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Why two given equations cannot both be true (fermat's for n=3)

  1. Nov 9, 2012 #1
    Okay, so I'm trying to figure something out.
    Why can (x+y)=integer^3 not be true while
    (x^2-x*y+y^2)=integer^3 is true?
    where x and y are also integers. The integer to the right is just an undefined integer.
    I'm trying to prove Fermat's last theorem x^n+y^n=z^n for n=3.
    I was able to define two factors of x^2+y^2 as:
    (x+y) and (x*y+(x-y)^2)) which reduces to (x^2-x*y+y^2).
    Also, z^3 must be the product of cubed integers. So, (x+y) must itself be a cubed integer as well as (x^2-x*y+y^2) in order for x^3+y^3 to be = to z^3. (Which we know it can't). However, I don't know enough maths to prove that (x+y) cannot be equal to a cubed integer while (x^2-x*y+y^2) is.

    Also, is this proof a novel one? I was just doing it for garbages and giggles.
    Thanks to any who may be able to help
  2. jcsd
  3. Nov 9, 2012 #2
    (x+y)*(x^2 -x*y +y^2) = z^3
    the problem is not that x+y can't be a cubed value, it can, or that x^2 -x*y +y^2 can't be a cubed value, it can. it's proving that there is no case where both are cubed values for the same x,y.

    let's take an example. 10^3 = 1000.
    10^3 = 2^3 *5^3
    so ether x+y = 8, or x+y = 125.
    let;s assume the smaller one. which means x^2 -x*y +y^2 needs to equal 125.
    x = 3 y= 5
    3^2 - 3*5 +5^2 = 9 -15 +25 = 19; no where close to 125.
    Last edited: Nov 9, 2012
  4. Nov 9, 2012 #3
    Yeah, I'm just not sure how (if it can be) to go about showing that. I'm trying to prove that the first and second terms cant both be equal to cubed integers at the same time. But im not sure how to set up or solve a proof of that. I'd assume modus tolens of some kind but dont really know
    Last edited: Nov 9, 2012
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