Editing Powerful number (section)

== Generalization ==

More generally, we can consider the integers all of whose prime factors have exponents at least ''k''. Such an integer is called a ''k''-powerful number, ''k''-ful number, or ''k''-full number.

:(2<sup>''k''+1</sup>&nbsp;&minus;&nbsp;1)<sup>''k''</sup>,&nbsp;&nbsp;2<sup>''k''</sup>(2<sup>''k''+1</sup>&nbsp;&minus;&nbsp;1)<sup>''k''</sup>,&nbsp;&nbsp;&nbsp;(2<sup>''k''+1</sup>&nbsp;&minus;&nbsp;1)<sup>''k''+1</sup>

are ''k''-powerful numbers in an [[arithmetic progression]]. Moreover, if ''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a''<sub>''s''</sub> are ''k''-powerful in an arithmetic progression with common difference ''d'', then

: ''a''<sub>1</sub>(''a''<sub>''s''</sub> + ''d'')<sup>''k''</sup>,&nbsp;&nbsp;
''a''<sub>2</sub>(''a''<sub>''s''</sub>&nbsp;+&nbsp;''d'')<sup>''k''</sup>, ..., ''a''<sub>''s''</sub>(''a''<sub>''s''</sub>&nbsp;+&nbsp;''d'')<sup>''k''</sup>, (''a''<sub>''s''</sub>&nbsp;+&nbsp;''d'')<sup>''k''+1</sup>

are ''s'' + 1 ''k''-powerful numbers in an arithmetic progression.

We have an identity involving ''k''-powerful numbers:

:''a''<sup>''k''</sup>(''a''<sup>''ℓ''</sup> + ... + 1)<sup>''k''</sup> + ''a''<sup>''k'' + 1</sup>(''a''<sup>''ℓ''</sup> + ... + 1)<sup>''k''</sup> + ... + ''a''<sup>''k'' + ''ℓ''</sup>(''a''<sup>''ℓ''</sup> + ... + 1)<sup>''k''</sup> = ''a''<sup>''k''</sup>(''a''<sup>''ℓ''</sup> + ... +1)<sup>''k''+1</sup>.

This gives infinitely many ''l''+1-tuples of ''k''-powerful numbers whose sum is also ''k''-powerful. Nitaj shows there are infinitely many solutions of ''x''&nbsp;+&nbsp;''y''&nbsp;=&nbsp;''z'' in relatively prime 3-powerful numbers(Nitaj, 1995). Cohn constructs an infinite family of solutions of ''x''&nbsp;+&nbsp;''y''&nbsp;=&nbsp;''z'' in relatively prime non-cube 3-powerful numbers as follows: the triplet

:''X'' = 9712247684771506604963490444281, ''Y'' = 32295800804958334401937923416351, ''Z'' = 27474621855216870941749052236511

is a solution of the equation 32''X''<sup>3</sup> + 49''Y''<sup>3</sup> = 81''Z''<sup>3</sup>. We can construct another solution by setting ''{{prime|X}}'' = ''X''(49''Y''<sup>3</sup>&nbsp;+&nbsp;81''Z''<sup>3</sup>), ''{{prime|Y}}'' = &minus;''Y''(32''X''<sup>3</sup>&nbsp;+&nbsp;81''Z''<sup>3</sup>), ''{{prime|Z}}'' = ''Z''(32''X''<sup>3</sup>&nbsp;&minus;&nbsp;49''Y''<sup>3</sup>) and omitting the common divisor.