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Is this a short, marvelous proof of Fermat's Last Theorem?

  1. Jul 21, 2012 #1
    Given: x, y, z, n are whole natural numbers, including the regular and irregular prime numbers; x and y < z. The, "Trichotomy Law", which states there are three relations possible between (x^2 + y^2 ) and z^2, i.e., < = > . Any square can be subdivided into an infinite number of 'sum of two squares' forming a Set, whose members are both rational and/or irrational, such that, ... = (a^2 + b^2) = (c^2 + d^2) = ... (x^2 + y^2) ... = z^2.
    Throughout, n > 2.
    Case 1:
    Part (a) - Let the equality relation, x^2 + y^2 = z^2, be given and assume, x^n + y^n = z^n, is true. Factor, x^n + y^n = z^n, thus: x^n-2(x^2) + y^n-2(y^2) = z^n-2(z^2). Substituting equals-for-equals, (x^2 + y^2, for, z^2), we have, x^n-2(x^2) + y^n-2(y^2) = z^n-2(x^2 + y^2). Distributing, we get, x^n-2(x^2) + y^n-2(y^2) = z^n-2(x^2) + z^n-2(y^2). However, it is given that x and y are < z, making the left hand side of the equation ≠ to the right hand side of the equation. Therefore, x^n + y^n ≠ z^n.
    Part (b) - Inversely, assume, x^n + y^n = z^n, and divide equals-by-equals thus: x^n/z^n-2 + y^n/z^n-2 = z^n/z^n-2 = z^2. Then the left hand side of the equation must be a member of the Set of two squares, rational and/or irrational, whose sum is equal to
    z^2. However, by Part (a), any members of the Set such that x^2 + y^2 = z^2, results in the inequality, x^n + y^n ≠ z^n.
    Consequently, whenever, x^2 + y^2 = z^2, then, x^n + y^n ≠ z^n, and FLT is true for Case 1.

    Case 2: Let the inequality relation, x^2 + y^2 < z^2, be given and assume, x^n + y^n = z^n. Let x^2 + y^2 + u^2 = z^2, where, u^2 = z^2 - (x^2 + y^2), and let x^2 + y^2 = q^2 < z^2. Then, since q^2 < z^2, therefore, q^n < z^n. And since, x^2 + y^2 < q^2, then by Case 1, x^n + y^n < q^n. Therefore, we have, x^n + y^n < q^n < z^n, or, x^n + y^n < z^n. Consequently, whenever x^2 + y^2 < z^2, then, x^n + y^n ≠ z^n, and FLT is true for Case 2.

    Case 3 - Let the inequality relation, x^2 + y^2 > z^2, be given and assume x^n + y^n = z^n. Let, x^2 + y^2 - u^2 = z^2, where, u^2 = (x^2 + y^2) - z^2. Then by the Trichotomy Law, x^n + y^n < = > z^n + u^n. If, x^n + y^n = z^n + u^n, then, x^n + y^n ≠ z^n. If, x^n + y^n < > z^n + u^n, then, by the Trichotomy Law, x^n + y^n < = >
    z^n. If, x^n + y^n < > z^n, there is agreement with FLT. If we assume, x^n + y^n = z^n, then by Case 1 the assumption is false and, x^n + y^n ≠ z^n.
    Consequently, FLT is true for Case 3.

    Therefore, since Case 1, 2, and 3 exhaust all whole number possibilities, it has been shown what is necessary and sufficient to prove FLT, namely, x^n + y^n ≠ z^n, n>2, in whole numbers, is always true.
    QED
    willacaleb (LKS)
     
  2. jcsd
  3. Jul 21, 2012 #2

    1) Learn how to type mathematics with LaTeX in this site, otherwise chances are that not many will have the stamina to begin deciphering all your stuff ;

    2) What you wrote above sure is short and perhaps it even is marvelous, but HUGE chances are it is not a proof of FLT, though I won't rule this possibility out until you re-write the above decently.

    DonAntonio
     
  4. Jul 21, 2012 #3

    Simon Bridge

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    Here - let me try:

    Given: [itex]x, y, z, n[/itex] are whole natural numbers, including the regular and irregular prime numbers; [itex]x,y < z[/itex]. The, "Trichotomy Law", which states there are three relations possible between [itex](x^2 + y^2 )[/itex] and [itex]z^2[/itex], i.e.,
    1. [itex]x^2 + y^2 < z^2[/itex]
    2. [itex]x^2 + y^2 = z^2[/itex]
    3. [itex]x^2 + y^2 > z^2[/itex]

    Any square can be subdivided into an infinite number of 'sum of two squares' forming a Set, whose members are both rational and/or irrational, such that,

    [itex]\cdots = (a^2 + b^2) = (c^2 + d^2) = \cdots (x^2 + y^2) \cdots = z^2[/itex].

    Throughout, n > 2.

    Case 1:

    Part (a)
    Let the equality relation, [itex]x^2 + y^2 = z^2[/itex], be given and assume, [itex]x^n + y^n = z^n[/itex], is true.

    Factor, [itex]x^n + y^n = z^n[/itex], thus: [tex]x^{n-2}x^2 + y^{n-2}y^2 = z^{n-2}z^2[/tex] Substituting equals-for-equals, i.e. [itex](x^2 + y^2, for, z^2)[/itex], we have, [tex]x^{n-2}x^2 + y^{n-2}y^2 = z^{n-2}(x^2 + y^2)[/tex]. Distributing, we get, [tex]x^{n-2}x^2 + y^{n-2}y^2 = z^{n-2}x^2 + z^{n-2}y^2[/tex] However, it is given that [itex]x, y < z[/itex], making the equation false. Therefore, [tex]x^n + y^n \neq z^n[/tex].

    And so on ... I'll leave the rest as an exercise for OP :)

    I suspect it just shows that any [itex]x,y,z \in \mathbb{Z}: x^2+y^2=z^2[/itex] will not satisfy the relation [itex]x^n+y^n=z^n[/itex] but has not shown that some other [itex]x,y,z \in \mathbb{Z}: x^2+y^2 \mathbf{\neq} z^2[/itex] cannot satisfy the relation.

    i.e. this has shown that the set of integer triples {(x,y,z)} that satisfy the pythogrean relation will certainly not satisfy Fermat's, more general, relation. But what of other triples?
     
    Last edited: Jul 21, 2012
  5. Jul 21, 2012 #4

    HallsofIvy

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    Basically, what you have proved here is that you have no idea what Fermat's Last Theorem is.
     
  6. Jul 21, 2012 #5
    Simon Bridge: Yes, Case 1 demonstrates the singular case that pythagorean triples cannot satisfy FLT.
    Case 2 and Case 3 demonstrate that the other possible triples when, x^2 + y^2 < > z^2, cannot satisfy FLT. These three cases exhaust all possible whole numbers.
    Sorry about the 'primitive' script, I am unfamiliar with LaTex.
     
  7. Jul 21, 2012 #6
    You say here "assume [itex]x^n + y^n = z^n[/itex]", then later say "[itex]x^n + y^n [/itex]< = > [itex]z^n + u^n[/itex]". As [itex]u^n[/itex] is positive, clearly [itex]x^n + y^n < z^n + u^n[/itex]. This makes nearly all of this case unneeded.

    And can you explain how Case 1 shows anything here is false? Case 1 only applies when [itex]x^2 + y^2 = z^2[/itex], which you've explicitly stated is not the case here.
     
  8. Jul 21, 2012 #7

    micromass

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    Right, you can't use Case 1 here. That's already one large gap in the proof.
     
  9. Jul 21, 2012 #8

    Simon Bridge

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    Technically - all pythagorean triples must satisfy FLT (if we accept case 1).

    Just quote my last post to see how I did it. If you want to communicate clearly in math here you absolutely need latex and what you want to do is not complicated.

    Otherwise you'll just get the, somewhat arrogant-sounding, reply: "Nope, it hasn't [proved FLT]." Without going into details. You are always free to pay a mathematician to show you where you went wrong.
     
  10. Jul 21, 2012 #9
    Well,cases 1 and 2 are correct, but trivial.


    I don't understand case three. Could you explain it again? It seems that you start by assuming that [itex]x^2 + y^2 > z^2[/itex] and that [itex]x^n + y^n = z^n[/itex] and then do a lot of stuff and assume, again, that [itex]x^n + y^n = z^n[/itex] and use the case when [itex]x^2 + y^2 = z^2[/itex] which, as has been pointed out, makes no sense because you assumed that [itex]x^2 + y^2 > z^2[/itex]
     
  11. Jul 21, 2012 #10
    Also, cases 1 and 2 can be done like this:

    Assume [itex]x \leq y < z[/itex] and [itex]x^2 + y^2 \leq z^2[/itex]. Fix a positive integer [itex]n[/itex]. Then [itex]x^n + y^n = x^{2+(n-2)} + y^{2+n-2} = x^{n-2}x^2 + y^{n-2}y^2 \leq y^{n-2} (x^2 + y^2) \leq y^{n-2}z^2 <z^{n-2}z^2=z^n[/itex]. That is, [itex]x^n + y^n < z^n[/itex].

    Also, I don't understand where this is used :
    Is this used somewhere in case 3?
     
  12. Jul 21, 2012 #11

    Simon Bridge

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    Oh well, if we are left with case 3, perhaps a bit of structure will help?

    -------------------------
    Case 3

    3.1 Let the inequality relation, [itex]x^2 + y^2 > z^2[/itex], be given and assume [itex]x^n + y^n = z^n[/itex].

    3.2 Let, [itex]x^2 + y^2 - u^2 = z^2[/itex], where, [itex]u^2 = (x^2 + y^2) - z^2[/itex]
    Then by the Trichotomy Law, [itex]x^n + y^n < = > z^n + u^n[/itex].

    3.3 If, [itex]x^n + y^n = z^n + u^n[/itex],
    then, [itex]x^n + y^n \neq z^n[/itex].

    3.4 If, [itex]x^n + y^n < > z^n + u^n[/itex],
    then, by the Trichotomy Law, [itex]x^n + y^n < = > z^n[/itex].

    3.5 If, [itex]x^n + y^n < > z^n[/itex],
    then there is agreement with FLT.

    3.6 If we assume, [itex]x^n + y^n = z^n[/itex],
    then by Case 1 the assumption is false
    and, [itex]x^n + y^n \neq z^n[/itex].

    Consequently, FLT is true for Case 3.

    ----------------------

    @3.4 line 2, equality is already assumed at 3.1 isn't it?
    @3.6, already made that assumption at 3.1 - why not just say it is false right at the top?
    Also - doesn't case 1 only say the assumption is false for {(x,y,z)} are pythagorean triples?

    Doesn't this just say:

    The possibilities are
    [itex]x^n + y^n < , = , > z^n[/itex]

    if [itex]x^n + y^n <, > z^n[/itex]
    then [itex]x^n + y^n \neq z^n[/itex]

    if [itex]x^n + y^n = z^n[/itex]
    then, by case 1, this is false.

    ?
     
  13. Jul 22, 2012 #12
    Simon Bridge, Abacus, Robert1986, Micromass: I am pressed for time today but I will respond tomorrow. Thank you and I appreciate your responses.

    Willacaleb
     
  14. Jul 23, 2012 #13
    I think this will clarify my attempt at proving FLT:
    Let or assume, x^2 + y^2 < = > z^2, and assume, x^n + y^n = z^n, n>2.
    Then, by the fundamental operation of dividing equals-by-equals, x^n/z^n-2 + y^n/z^n-2
    = z^n/z^n-2 = z^2, would be true for the assumption we have made for the < = > relations.
    However, by Case 1, the equality of the 'sum of two squares' on the left hand side of the equation always precludes the equality of, x^n + y^n = z^n.
    Therefore, the assumed equality of, x^n + y^n = z^n, is false and FLT is true.

    willacaleb
     
  15. Jul 23, 2012 #14


    First: please do learn how to type with LaTeX in this site. This will greatly improve your presentation and your chancea to be read and understood.

    Second: what do you mean by [itex]\,x^2+y^2 <=>z^2\,[/itex] ?? And whatever this is, how from it follows that
    [tex]x^n+y^n=z^n\Longrightarrow \frac{x^n}{z^{n-2}}+\frac{y^n}{z^{n-2}}=z^2\,\,?[/tex]

    The above relation follows only from the relation [itex]x^n+y^n=z^n\,[/itex] , nothing else.






    Even assuming case 1, how do you apply it to the above? The relation
    [tex]\frac{x^n}{z^{n-2}}+\frac{y^n}{z^{n-2}}=z^2[/tex]
    is NOT a "sum of squares equals a square" , as for this being true you'd have to prove first that each of
    [tex]\frac{x^n}{z^{n-2}}\,,\,\frac{y^n}{z^{n-2}}[/tex]
    is a square...

    DonAntonio

     
  16. Jul 23, 2012 #15

    micromass

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    How is

    [tex]\frac{x^n}{z^{n-2}}+\frac{y^n}{z^{n-2}}=z^2[/tex]

    a sum of two squares?? You don't even know that it's an integer.
     
  17. Jul 23, 2012 #16

    Simon Bridge

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    seconded, thirded, and fourthed... it's not hard to do and, in your case, easy to figure out just by hitting the "quote" button from any post that has latex in it and comparing with the rendered text.
    Yeah - I have been guessing that this means that there are three possibilities vis:
    1. [itex]\,x^2+y^2 < z^2\,[/itex],
    2. [itex]\,x^2+y^2 = z^2\,[/itex], or
    3. [itex]\,x^2+y^2 > z^2\,[/itex]

    ... and he does not want to type them out each time. That's why I put commas in the rendering above (an attempt to show that only one of the signs should apply at a time).
    I'd have appreciated some feedback on whether my attempts to convert the reasoning into something more intelligible were any good :(
     
  18. Jul 23, 2012 #17
    Which gives you [itex]\sqrt{\frac{x^{n}}{z^{n-2}}}^{m}+\sqrt{\frac{y^{n}}{z^{n-2}}}^{m}≠z^{m}[/itex] for every [itex]m>2[/itex]. So what?
     
  19. Jul 23, 2012 #18
    Simon Bridge: Your conversions are helpful and accurate. I would like to use LaTex, however, when I hit the "quote button" nothing happens. I am in "Quick Reply."
    I will send the answers I wrote today tomorrow. The site kicked them out when I tried to send them.

    willacaleb
     
  20. Jul 23, 2012 #19
    Sending to check the quote fuction.
     
  21. Jul 23, 2012 #20
    I think he means to press the "QUOTE" button. That way, you will see the raw tex code and can use it as a template and as examples to do your own stuff.
     
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