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ashur
 3 years ago
How do I prove that every powerful number can be written as the product of a perfect square and a perfect cube?
ashur
 3 years ago
How do I prove that every powerful number can be written as the product of a perfect square and a perfect cube?

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ilikephysics2
 3 years ago
Best ResponseYou've already chosen the best response.0Think about it, what do you know about a square?

ilikephysics2
 3 years ago
Best ResponseYou've already chosen the best response.0there you go thats all it is

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Just curious, but what class is this for?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0how does that prove it? I didn't know what "powerful numbers" were until I just looked them up. if I understand it correctly then a powerful number is a positive integer m such that for every prime number p dividing m, p^2 also divides m.

ilikephysics2
 3 years ago
Best ResponseYou've already chosen the best response.0I didn't really understand the question

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I was curious because I helped out on the same question yesterday. See http://openstudy.com/study#/updates/50552fcde4b02986d370aedd

ashur
 3 years ago
Best ResponseYou've already chosen the best response.0The first part makes sense but why are u subtracting 3 form ei when ei is odd?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1So that I get \[\large p_i^{e_i}=p_i^{f_i+3}=p_i^{f_i}\cdot p_i^3\]Note that since \(e_i\) is odd, \(e_i3\) is even, so \(\displaystyle p_i^{e_i3}=p_i^{f_i}\) is a perfect square.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Additionally, \(p_i^3\) is a perfect cube.

ashur
 3 years ago
Best ResponseYou've already chosen the best response.0oh that makes perfect sense, thanks alot
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