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FLT attempt

  1. Oct 28, 2012 #1
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    This is what the theorem states:

    "No three positive integers a, b, and c can satisfy the equation a^p+b^p=c^p, for any integer values of p >2"

    My attempt (p = n+2 > 2):

    ap+bp=cp
    a(n+2)+b(n+2)=c(n+2)
    (an*a2 )+(bn*b2 )=(cn*c2 )

    (an*a2)/(an*bn*cn )+(bn*b2)/(an*bn*cn)=(cn*c2)/(an*bn*cn)

    a2 (1/bc)n + b2 (1/ac)n = c2 (1/ab)n

    The equation is now similar to the Pythagorean sum, and in order for it to remain true we must have the following:

    a2 (1) + b2 (1) = c2 (1)

    (1/bc)n=(1/ac)n=(1/ab)n=1

    ∴n=0 or a=b=c=1

    If a=b=c=1, we will have the following:

    1p+1p=1p
    2=1

    Therefore, a=b=c=1 cannot be true, hence n = 0 is the remaining option which means:

    p=n+2
    p=0+2
    p=2
    ∴p cannot be greater than 2
     

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  3. Oct 28, 2012 #2

    Simon Bridge

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    You know it's been done right?
    Be suspicious of simple proofs for this thing - if it were that easy, it would have been found by someone else long ago.
    Faulty logic here.
    You have not demonstrated that those are the only conditions that would satisfy the equation.
    It could be that some integer a,b,c could be found to satisfy the equation when the parts are multiplied out. The proof is to show that there are none.
     
  4. Oct 29, 2012 #3
    I disagree... let's assume there could be integers a, b, and c not equal to each other that will satisfy the equation when multiplied out.

    Let's take for example the first part: a2/(bc)n = xα
    xα will never be an integer, do you agree? This would also be the case for b2/(ac)n = yβ, and c2/(ab)n = zγ

    Since xα, yβ, and zγ are not integers, then the following would not be part of FLT: xα+yβ=zγ

    I also understand that this is too simple not to have been brought up before, but I do not see any flaws to it yet.
     
    Last edited: Oct 29, 2012
  5. Oct 29, 2012 #4
    No your wrong! Whether or not the values you created are integers or not has nothing to do with it as the question is whether x,y and z could be integers not whether the values you created are integers. The fact that the values you created are not integers does not mean the only way that expression would be true is the one you chose to examined. Your logic if extended to the equation x^2 + y^2 = z^2 => x/yz + y/xz = z/xy would be to say that each of the latter expressions must be integers which is not true.
     
    Last edited: Oct 29, 2012
  6. Oct 29, 2012 #5


    You haven't proved this lot:

    $$"\,a^2 \left(\frac{1}{bc}\right)^n + b^2\left (\frac{1}{ac}\right)^n = c^2\left (\frac{1}{ab}\right)^n$$

    The equation is now similar to the Pythagorean sum, and in order for it to remain true we must have the following:

    $$a^2 (1) + b^2 (1) = c^2 (1)\,\,"$$

    DonAntonio
     
  7. Oct 29, 2012 #6

    Simon Bridge

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    with what? This?
    ???
    Nope.

    I can find a ratio composed like that which is also an integer. eg. a=10, b=2, c=5, n=2 ... then ##x^\alpha##=1 - an integer.

    I believe there are histories of attempts at this theorem around - you'll probably find this example in one of them since they usually include examples of past attempts as a way to demonstrate how hard it is to do.

    I agree also with DonAntonio (post #5) and ramsey2879 (post #4).
    Perhaps you have been a bit glib - go back and work through the "proof" with more rigor. The trouble with these attempts has always been making sure you have included everything.

    OTOH: you sound pretty sure of yourself - so maybe you should just submit it for publication and see how far you get?
     
  8. Oct 29, 2012 #7
    Well first of all, thank you to all for making it clear to me that the ratio doesn't necessarily have to be an integer. I actually showed this to a mathematics professor at my school, and my argument actually convinced them! That's why I thought that part was correct...

    And by the way Simon...

    Isn't 'c' supposed to be larger than 'a' or 'b'? Because 10n+2n=5n can never be true for any positive integer of 'n'...

    And there is no reason for saying:
    I came here initially with the idea that I had made a mistake somewhere.
     
    Last edited: Oct 29, 2012
  9. Oct 29, 2012 #8

    Simon Bridge

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    Well - part of it is :)
    By "professor" and "school" do you mean a "professor of mathematics" at a college-level "school of mathematics"?
    That wasn't the question - but yes, I know: in context ... what I am trying to show you is that you are being less than rigorous in your statements... which is likely to be where the error lies.

    It will be difficult to pinpoint exactly where the error lies without that rigor.
    But there are signs where more work is needed and these have been pointed out to you.
    Have you seen the actual proof we have already?
     
  10. Oct 30, 2012 #9

    micromass

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    Please report such threads. Locked.
     
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