Perfect power

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In mathematics, a perfect power is a positive integer that can be expressed as a power of another positive integer. More formally, n is a perfect power if there exist natural numbers m > 1, and k > 1 such that m^k = n. In this case, n may be called a perfect kth power. If k = 2 or k = 3, then n is called a perfect square or perfect cube, respectively. Sometimes 1 is also considered a perfect power (1^k = 1 for any k).

[edit] Examples and sums

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 OEIS):

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

The sum of the reciprocals of the perfect powers (including duplicates) is 1:

$\sum_{m=2}^{\infty} \sum_{k=2}^{\infty}\frac{1}{m^k}=1.$

which can be proved as follows:

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

The first perfect powers without duplicates are (A001597):

(sometimes 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, ...

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:

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

This is sometimes known as the Goldbach-Euler theorem.

[edit] Detecting perfect powers

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 $k \leq \log_2 n$ . So if the divisors of $n$ are $n_1, n_2, \dots, n_j$ then one of the values $n_1^2, n_2^2, \dots, n_j^2, n_1^3, n_2^3, \dots$ 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 $n = m k$ for a composite $k = a p$ where p is prime, then this can simply be rewritten as $n = m k = m a p = (m a) p$ . Because of this result, the minimal value of k must necessarily be prime.