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I A surprising discovery?

  1. Mar 20, 2016 #1
    A SURPRISING DISCOVERY:

    I have only recently discovered the following equations of a certain function of a pair of two integers, which surprised me:-

    f(x, y) = {f(x+z, y+z) + f(x-z, y-z)}/2 ----------------1

    f(x, y) = {f(x+1, y+1) + f(x-1, y-1)}/2-2 .-----------------2

    Where z = any value, positive or negative.

    Of course the first equation is true for f(x,y) = x+/-y, and the second is true for f(x,y) = x+/-y-2, but there is another surprising function (non linear) which I will leave the reader to discover for now.
     
  2. jcsd
  3. Mar 20, 2016 #2

    mfb

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    You are looking for functions satisfying one of those equations?
    f(x,y)=a*x+b*y+c+g(x-y) for the first, with arbitrary a,b,c and an arbitrary function g. Every linear function will work, and you can even make nonlinear functions via g.

    f(x,y) = x+y-2 (same with "-") does not work for the second equation. Take x=0, y=0 for example, then f(x,y)=-2 but {f(x+1, y+1) + f(x-1, y-1)}/2-2 = (0-4)/2-2 = -4.
    Actually, no linear function satisfies the second equation. Something like f(x,y)=-(x+y)^2 should work. You can then add any function that satisfies equation 1.
     
  4. Mar 20, 2016 #3
    Sorry I did not mean to include the second equation as applicable to a linear function. What surprised me was the non linear function that you have probably known of a long before me. In particular it is fairly simple to show that a function of order higher than 2 is not possible, so is this relevant to FLT?
     
  5. Mar 20, 2016 #4

    mfb

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    FLT?

    If you look at the one-dimensional case, it is easier to find the function. What kind of function satisfies h(x)=(h(x+1)+h(x-1)/2 - 2?
    This can be written as h(x+1)-h(x) = (h(x)-h(x-1)) - 2. No linear function can satisfy this, but the expression suggests that the second derivative (related to the difference of differences of function values) is constant. So you can test h(x)=c*x^2. Plugging it in gives you c=-1/2.
    Higher powers of x would not lead to a constant difference of 2.

    Going back to the original function: you can always write f(x,y) as g(x+y,x-y)=g(u,v). We need g(u,v)=(g(u+2,v)+g(u-2,v))/2 - 2. So in the first parameter, it has to behave like the function h from above, while the function can do anything it wants in the second parameter. All that is left is finding the prefactor, which you can do with an example (like x=0, y=0).
     
  6. Mar 20, 2016 #5
    Thanks for your input. The particular function I am interested in that fits the first equation is simply x^p -y^p where p is either one or two which will include all pythagoras triples. And with power above two the equation cannot be true. (The second equation relates to x^p + y^p)
    So does this prove, or point to a proof of Fermat's last theorem?
     
  7. Mar 20, 2016 #6

    micromass

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    No.
     
  8. Mar 20, 2016 #7

    mfb

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    It has absolutely no relation to Fermat's last theorem.
     
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