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BrinWolf

  • 2 years ago

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.

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

  2. KingGeorge
    • 2 years ago
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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.

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

  4. KingGeorge
    • 2 years ago
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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.

  5. BrinWolf
    • 2 years ago
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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?

  6. KingGeorge
    • 2 years ago
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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.

  7. BrinWolf
    • 2 years ago
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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!

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

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