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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?
 2 years ago
 2 years 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?
 2 years ago
 2 years 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?
 2 years ago

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

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

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

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

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

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

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