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Perfect power

From Wikipedia, the free encyclopedia
Positive integer that is an integer power of another positive integer
For the racehorse, seePerfect Power.
Demonstration, withCuisenaire rods, of the perfect power nature of 4, 8, and 9

Inmathematics, aperfect power is anatural number that is a product of equal natural factors, or, in other words, aninteger that can be expressed as a square or a higher integerpower of another integer greater than one. More formally,n is a perfect power if there existnatural numbersm > 1, andk > 1 such thatmk =n. In this case,n may be called aperfectkth power. Ifk = 2 ork = 3, thenn is called aperfect square orperfect cube, respectively. Sometimes 0 and 1 are also considered perfect powers (0k = 0 for anyk > 0, 1k = 1 for anyk).

Examples and sums

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Asequence of perfect powers can be generated by iterating through the possible values form andk. The first few ascending perfect powers in numerical order (showing duplicate powers) are (sequenceA072103 in theOEIS):

22=4, 23=8, 32=9, 24=16, 42=16, 52=25, 33=27,{\displaystyle 2^{2}=4,\ 2^{3}=8,\ 3^{2}=9,\ 2^{4}=16,\ 4^{2}=16,\ 5^{2}=25,\ 3^{3}=27,}25=32, 62=36, 72=49, 26=64, 43=64, 82=64,{\displaystyle 2^{5}=32,\ 6^{2}=36,\ 7^{2}=49,\ 2^{6}=64,\ 4^{3}=64,\ 8^{2}=64,\dots }

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

m=2k=21mk=1.{\displaystyle \sum _{m=2}^{\infty }\sum _{k=2}^{\infty }{\frac {1}{m^{k}}}=1.}

which can be proved as follows:

m=2k=21mk=m=21m2k=01mk=m=21m2(mm1)=m=21m(m1)=m=2(1m11m)=1.{\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, ... (sequenceA001597 in theOEIS)

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

p1p=k=2μ(k)(1ζ(k))0.874464368{\displaystyle \sum _{p}{\frac {1}{p}}=\sum _{k=2}^{\infty }\mu (k)(1-\zeta (k))\approx 0.874464368\dots }

where μ(k) is theMöbius function and ζ(k) is theRiemann zeta function.

According toEuler,Goldbach showed (in a now-lost letter) that the sum of1/p − 1 over the set of perfect powersp, excluding 1 and excluding duplicates, is 1:

p1p1=13+17+18+115+124+126+131+=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 theGoldbach–Euler theorem.

Detecting perfect powers

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Detecting whether or not a given natural numbern is a perfect power may be accomplished in many different ways, with varying levels ofcomplexity. One of the simplest such methods is to consider all possible values fork across each of thedivisors ofn, up toklog2n{\displaystyle k\leq \log _{2}n}. So if the divisors ofn{\displaystyle n} aren1,n2,,nj{\displaystyle n_{1},n_{2},\dots ,n_{j}} then one of the valuesn12,n22,,nj2,n13,n23,{\displaystyle n_{1}^{2},n_{2}^{2},\dots ,n_{j}^{2},n_{1}^{3},n_{2}^{3},\dots } must be equal ton ifn is indeed a perfect power.

This method can immediately be simplified by instead considering onlyprime values ofk. This is because ifn=mk{\displaystyle n=m^{k}} for acompositek=ap{\displaystyle k=ap} wherep is prime, then this can simply be rewritten asn=mk=map=(ma)p{\displaystyle n=m^{k}=m^{ap}=(m^{a})^{p}}. Because of this result, theminimal value ofk must necessarily be prime.

If the full factorization ofn is known, sayn=p1α1p2α2prαr{\displaystyle n=p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}}\cdots p_{r}^{\alpha _{r}}} where thepi{\displaystyle p_{i}} are distinct primes, thenn is a perfect powerif and only ifgcd(α1,α2,,αr)>1{\displaystyle \gcd(\alpha _{1},\alpha _{2},\ldots ,\alpha _{r})>1} where gcd denotes thegreatest common divisor. As an example, considern = 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).

Gaps between perfect powers

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

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

See also

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References

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  1. ^Weisstein, Eric W."Perfect Power".MathWorld.
  2. ^Weisstein, Eric W."Pillai's Conjecture".MathWorld.

External links

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