Question

While reading Peter Shor's paper Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer, I came across the following quote:

"This scheme will thus work as long as $n$ is odd and not a prime power; finding factors of prime powers can be done efficiently with classical methods."

I have two questions:

(1) How does one efficiently determine if a given number $N$ is a prime power, and

(2) How does one efficiently determine the factorization of a prime power?

(Note: I have included these as separate questions since I am aware that many of the standard algorithms for determining a number is composite will not necessarily produce a proper factor.)

Answer 1

If N is a prime power it is of the form $p^i$ where $i \leq \log_2(N)$. One can thus compute each of the first $\log_2(N)$ roots of $N$, and test if the resulting number is (first an integer) and then if it is a prime using the AKS algorithm (although perhaps Shor meant to use a randomized test, such as Miller–Rabin, as the paper pre-dates AKS). Obviously if $N$ is a prime power this will produces the factorization.

Answer 2

See Dan Bernstein's paper, ``Detecting perfect powers in essentially linear time.'' Mathematics of Computation 67 (1998), 1253--1283, available at http://cr.yp.to/papers.html#powers. Here "linear" means "linear in log n".

Answer 3

If $n = p^k$ with $p \geq 2$ then $k \leq \log_2 n$. For each candiate $k$, we can compute the integral part of $m = \lfloor n^{1/k} \rfloor$ in time polynomial in $\log_2 n$, and then test whether the $n = m^n$. This gives a time complexity polynomial in $\log_2 n$, also known as "efficient".