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jennwatso15

Prove that there are no positive perfect cubes less than 1000 that are the sum of the cubes of two positive integers? I know all the cubes greater than zero and less that 1000.

  • one year ago
  • one year ago

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  1. gianarpt
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    The perfect cubes less than 1000 are: 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 7^3 = 343 8^3 = 512 9^3 = 729 What we want to prove is that none of these is the sum of any other two. This involves a simple (but tedious) checking of the list. Clearly a number could only be the sum of two numbers unless both were strictly smaller than it, so we only need to consider smaller terms. Thus 1 can't be the sum of any other terms. 2^3 = 8 could only be 1 + 1, but this is 2, not 8. 3^3 = 27 could be 1 + 1, 1 + 8 or 8 + 8, but all of these are too small 4^3 = 64 could be 1 + 1, 1 + 8, 1 + 27, 8 + 8, 8 + 27, 27 + 27... but all of these are too small Continuing in this way you can verify the proposition. There is a general proof that a^3 + b^3 = c^3 holds for no positive integers (and no, I don't mean the full proof of Fermat's last theorem) but it's too involved to post here.

    • one year ago
  2. jennwatso15
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    alright that's a little bit clearer than what the book gave. thank you

    • one year ago
  3. gianarpt
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    sorry if its too much, i was just trying to help you understand it better! hope it helped, and you're welcome :)

    • one year ago
  4. jim_thompson5910
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    to sum up what gianarpt said there are no integer or whole number solutions a, b, c that satisfy the equation a^3 + b^3 = c^3 and this is based off of fermat's last theorem http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem

    • one year ago
  5. jim_thompson5910
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    so there's no way to find two integers, say x and y, that make the equation x^3 + y^3 = 8^3 true for instance

    • one year ago
  6. jennwatso15
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    it kinda did. it would help if we went over this in class yet but we have not so i am hoping he doesn't take up the homework. what's not making sense is why your doing the one plus one and stuff. because the example is different i'm not proving that there isn't any

    • one year ago
  7. gianarpt
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    Well I hope you get better at this and figure it out and pass your quiz/homework :) But sadly I have to get off but thanks for the medal also Jim!

    • one year ago
  8. jennwatso15
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    thank you for your help!!!

    • one year ago
  9. jim_thompson5910
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    yw gianarpt the 1+1, and all that is adding up the cubes 1, 8, 27, etc to try to get another cube 1+1 (sum of 2 cubes) = 2 but 2 is NOT a perfect cube

    • one year ago
  10. jennwatso15
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    and sorry if i'm becoming difficult I don't know much about proofs at all I was thrown in this class in college without taking calc 1 first so sorry if i'm slow at this.

    • one year ago
  11. jim_thompson5910
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    1+8 = 9, but 9 isn't a perfect cube you'd be stuck looking forever trying to find a perfect cube that breaks down into x^3 + y^3 because there aren't any

    • one year ago
  12. jennwatso15
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    oh alright i see how that works

    • one year ago
  13. jim_thompson5910
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    the justification I would use is fermat's last theorem the proof is very hard to understand...it took about 400 years (or so) for someone to figure it out...and it was probably the most famous and well known problems in math

    • one year ago
  14. jennwatso15
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    i'm debating on if I should write all of these out in the homework it now makes sense on what she first responded

    • one year ago
  15. jim_thompson5910
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    then write whatever makes the most sense to you, that's what I would do

    • one year ago
  16. jennwatso15
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    alright! well thank you! the question makes alot of sense now

    • one year ago
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