May 6th, 2019, 05:46 PM  #21  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
I'd still appreciate it if you'd briefly just say how your Y calculation proves FLT. I did look at the end of page three and it just says that FLT is proven. But I don't see how that follows from the ideas you've been presenting.  
May 7th, 2019, 07:40 AM  #22 
Member Joined: Mar 2019 From: california Posts: 40 Thanks: 0 
Quickly, if the only value of Y that satisfies the equations is less than 1, and is false for every other value of Y, there is no possible set of integers that can be used to satisfy original equation.

May 7th, 2019, 08:01 AM  #23 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  
May 7th, 2019, 01:01 PM  #24  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
Of course it is perfectly true that if you keep constructing triangles with smaller and smaller sides, as you are doing, that eventually the sides will be smaller than any given value (unless your particular construction method approaches a nonzero limit, but I don't think that's the case here). But what of it? It is perfectly true that there are no integersided triangles at all where all the sides are constrained to be less than 1. Right? Of course. You have a geometric construction that keeps making smaller triangles, and eventually the sides are too short to be positive integers. Now that is true. But it is not a proof of FLT. It's a proof that your construction eventually produces a right triangle whose sides can't possibly be integers. But what of it? That doesn't prove FLT. Is your algebraic proof a different form of the same idea? If so, it's subject to the same criticism. Last edited by Maschke; May 7th, 2019 at 01:04 PM.  
May 9th, 2019, 12:33 PM  #25 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706 
Ok, I see what you're doing. You are NOT making an infinite descent argument. This is a NEW argument. That's was a little confusing. So your argument here is as follows: * If we have $x^n + y^n = z^n$ for $n > 2$ then $y$ can't be an integer, therefore FLT is established. It's always good to say up front what is the point of the next three pages of calculations. Puts the reader in a better frame of mind. Just a suggestion in case you are writing this up for other audiences. Right at the beginning I have some questions. First, you don't say that $n > 2$ but in fact that must be the case since any Pythagorean triple violates your equation (2). Again, it's helpful to explicitly state all your assumptions for clarity. Regarding your diagram, I don't understand it. It doesn't do much for the clarity of your presentation. Not for me anyway. Now I passed on this earlier, but we have to take a close look at your equations (1) and (2). You have not presented a proof of these inequalities and they are the heart of the matter. The equations are: (1) $x + y = \frac{c}{d} z$ where $\frac{c}{d} > 1$ and (2) $x^2 + y^2 = \frac{e}{f} z^2$ where $\frac{e}{f} > 1$ Earlier I ignored (1) and sort of provisionally agreed to (2). But now I have to wonder if that's correct. In fact, what if these equations depend on FLT in the first place? That's a concern I've expressed to some of your earlier attempts. I find it helpful to rewrite your equations as inequalities, since that's what I think you're trying to express. Since $\frac{c}{d} > 1$ and $\frac{e}{f} > 1$, your equations (1) and (2) can be written, respectively, as (3) $x + y > z$ and (4) $x^2 + y^2 > z^2$ Now since by assumption $z = (x^n +y^n)^\frac{1}{n}$, we have the inequalities (5) $x + y > (x^n +y^n)^\frac{1}{n}$ and (6) $x^2 + y^2 > (x^n +y^n)^\frac{2}{n}$ I would like to see proofs of both of these. I would not be surprised if they're harder than they look; and if they don't in fact ultimately depend on FLT, making your argument circular. Last edited by Maschke; May 9th, 2019 at 12:37 PM. 
May 10th, 2019, 02:54 AM  #26 
Senior Member Joined: Mar 2019 From: iran Posts: 147 Thanks: 9 
do you want to prove x + y > (x^n + y^n)^(1/n) ? we know (x + y)^n > x^n + y^n (binomial theorem) then ((x + y)^n)^(1/n) > (x^n + y^n)^(1/n) then x + y > (x^n + y^n)^(1/n) 
May 10th, 2019, 11:51 AM  #27 
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  This is false. Take $x = 5$, $y = 5$, $n = 4$. Then the LHS is zero and the RHS is positive. Or if you prefer, take $x = 5$, $y = 6$, $n = 4$. FLT applies to integers, not just positive integers. @michaelcweir's latest attempt is wrong because equation 1 is false. 
May 10th, 2019, 12:57 PM  #28 
Senior Member Joined: Mar 2019 From: iran Posts: 147 Thanks: 9 
proving FLT for positive integers proves it for negative integers too!

May 10th, 2019, 01:33 PM  #29  
Senior Member Joined: Aug 2012 Posts: 2,306 Thanks: 706  Quote:
But even if that's true, @michaelweir's equation (1) is false, as witnessed by the counterexamples I gave, hence his alleged proof of FLT is wrong. You agree? Disagree? ps  Wikipedia says that FLT applies ONLY to positive integers, in which case @michaelweir's equation (1) is saved for the moment. https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem This is news to me but I'm no specialist. Are there FLT counterexamples involving negative integers? I'm curious on this point. pps ... found this ... http://www.researchpipeline.com/word...tiveintegers/ Evidently FLT applies to negative integers with the trivial exception of $x = y$. Last edited by Maschke; May 10th, 2019 at 01:41 PM.  
May 10th, 2019, 07:33 PM  #30 
Senior Member Joined: Mar 2019 From: iran Posts: 147 Thanks: 9 
ok the first equation is true for positive x and y and positive n > 1 the second equation is true for positive x and y and positive n > 2 

Tags 
analysis, diophantine, diopphantine, fermat, theorem 
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