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Simple Proof

  1. Dec 18, 2008 #1

    Noo

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    I'm studying pure maths/numb theory for the first time (independantly, and from a shallow, brief-ish book with no given solutions, so i have noone else to ask :P ). I just started a day or two ago. The book leaves Induction till a little later, so this should be proved directly, or maybe by contradiction - and only with basic high-school maths.

    Even to my novice eyes this problem seems very simple, not simple enough for me, yet, though.

    [tex]\forall n \in Z, \exists a,b,...,h \in Z[/tex] such that [tex]n = a^{3}+b^{3}+. . .+h^{3}[/tex]

    I must either prove the above is true, or prove that its negation is true.

    I'm not sure where to start, and havent been for 1 or 2 other problems (such as proving 'a^3 -a' is always divisible by 6). I have been thinking of something along the lines of;

    Showing [tex]\sqrt[3]{a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}+g^{3}-n}[/tex] cant always equate to an integer. But that is random and useless. I'm noy even sure whether the original statement is true or not.

    What should i be looking for in trying to prove these things? Or is it mainly that i am not familiar enough with properties of numbers yet? In any case, suggestions/hints/advice or even solutions will all help me in starting out.
     
  2. jcsd
  3. Dec 18, 2008 #2

    HallsofIvy

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    First, what exactly is the problem? "Such that" is usually a condition, not a conclusion. Are you trying to prove "If n is an integer, then there exist integers a,b, c..., h that n= a3+ b3+ ... h3" And a, b, ... can be any number of integers, not specifically 8? That probably is NOT the case since then it would be trivial- any number can be written as a sum of "1"s.
     
  4. Dec 18, 2008 #3

    Noo

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    Yes, Halls, its as you have assumed (or maybe this isnt what you assumed). "For all integers n, there exists 8 integers cubed which sum to n" (That includes 0's). So, it is to prove that statement true, or to prove its negation ("there exists an integer n, for which 8 integers cubed do not sum to n").
     
  5. Dec 18, 2008 #4

    Dick

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    Proving something like that would be extremely difficult. You'd better pin your hopes on finding a fairly easy counterexample. There is one. Start expressing small numbers as the sum of cubes and you should find it pretty quickly.
     
  6. Dec 18, 2008 #5
    I could have missed something, but I don't think it's so easy. All integers between -100 and 100 can be represented as the sum of cubes of at most 8 integers, for example.
     
  7. Dec 18, 2008 #6

    Dick

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    I'll take your word for it, but I think the problem is supposed to be the sum of cubes of nonnegative integers. At least that's the way I took it. Is it Noo? In which case there is an exception.
     
  8. Dec 18, 2008 #7
    Yes, you're right. If the problem is about nonnegative integers, then there is a small exception :smile:
     
  9. Dec 18, 2008 #8

    Noo

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    I took your advise and have checked all |n| up to 150 so far, i have no counter-example yet. Just to clarify, as i am sure you realise, the sum can include subtractions; so for example the interval [157, 167] is covered by "[tex]5^{3} + 4^{3} + (-3)^{3} ^{+}_{-} m [/tex], where, since there are 5 remaining components of the sum, each of which can be at 0 or 1, [tex](m=0,1..,5) [/tex] That is: 125+64-27 = 162, so [(162-5), (162+5)].


    Edit: Oh, sorry. I took so long writing this, working LaTex out, that 3 posts arrived before mine. It is for all integers, negative included. And yes, using the interval method i hinted at above i am effectively checking off intervals of the number line. But after i hit 200 i'm giving up hope of finding a counter example:D

    P.S. All |n|<205 confirmed.
     
    Last edited: Dec 18, 2008
  10. Dec 18, 2008 #9

    Dick

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    You can give up the search then. If the cubes are allowed to be signed then it is true. In fact you only need 5. I know this because I looked it up, not because I know the proof. I'm pretty sure it's beyond the range of high school algebra. But maybe there is a trick for 8 cubes that makes it accessible. If you are SURE that's the intention, then it's back to the drawing board.
     
  11. Dec 18, 2008 #10

    Noo

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    Alright, and thanks.

    It seems very interesting to me that you only ever need 5. Although in googling just now i learned that all cubes can be expressed as a sum of consecutive odd integers, which i also found very interesting, so perhaps i am easily interested.
     
  12. Dec 18, 2008 #11

    Dick

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    Since you are so easily interested here's a the '5' proof. I found this in a pdf titled 'Waring's problem, taxicab numbers and other sums of powers'. Google around for it. Start with the identity,

    6x=(x+1)^3+(x-1)^3-2*x^3

    That proves it for all numbers divisible by six. Now just create some variations on that formula, e.g.

    6x+1=(x+1)^3+(x-1)^3-2*x^3+1
    6x-1=(x+1)^3+(x-1)^3-2*x^3-1
    6x+2=x^3+(x+2)^3-2(x+1)^3-2^3
    6x-2=x^3+(x-2)^3-2(x-1)^3+2^3
    6x+3=(x-3)^3+(x-5)^3-2(x-4)^3+3^3

    There may be some typos in there. Try to find the paper, it might have other interesting stuff in it. I can't find a author on the paper.
     
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