# P=NP implies a deterministic algorithm to find primes

Consider the decision problem of determining whether there is a prime in the range $[m,n]$, where $m \lt n$. This problem is in NP, since if there is a prime in the range, we can efficiently verify that it is prime. Therefore, if P=NP, then this decision problem is in P, i.e., there is a polynomial-time algorithm to determine whether there is a prime in the range $[m,n]$.
We can use this algorithm as a subroutine to quickly find a prime in the range [n,2n] for large n. We use the well-known fact that such a prime always exists. We then use the polynomial-time algorithm described above to do a binary search to locate such a prime. This terminates after at most $\log_2 n$ applications of the subroutine, and thus is a polynomial-time algorithm to generate a prime.