution of A (a polynomial operation because B is in NPC), and that gives a polynomial solution of A.

Proving that any single NPC problem is not in P is enough to prove that all NPC problems are not in P. It would will leave in question those problems in NP, but not NPC. Currently it is generally believed, but not known, that factoring is such a problem.

Added in edit, to reply to your edit:

    RoG> It would will leave in question those problems
    RoG> in NP, but not NPC. Currently it is generally 
    RoG> believed, but not known, that factoring is such
    RoG> a problem.

    dejb> This is exactly my point. P = NPC doesn't imply
    dejb> anything about factorisation. Since factorisation
    dejb> is in NP we haven't shown that P = NP

    Proving that any single NPC problem is in P will be
    enough to prove that every NP problem is in P.

Let me add a little more.

   1. Suppose 3-SAT is in P.
   2. Let A be any problem in NP.
   3. 3-SAT is in NPC.
   4. Therefore A can be converted to A', an instance of 3-SAT.
   5. By (1), A', as an instance of 3-SAT, can be solved in polynomial time
   6. Hence A has been solved in polynomial time.

This was proved for SAT first: http://en.wikipedia.org/wiki/Cook%E2%80%93Levin_theorem

http://en.wikipedia.org/wiki/NP_complete

a. P is in NP, and

b. Every problem in NP can be converted to an instance of P.

Consider any two problems in NPC, P and Q. Then each is in NP, and each can be converted to the other. They are therefore equivalent.

QED.

Every NP problem is polynomial-time reducible to SAT. I do not recall the title of the paper off the top of my head, but it has been proven that the definition of an NP problem (a question that can be answered given some additional piece of information, a "certificate") is polynomial-time reducible to SAT.

The methodology was basically an algorithm taking as input a description of such a solution, and producing a boolean expression on the certificate and some other values needed to maintain integrity. The expression is satisfiable if and only if the answer is affirmative.

The reason that factorization is not known to be NP-complete is that the reverse is not known. That is, no NP-complete problem is known to be polynomial-time reducible to factorization.

Yes, it is. Any NP problem can be many-one reduced to SAT, which means that there's a function that transforms an instance of the NP problem to an instance of SAT, and the answer to whether that formula is satisfiable is the same as whether that instance is a member of the NP problem (in its decision version).

So who is still standing as a contender for the hardest problem in NP?

If P = NP, then all problems in NP are as easy as each other, so this question is meaningless.

The provable lower bound for the worst-case performance of searching is O(n log n). Finding the median takes linear time i.e. O(n).

Sorting and finding a median reduce to each other polynomially, of course.

Interesting subsets of polynamial problems are e.g. linear problems, or highly parallelizeable problems whose runtime tends to O(log n) as the number of processors tends to infinity. Adding a list of numbers is not only linear but also highly parallelizeable. Proving a problem to be not parallelizeable like that, is also interesting.

There's also the question whether randomness helps. It does in practice, but we don't know whether access to random bits helps in theory.

'It is suspected to be outside of all three of the complexity classes P, NP-complete, and co-NP-complete'

So even if this showed 'P = NP-complete' it may not imply imply factorization is in P. I'm at the limits of my rusty complexity knowledge here so if any knows better please correct me.

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 …”.