NP Solvable in Polynomial Time I just took an exam and am a little curious about this question (it may not be verbatim, but the idea is clear):
TRUE/FALSE: If an NP complete problem can be solved in polynomial time, then P = NP.
My thought was FALSE.  A single NP-complete problem being solved in polynomial time doesn't prove that P = NP, it just proves that this one problem $\in$ P.
It made sense during the exam, but I'm not too confident now.

You're incorrect. If some NP-complete problem $A$ can be solved in polynomial time, then given any other NP problem $B$ we can solve it in polynomial time by first reducing $B$ to $A$ in polynomial time and then running the polynomial algorithm for $A$. A nearby statement is false, namely, that if a problem in NP can be solved in polynomial time then $P=NP$. Any problem in $P$ is a counterexample-unless, of course, $P=NP$!
