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ashur
Group Title
How do I prove that every powerful number can be written as the product of a perfect square and a perfect cube?
 one year ago
 one year ago
ashur Group Title
How do I prove that every powerful number can be written as the product of a perfect square and a perfect cube?
 one year ago
 one year ago

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ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
Think about it, what do you know about a square?
 one year ago

ashur Group TitleBest ResponseYou've already chosen the best response.0
they are even
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
there you go thats all it is
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Just curious, but what class is this for?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
how 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.
 one year ago

ashur Group TitleBest ResponseYou've already chosen the best response.0
Proofs class
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
I didn't really understand the question
 one year ago

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
So 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.
 one year ago

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

ashur Group TitleBest ResponseYou've already chosen the best response.0
oh that makes perfect sense, thanks alot
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You're welcome.
 one year ago
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