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Fermat's Last Theorem: an amateur proof

  1. Nov 6, 2008 #1
    Below is the address of an attempted proof of FLT that I have not been able to find a flaw in. It works by increasing x, y, and z by a common integer factor so that, if x^p +y^p = z^p, with x + y = a^p and z - y = b^p, then (z/a) = (x/b)^((1 + 1/(p-1)) (mod g), with g a prime factor of y that is coprime to z - x. The total argument is 23 lines long.

    http://groups.google.com/group/fermats-last-theorem-flt-2008
     
    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 6, 2008 #2

    HallsofIvy

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    I don't have time to go through it all but I notice right off that you say
    "It can be seen from the equation xp + ((x + y) - x)p = zp that x+ y= ap".

    That makes no sense because you don't say what a is! Are you saying that x+ y must be the pth power of an integer? That's obviously not true in the case that p= 2. 32+ 42= 52 but 3+ 4 is not a square.

    In fact, one way to test your proof is this: where does it fail for p= 2?
     
  4. Nov 6, 2008 #3
    zp = p(x + y)xp-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p

    If z is coprime to p then the RHS can be divided by (x + y) to leave an integer coprime to (x + y), therefore (x + y) is a perfect pth power.

    I've corrected the proof to mention that the variables are all nonzero integers.

    The characteristics of adding perfect squares is quite different.
     
  5. Nov 6, 2008 #4

    HallsofIvy

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    Then 5 is coprime to 2- why is 3+ 4= 7 not a "perfect square"?

    And, again, where does your proof fail for p= 2?
     
  6. Nov 6, 2008 #5

    In what context? If zp can be divided by (x + y) to leave an integer coprime to (x + y), then (x + y) must be a perfect pth power.

    The proof makes assumptions (hopefully correct) about z/a and x/b. I see no reason why those assumptions should be correct for p = 2. Also, operations to make x a perfect (p-1)st power become nonsense.
     
    Last edited: Nov 6, 2008
  7. Nov 6, 2008 #6

    HallsofIvy

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    We are talking about NUMBERS. What you said was that "If z is coprime to p then the RHS can be divided by (x + y) to leave an integer coprime to (x + y), therefore (x + y) is a perfect pth power." Okay, I am looking at xp+ yp= zp with x= 3, y= 4, z= 5, and p= 2. The right hand side you are referring to is "p(x + y)xp-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p" which, in this case, is 2(3+4)- (1/2)(2)(1)(3+ 4)2= 14+ 72= 14+ 49= 63. According to you "the RHS can be divided by (x + y) to leave an integer coprime to (x + y)". Yes, 63 can be divided by 7 to leave 9 which is coprime to 7. You then say "therefore (x + y) is a perfect pth power." No, that does not follow: 7 is not a perfect square!

    You then say "The proof makes assumptions (hopefully correct) about z/a and x/b. I see no reason why those assumptions should be correct for p = 2."

    Well, a proof shouldn't include statements that are "hopefully correct". If you understand your own proof you should be able to say exactly what those "assumptions about z/a and x/b" are, and why they are true for p>2 but not for p= 2. That is what I am asking.
     
    Last edited: Nov 6, 2008
  8. Nov 6, 2008 #7
    One thing I see (or so I think) is that the signs in the binomial expansion for ((x + y) - x)^p (later subtracting -x^p) are drawn from the assumption that p is odd. Which is why this particular equation won't work for p=2. (Unfortunately, there is a lack of counterexamples for higher powers.)

    What I don't get is why a) after division by (x + y), the quotient is coprime to (x + y), and b) why a) implies that (x + y) is a p-th power.

    And yes, a clearer exposition would make a difference.
     
    Last edited: Nov 6, 2008
  9. Nov 7, 2008 #8
    In the RHS of

    zp = p(x + y)xp-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p

    both p and x are coprime to (x + y), while the second term is a multiple of (x + y)2, the third term is a multiple of (x + y)3, and so on. Therefore dividing the first term by (x + y) leaves an integer coprime to (x + y), while dividing all remaining terms leaves multiples of (x + y). Therefore the RHS can be divided by (x + y) to leave an integer coprime to (x + y). Therefore zp can be divided by (x + y) to leave an integer coprime to (x + y).

    Write zp as the product of all its prime factors, showing each prime factor the appropriate number of times:

    zp = j1 * j2 * ... * jn

    The product of all the j's found in (x + y) is a perfect pth power, as is the product of all the j's coprime to (x + y), therefore (x + y) is a perfect pth power.

    When I spoke of assumptions about z/a and x/b, I was thinking about (x + y) and (z - y) being perfect pth powers.

    http://groups.google.com/group/fermats-last-theorem-flt-2008
     
    Last edited: Nov 7, 2008
  10. Nov 9, 2008 #9
    I have another question. If p, x and y are coprime, this doesn't mean that p and x are coprime to (x + y). How do you justify so?
     
  11. Nov 9, 2008 #10
    The two cases of Fermat's Last Theorem are the case where p does or does not divide (X+Y). Should P divide Z, then it will divide (X+Y) p-1 times and divide the remaining quotient exactly once. Since X==-Y Mod p implies that (X)^(p-1) -(-X)^p-2(X) + - +...==P(x^(p-1)) ==0 Mod p.

    It would seem that anybody working on this problem would have realized that the two cases exist.

    So, in your first line: It can be seen from the equation z p = x p + ((x + y) - x) p that in the RHS of zp = p(x + y)x p-1 - 1/2 p(p-1)(x + y)2xp-2 + ... + (x + y)p both p and x are coprime to (x + y),

    There is already a mistake and one that you should not have made if you had done any research at all! This problem has been worked on for hundreds of years and something so simple as the transformation you employ will not work! All that is occurring is sleight of hand and self-deception.
     
    Last edited: Nov 9, 2008
  12. Nov 10, 2008 #11
    I found the flaw. Line [14] does not follow from [13].

    If z is a multiple of p then p divides (x + y) (p -1 +pn) times, with n a positive integer. Both (xp)/(bp) and (yp/cp) have the form (pk + 1)p, so in the equation (x + y) = bp + cp + 2pabch, a is a multiple of p.

    Both z and x are defined in the proof as being coprime to p, so, since (x + y) is a factor of zp, therefore (x + y) is coprime to p and x.
     
    Last edited: Nov 10, 2008
  13. Nov 11, 2008 #12
    You eliminate my concern with “The variables x and z are here defined as being coprime to p.” (And you thus, it seems to me, eliminating the most important case historically)

    But, then you bring the whole problem in the back door: "z - x = p^ (p - 1)*c^ p or c^ p [4]" So that nothing is gained by saying p does not divide z or x. Since the prime is odd, the equation could have been written as Y^p = Z^p + (-X)^p. (Usually this matter is handled by saying p does not divide xyz, if it divides one of them we have the two cases, though I did not make this clear.)

    I don’t see x+y+z =pabch. [5] Is this being left to the imagination, or what?

    Increase x, y, and z by a factor of (x^ p)^ (p - 2) so that x becomes a perfect p - 1 power,
    “Increase” is one of those words that can be left to imagination, so that confusion and errors can follow. Anyway, what you have written makes X a perfect p-2 power, not a p-1 power! This is simple arithmetic.

    Then, “And x + y and z - y continue to be perfect pth powers.” I don’t know what to make of that. Did you happen to “increase” anything here?
     
    Last edited: Nov 11, 2008
  14. Nov 12, 2008 #13
    The idea is that if xyz is a multiple of p then y is a multiple of p. When the RHS of

    y^p = ((z - x) + x)^p - x^p

    is expanded, the first term can be divided by p(z - x) to leave an integer coprime to (z - x) while all remaining terms are multiples of p(z - x)^2, leading to the conclusion that z - x = (p^(p - 1))c^p.

    It says x + y - z = pabch. Write x as ((z - y) + (x + y - z)) If x^p can be divided by b^p to leave an integer coprime to (z - y) then x and x + y - z can be divided by b to leave an integer coprime to (z - y). From this it follows that x + y - z is a multiple of pabc.

    Okay. "Multiply" would have been a better choice. Multiply x^(p(p - 2)) by x^1 and you get x^((p - 1 )(p - 1)).
    If they are already perfect pth powers, and are then multiplied by perfect pth powers...
     
  15. Nov 15, 2008 #14
    Looking over how somebody else wrote it up, since your way can not be found, I see:

    Given x^p + y^p = z^p
    x+y = a^p
    z-y = b^p
    z-x = p^{p-1}c^p or c^p
    x+y-z = pabch
    x' := x^{p^2-2p+1}
    y' := y * x^{p^2-2p}
    z' := z * z^{p^2-2p}
    we find
    x'^p + y'^p = z'^p
    x' + y' = (a*x^{p-1})^p. BUT SHOULD BE: (a^p)(x^(P^2-2p)
    which makes sense since both x and y were multiplied by x^(p^2-2p).
     
  16. Dec 5, 2008 #15
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