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Sums of 6th and 7th degree powers

  1. Jan 31, 2012 #1
    This is a very similar question to what I posted earlier.

    Basically I am trying to find when (x+y)6 = x6 + y6 assuming that xy≠0

    I am trying to play with it algebraically to find a contradiction, but have been unsuccessful

    I'm also working on (x+y)7 = x7 + y7 assuming xy≠0

    I'm trying to play with it algebraically to show the only other case is when y=-x
     
  2. jcsd
  3. Jan 31, 2012 #2
  4. Jan 31, 2012 #3

    micromass

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    1) We can take y=1. If an (x,y) exists such that the equality holds (with xy nonzero), then there also exists an x' such that the equality holds for (x',1).

    2) Take y fixed. The goal is to prove that

    [tex]f(x)=(x+1)^6-x^6-1^6[/tex]

    has a zero. In order to do this, it suffices to show that f is monotonically increasing. Show this with derivatives.
     
  5. Jan 31, 2012 #4
    [tex]f(x)=6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x[/tex]

    Guess where the zero is... anyways, the poster wants a non zero x in which x != -y. Really- remember that (x+y)^n has a simple binomial expansion..... and subtracting (x^n+y^n) from it will leave you with the middle part....
     
  6. Jan 31, 2012 #5

    micromass

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    You're making it a lot harder then it needs to be really.
     
  7. Feb 1, 2012 #6
    I understand the approach you are taking, but I don't understand how setting y as a constant allows us to show that there is no solution to (x+y)6 = x6 + y6 other than x=0 or y=0 for all possible values of x and y
     
  8. Feb 1, 2012 #7

    micromass

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    I just take an arbitrary y. And I try to prove that

    [tex]f(x)=(x+y)^6-x^6-y^6[/tex]

    has only one zero. That proves it because I took y arbitrary but fixed.
     
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