perfect power

The power $m^n$ is called a perfect power if $m$ and $n$ are integers both greater than 1. The perfect powers form the ascending order sequence (cf. Sloane’s http://oeis.org/classic/A001597A001597)

$4,\mathrm{\hspace{0.17em}8},\mathrm{\hspace{0.17em}9},\mathrm{\hspace{0.17em}16},\mathrm{\hspace{0.17em}25},\mathrm{\hspace{0.17em}27},\mathrm{\hspace{0.17em}32},\mathrm{\hspace{0.17em}36},\mathrm{\hspace{0.17em}49},\mathrm{\hspace{0.17em}64},\mathrm{\hspace{0.17em}81},\mathrm{\hspace{0.17em}100},\mathrm{\hspace{0.17em}121},\mathrm{\hspace{0.17em}125},\mathrm{\dots },$ (1)

i.e.

$$2^2,{\mathrm{\hspace{0.33em}2}}^3,{\mathrm{\hspace{0.17em}3}}^2,{\mathrm{\hspace{0.33em}2}}^4=4^2,{\mathrm{\hspace{0.33em}5}}^2,{\mathrm{\hspace{0.33em}3}}^3,{\mathrm{\hspace{0.33em}2}}^5,{\mathrm{\hspace{0.17em}6}}^2,{\mathrm{\hspace{0.33em}7}}^2,{\mathrm{\hspace{0.33em}2}}^6=4^3=8^2,{\mathrm{\hspace{0.33em}3}}^4=9^2,{\mathrm{\hspace{0.33em}10}}^2,{\mathrm{\hspace{0.33em}11}}^2,{\mathrm{\hspace{0.33em}5}}^3,\mathrm{\dots }$$

S. S. Pillai has conjectured in 1945, that if the $i^{\mathrm{th}}$ member of the sequence (1) is denoted by $a_i$, then

$\underset{i\to \mathrm{\infty }}{lim\; inf}(a_{i+1}-a_i)=\mathrm{\infty }.$ (2)

This does not necessarily that one had  ${lim}_{i\to \mathrm{\infty }}(a_{i+1}-a_i)=\mathrm{\infty }$,  since there may always exist little differences $a_{i+1}-a_i$ arbitrarily far from the begin of the sequence (1).

The equation (2) is equivalent (http://planetmath.org/Equivalent3) to the

Pillai’s conjecture.  For any positive integer $k$, the Diophantine equation

$$x^m-y^n=k$$

has only a finite number of solutions  $(x,y,m,n)$  where the integers $x,y,m,n$ all are greater than 1.

Pillai’s conjecture generalises the Catalan’s conjecture ($k=1$) in which the number of solutions is 1.

The series formed by the inverse numbers of the perfect powers converges absolutely, and its sum may be calculated easily:

${\displaystyle \sum _{m,n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^n}}$ $\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^n}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}\cdot {\displaystyle \frac{1}{m^{n-2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}{\displaystyle \sum _{n=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^{n-2}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\left({\displaystyle \frac{1}{m}}\right)}^n$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^2}}\cdot {\displaystyle \frac{1}{1-\frac{1}{m}}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m(m-1)}}$

$\mathrm{\hspace{0.33em}}={\displaystyle \sum _{m=2}^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{m-1}}-{\displaystyle \frac{1}{m}}\right)$

The sum of this telescoping series (http://planetmath.org/TelescopingSum) is equal to 1.

Title	perfect power
Canonical name	PerfectPower
Date of creation	2013-03-22 19:15:51
Last modified on	2013-03-22 19:15:51
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Definition
Classification	msc 11D61
Classification	msc 11D45
Classification	msc 11B83
Synonym	Pillai’s conjecture
Related topic	PerfectSquare
Related topic	LimitInferior
Related topic	DoubleSeries
Related topic	SolutionsOfXyYx
Related topic	MarshallHallsConjecture