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Simple proof of Fermat's theorem?

  1. Aug 18, 2011 #1
    Can someone point out the error in the following "proof":

    Prove a^n + b^n =/ c^n for n>2, a,b,c>1 (=/ means not equal to)

    Let b=xa where x>1 and is from the set of real numbers generated by fractions, such that b is an integer


    a^n + (xa)^n =/ c^n


    a^n + x^n.a^n =/ c^n

    then, taking the common factor a^n out

    a^n(1+x^n) =/ c^n

    then dividing through by a^n

    1+x^n =/ c^n/a^n

    Substitute y from the set of real numbers given by the fraction c/a, then

    1+x^n =/ y^n


    y^n - x^n =/ 1


    x^n - y^n = (x - y)*(x^[n-1] + x^[n-2]*y + ... + y^[n-1]).

    If the left side equals 1, then x > y, and x - y must be a positive
    divisor of 1, namely 1. Then x = y + 1. Substitute that into the
    second factor above, and set that second factor also equal to 1. That
    should give you a contradiction. The contradiction means that the
    assumption that the left side equals 1 must be false, and then you're
  2. jcsd
  3. Aug 18, 2011 #2
    a and b are integers by hypothesis, so b = xa implies that x is an integer, so that b is actually an integer multiple of a. Is it okay to assume this?
  4. Aug 18, 2011 #3
    Huh? If b = 6 and a = 4, then x = 1.5.
  5. Aug 18, 2011 #4


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    Why must x-y be an integer?
  6. Aug 18, 2011 #5
    Isn't that the assumption made in Fermat's Last Theorem?
    The statement is that no solutions (x, y, z) exist for the Diophantine equation [itex]x^n + y^n = z^n[/itex] if n > 2.
  7. Aug 18, 2011 #6
    To prove the x^n-y^n is not equal to 1, you assume the opposite, so x^n-y^n=1, then you show this won't work... x (any y) are not integers, and there's no reason to suppose they are simply because a and b are, as JG89 shows by example. x is from the group of real numbers formed by fractions of two integers. y is from the same group (formed by c/a).

    There must be an error somewhere - this is too easy a proof that took mathematicians many years to arrive at. The main difference I see is that we move from an integer problem to a real number problem (with an integer result - 1).
  8. Aug 19, 2011 #7


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    x = b/a
    y = c/a
    x-y = (b-c)/a

    That is not an integer, so you can't talk about divisibility.
  9. Aug 19, 2011 #8
    Yes, thinking some more about this, I realise the issue is that the re-arranged formula may actually have solutions for ANY real number x, y (i.e. there are an infinite number of solutions that can satisfy x^n-y^n=1), but Fermat's problem only allows a certain subset of the real numbers - those that can be produced by the ratio of two integers. So, IF you could prove that there were no solutions for all the real numbers, x and y, then you would also prove Fermats theorem, but this is probably not the case...
  10. Aug 19, 2011 #9


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    This is your error. Throughout, you have only required x and y to be rational numbers, not integers. So it does NOT follow that x- y is an integer divisor of 1.

    Added: that's pretty much what pwsnafu and barryn56 said before.
  11. Aug 19, 2011 #10
    OK, so it comes down to - given real numbers x and y, is there a (simple) proof that x^n-y^n =/ 1 for all n>2 or not?
  12. Aug 19, 2011 #11

    Well... assuming x,y are positive anyway x must be at least 1 and for any x>=1 , let y be the (positive, real) nth root of x^n-1. Then x^n-y^n=1.
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