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BrinWolf Group Title

I just need a little help with the concept. A positive integer is calledpowerful if all exponents in its prime number factorization are greater than 1. Prove that every powerful number can be written as the product of a perfect square and a perfect cube.

  • 2 years ago
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  1. NoelGreco Group Title
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    1 is the perfect cube.

    • 2 years ago
  2. KingGeorge Group Title
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    Let's see if I get anywhere. Suppose \[n=p_1^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_r^{e_r}\]Where \(e_i\ge 2\). Now we have two cases for each \(e_i\). Case 1: \(e_i\) is even. If this is the case, it is a perfect square, and we can factor it out. Case 2: \(e_i\) is odd. In this case, look at \(e_i-3=f_i\). Here, \(f_i\) is even, so we have a perfect square, and we are left with \(p_i^{f_i}\cdot p_i^3\). Note that \(p_i^{f_i}\) is a perfect square, and \(p_i^3\) is a perfect cube. Hopefully you'll be able to finish it. This is the meat of the proof.

    • 2 years ago
  3. NoelGreco Group Title
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    \[36=3^{2}\times2^{2}\times1^{3}\]

    • 2 years ago
  4. KingGeorge Group Title
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    To finish it, you just need to see why those two cases imply that it can be factored into a perfect square and a perfect cube.

    • 2 years ago
  5. BrinWolf Group Title
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    ohhhhhhh!! makes more sense! so basically if fi ends up being 4 we can factor it out into two perfect square. If fi is odd then is it that that means that it must be the multiplication of both an exponent of two and three?

    • 2 years ago
  6. KingGeorge Group Title
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    That's the general idea. If \(e_i\) is even, then \(p_i^{e_i}\) is already a perfect square, and if \(e_i\) is odd, we can make it so that \(p_i^{e_i}\) is a product of a perfect square and perfect cube. Hence, we can just factor the parts out, and there we go.

    • 2 years ago
  7. BrinWolf Group Title
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    ok!! that makes perfect sense! thank you so much! I can write a proof now since i understand the concept! thank you!

    • 2 years ago
  8. KingGeorge Group Title
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    You're welcome.

    • 2 years ago
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