Frequently called the most important outstanding question in theoretical computer science, the equivalency of P and NP is one of the seven problems that the Clay Mathematics Institute will give you a million dollars for proving — or disproving. Roughly speaking, P is a set of relatively easy problems, and NP is a set of what seem to be very, very hard problems, so P = NP would imply that the apparently hard problems actually have relatively easy solutions. But the details are more complicated.
We just covered P vs. NP in my theory of computation class on Thursday. My professor joked that one of the consequences of P=NP would be that a lot of mathematicians would lose their jobs because a lot of hard problems would be doable.
