American Mathematical Monthly, Vol.111, March 2005

Problem 11117
(American Mathematical Monthly, Vol.111, March 2005)
Proposed by Michael Nyblom, RMIT University, Melbourne, Australia.
An integer n is a positive power if there exist integers a and k such that a ≥ 1,
m ≥ 2, and n = am . Let N (x) denote the number of positive powers n such
that 1 ≤ n ≤ x. For real x ≥ 4 and with L = blog2 xc, show that
N (x) =
L−1
X
(−1)k+1
X
bx1/lcm(i1 ,...,ik ) c.
2≤i1 <···<ik ≤L
k=1
Solution proposed by Giulio Francot and Roberto Tauraso,
Dipartimento di Matematica, Università di Roma “Tor Vergata”, via della
Ricerca Scientifica, 00133 Roma, Italy.
For m ≥ 2 and for x ≥ 4, let Am (x) be the set of m-powers contained in
the interval [1, x]. Since |Am (x)| = bx1/m c then Am (x) = {1} for m > L =
blog2 xc ≥ 2 and by the inclusion-exclusion principle
L
[
L−1
X
X
Am (x) =
(−1)k+1
|Ai1 (x) ∩ · · · ∩ Aik (x)|
N (x) = m=2
k=1
2≤i1 <···<ik ≤L
αr
1 α2
Let pα
1 p2 · · · pr be the prime factorization of n then n ∈ Am (x) if and only if
1 ≤ n ≤ x and m | αs for all s = 1, . . . , r. Hence
Ai1 (x) ∩ · · · ∩ Aik (x) = Alcm(i1 ,...,ik ) (x),
and therefore
N (x) =
L−1
X
k=1
(−1)k−1
X
Alcm(i
2≤i1 <···<ik ≤L
X
X
L−1
=
(−1)k+1
(x)
1 ,...,ik )
k=1
bx1/lcm(i1 ,··· ,ik ) c.
2≤i1 <···<ik ≤L