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When does (x+y)^4 =

  1. Nov 22, 2004 #1

    JasonRox

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    [tex](x+y)^4=x^4+y^4[/tex]

    We know this is true if x and/or y = 0, but what other possibilities are there?

    I solved for (x+y)^2=x^2+y^2, which has x=-y as a possibility.

    The solution isn't necessary, a couple tips should lead me in the right direction.

    I broke it down in several ways, but a hint for something to look for would be great. I happen to always be stuck with something along the line like...

    STUFF+x^2y^2

    I can get the stuff area to equal to zero, but because the are both odd or even numbers that are equal I am always stuff with the odd one at the end.

    I lost my paper that shows what STUFF is (stupid I know), but I'll work out a second time and maybe then I'll find a solution.

    Any advice would be great.

    Thanks.
     
  2. jcsd
  3. Nov 22, 2004 #2

    Gokul43201

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    Expand the LHS, and cancel off the common terms with the RHS. You have 3 remaining terms =0. Factoring out xy (stipulating that they be not zero) from these gives you a quadratic, which has no real solutions, and this, you'll figure when you get to it...
     
  4. Nov 22, 2004 #3
    I solved for (x+y)^2=x^2+y^2, which has x=-y as a possibility.

    Are u sure?
    Substitute x=-y on the LHS and then in the RHS .. are they equal?
    Simple analysis ,
    (x+y)^2 = x^2+2xy+y^2
    if (x+y)^2=x^2+y^2
    implies
    x^2+2xy+y^2 = x^2+y^2
    implies
    2xy = 0
    implies
    x or y or both = 0

    A similar analysis has been outlined by Gokul in his last post for the power 4 case.

    -- AI
     
  5. Nov 22, 2004 #4

    HallsofIvy

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    (x+y)4= x4+ 4x3y+6x2y2+4xy3+ y4.

    Setting that equal to x4+ y4, we can cancel the x4 and y4 terms on both sides leaving

    4x3y+ 6x2y2+ 4xy3= 0 or
    2xy(2x2+ 3xy+ 2y2)= 0

    Clearly x= y= 0 works. It should be evident that 2x2+ 3xy+ 2y2= 0 has no real solutions so x= y= 0 is again the only solution.
     
  6. Nov 22, 2004 #5

    JasonRox

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    Nevermind about x=-y. I was looking at the wrong solution.
     
    Last edited: Nov 22, 2004
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