Detecting perfect powers by factoring into coprimes

This paper presents an algorithm that, given an integer n > 1, finds the largest integer k such that n is a kth power. A previous algorithm by the first author took time b1+o(1) where b = lg n; more precisely, time b exp(O( √ lg b lg lg b)); conjecturally, time b(lg b)O(1). The new algorithm takes time b(lg b)O(1). It relies on relatively complicated subroutines—specifically, on the first author’s fast algorithm to factor integers into coprimes—but it allows a proof of the b(lg b)O(1) bound without much background; the previous proof of b1+o(1) relied on transcendental number theory. The computation of k is the first step, and occasionally the bottleneck, in many number-theoretic algorithms: the Agrawal-Kayal-Saxena primality test, for example, and the number-field sieve for integer factorization. Here is an algorithm that, given an integer n > 1, finds the largest integer k such that n is a kth power: 1. For each prime power q such that 2 ≤ n, write down a positive integer rq such that if n is a qth power then n = r q . 2. Find a finite coprime set P of integers larger than 1 such that each of n, r2, r3, r4, r5, r7, . . . is a product of powers of elements of P . (In this paper, “coprime” means “pairwise coprime.”) 3. Factor n as ∏ p∈P p np , and compute k = gcd{np : p ∈ P}. It is easy to see that the algorithm is correct. Say n is an `th power. Take any prime power q dividing `. Then n is a qth power, so n = r q ; but rq is a product ∏ p∈P p ap for some exponents ap, so n is a product ∏ p∈P p qap . Factorizations over P are unique, so np = qap for each p. Thus q divides gcd{np : p ∈ P} = k. This is true for all q, so ` divides k. Conversely, n is certainly a kth power. Take, for example, n = 49787136 < 2. Compute approximations r2 = 7056 ≈ n r8 = 9 ≈ n r17 = 3 ≈ n r3 = 368 ≈ n r9 = 7 ≈ n r19 = 3 ≈ n r4 = 84 ≈ n r11 = 5 ≈ n r23 = 2 ≈ n r5 = 35 ≈ n r13 = 4 ≈ n r25 = 2 ≈ n r7 = 13 ≈ n r16 = 3 ≈ n where ≈ means “within 0.6.” Factor {49787136, 7056, 368, 84, 35, 13, 9, 7, 5, 4, 3, 2} into coprimes: each of these numbers is a product of powers of elements of P = Date: 2004.11.13. Permanent ID of this document: bbd41ce71e527d3c06295aadccf60979. 2000 Mathematics Subject Classification. Primary 11Y16. Initial work: Lenstra was supported by the National Science Foundation under grant DMS– 9224205. Subsequent work: Bernstein was supported by the National Science Foundation under grant DMS–0140542. The authors thank the University of California at Berkeley and the Fields Institute for Research in Mathematical Sciences.