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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?

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  1. ilikephysics2
    • 3 years ago
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    Think about it, what do you know about a square?

  2. ashur
    • 3 years ago
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    they are even

  3. ilikephysics2
    • 3 years ago
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    there you go thats all it is

  4. KingGeorge
    • 3 years ago
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    Just curious, but what class is this for?

  5. asnaseer
    • 3 years ago
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    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.

  6. ashur
    • 3 years ago
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    Proofs class

  7. ilikephysics2
    • 3 years ago
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    I didn't really understand the question

  8. asnaseer
    • 3 years ago
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    :/

  9. KingGeorge
    • 3 years ago
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    I was curious because I helped out on the same question yesterday. See http://openstudy.com/study#/updates/50552fcde4b02986d370aedd

  10. ashur
    • 3 years ago
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    The first part makes sense but why are u subtracting 3 form ei when ei is odd?

  11. KingGeorge
    • 3 years ago
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    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, \(e_i-3\) is even, so \(\displaystyle p_i^{e_i-3}=p_i^{f_i}\) is a perfect square.

  12. KingGeorge
    • 3 years ago
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    Additionally, \(p_i^3\) is a perfect cube.

  13. ashur
    • 3 years ago
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    oh that makes perfect sense, thanks alot

  14. KingGeorge
    • 3 years ago
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    You're welcome.

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