What impact would P=BQP have on NP?

Assuming P=BQP (ie we have polynomial time algorithms to solve all BQP problems) can we use it to prove that P=NP?

The argument is that since we have the Grover's algorithm which can solve NP complete problems with a quadratic speedup and since we have assumed that P=BQP, we can apply the Grovers algorithm repeatedly until it is reduced to a polynomial time problem.

• The argument makes no sense to me. Grover’s algorithm does not solve any NP-complete problem, in fact it solves a P-problem, hence assuming P=BQP does not tell us anything about Grover’s algorithm that we don’t already know. More generally, P=BQP might mean that every problem solvable in time $n^c$ on quantum computer is solvable in time, say, $n^{42c}$ on a classical computer, hence it doesn’t imply anything about problems where we only have polynomial speed-up. – Emil Jeřábek Aug 9 '14 at 20:04