Perfect power

A sequence of perfect powers can be generated by iterating through the possible values for m and k. The first few ascending perfect powers in numerical order (showing duplicate powers) are (sequence A072103 in the OEIS):

${\displaystyle 2^{2}=4,\ 2^{3}=8,\ 3^{2}=9,\ 2^{4}=16,\ 4^{2}=16,\ 5^{2}=25,\ 3^{3}=27,}$ ${\displaystyle 2^{5}=32,\ 6^{2}=36,\ 7^{2}=49,\ 2^{6}=64,\ 4^{3}=64,\ 8^{2}=64,\dots }$

The sum of the reciprocals of the perfect powers (including duplicates such as 3⁴ and 9², both of which equal 81) is 1:

${\displaystyle \sum _{m=2}^{\infty }\sum _{k=2}^{\infty }{\frac {1}{m^{k}}}=1.}$

which can be proved as follows:

${\displaystyle \sum _{m=2}^{\infty }\sum _{k=2}^{\infty }{\frac {1}{m^{k}}}=\sum _{m=2}^{\infty }{\frac {1}{m^{2}}}\sum _{k=0}^{\infty }{\frac {1}{m^{k}}}=\sum _{m=2}^{\infty }{\frac {1}{m^{2}}}\left({\frac {m}{m-1}}\right)=\sum _{m=2}^{\infty }{\frac {1}{m(m-1)}}=\sum _{m=2}^{\infty }\left({\frac {1}{m-1}}-{\frac {1}{m}}\right)=1\,.}$

The first perfect powers without duplicates are:

(sometimes 0 and 1), 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, ... (sequence A001597 in the OEIS)

The sum of the reciprocals of the perfect powers p without duplicates is:[1]

${\displaystyle \sum _{p}{\frac {1}{p}}=\sum _{k=2}^{\infty }\mu (k)(1-\zeta (k))\approx 0.874464368\dots }$

where μ(k) is the Möbius function and ζ(k) is the Riemann zeta function.

According to Euler, Goldbach showed (in a now-lost letter) that the sum of 1/p − 1 over the set of perfect powers p, excluding 1 and excluding duplicates, is 1:

${\displaystyle \sum _{p}{\frac {1}{p-1}}={{\frac {1}{3}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{15}}+{\frac {1}{24}}+{\frac {1}{26}}+{\frac {1}{31}}}+\cdots =1.}$

This is sometimes known as the Goldbach–Euler theorem.

Detecting whether or not a given natural number n is a perfect power may be accomplished in many different ways, with varying levels of complexity. One of the simplest such methods is to consider all possible values for k across each of the divisors of n, up to ${\displaystyle k\leq \log _{2}n}$. So if the divisors of ${\displaystyle n}$ are ${\displaystyle n_{1},n_{2},\dots ,n_{j}}$ then one of the values ${\displaystyle n_{1}^{2},n_{2}^{2},\dots ,n_{j}^{2},n_{1}^{3},n_{2}^{3},\dots }$ must be equal to n if n is indeed a perfect power.

This method can immediately be simplified by instead considering only prime values of k. This is because if ${\displaystyle n=m^{k}}$ for a composite ${\displaystyle k=ap}$ where p is prime, then this can simply be rewritten as ${\displaystyle n=m^{k}=m^{ap}=(m^{a})^{p}}$. Because of this result, the minimal value of k must necessarily be prime.

If the full factorization of n is known, say ${\displaystyle n=p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}}\cdots p_{r}^{\alpha _{r}}}$ where the ${\displaystyle p_{i}}$ are distinct primes, then n is a perfect power if and only if ${\displaystyle \gcd(\alpha _{1},\alpha _{2},\ldots ,\alpha _{r})>1}$ where gcd denotes the greatest common divisor. As an example, consider n = 2⁹⁶·3⁶⁰·7²⁴. Since gcd(96, 60, 24) = 12, n is a perfect 12th power (and a perfect 6th power, 4th power, cube and square, since 6, 4, 3 and 2 divide 12).

In 2002 Romanian mathematician Preda Mihăilescu proved that the only pair of consecutive perfect powers is 2³ = 8 and 3² = 9, thus proving Catalan's conjecture.

Pillai's conjecture states that for any given positive integer k there are only a finite number of pairs of perfect powers whose difference is k. This is an unsolved problem.[2]

• Prime power

• Daniel J. Bernstein (1998). "Detecting perfect powers in essentially linear time" (PDF). Mathematics of Computation. 67 (223): 1253–1283. doi:10.1090/S0025-5718-98-00952-1.

• Lluís Bibiloni, Pelegrí Viader, and Jaume Paradís, On a Series of Goldbach and Euler, 2004 (Pdf)