March 18th, 2019, 12:45 PM  #11 
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760  
March 24th, 2019, 09:45 AM  #12  
Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0  Quote:
Last edited by skipjack; March 24th, 2019 at 03:58 PM.  
March 24th, 2019, 10:33 AM  #13 
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760 
I prefer not to keep looking at videos. You can do what you like on your own Youtube channel. But here, this is a discussion forum. Please type in any mathematical points you'd like to make. You can link graphic images if you like. I did glance at your video and got as far as your claiming that FLT applies for $n \geq 2$. You stated this TWICE, and ended up (as you did earlier) erroneously disproving the existence of Pythagorean triples. Last edited by Maschke; March 24th, 2019 at 10:55 AM. 
March 25th, 2019, 12:38 PM  #14 
Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0  Pythagorean triple must be assumed to be in set
Of necessity, when you construct the right triangle given the initial conditions where X,Y<Z are whole numbers, then the set of all solutions must contain Pythagorean triples. Until you show otherwise,, then the assumption must include the triples. The proof shows that Pythagorean triples are only possible when a = 0. When a =0, only then is the absolute values of x(n/2) squared, and absolute value of Y(n/2) squared when added equal to Z(n). In every other instance, when a >0 , then Z(n) > X(n) + Y(n).. 
March 25th, 2019, 05:13 PM  #15  
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760  Quote:
Also I haven't thought about this in detail yet, but taking the greatest integer of n/2 must introduce some inaccuracies you haven't thought about. You're trying to reduce every Fermat triple to an "almost" Pythagorean triple. I don't see any analysis that shows that $n > 2$ forces $0 < \epsilon < 1$. But just to help me understand your idea, perhaps you can walk through your proof for n = 3 and show that there is no Fermat triple; and show how your proof DOES allow for Pythagorean triples. And again, same question as before, where does your proof fail for n = 2? You never actually use the fact that n > 3. It doesn't appear anywhere in your written proof. You write $n \geq$ twice and never write $n > 2$. That's a big red flag in any alleged FLT proof. Last edited by Maschke; March 25th, 2019 at 05:17 PM.  
March 26th, 2019, 02:25 PM  #16 
Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0  Initial assumption is that there exist X,Y,Z which are integer.
The purpose of the graph is to show that when n =2, then all the Pythagorean triples are present. for example when n/2 is a whole number, b and Z(2b) = X(2b) + Y(2b) and all Z,Y,X are integer, then equation can be rewritten as (Z(b))(2) = (Y(b))(2) + (X(b)) (2) So what you have is a nonprimitive Pythagorean triple. When a = 0 then the every true statement of the equation means that it must be a Pythagorean triple with n=2, in some form or another.I think that this demonstrates both where the Pythagorean triples must only exist when n =2. Initially, I THOUGHT THAT a MUST BE LESS THAN 1, BUT i FOUND THAT IT WAS ONLY NECESSARY TO HAVE a GREATER THAN 0. For the analysis of the value of Z ,all that is necessary is Z be Y+1 ( you want to examine the inequality created when the square root of X(n) and Y(n) have a remainder.) . So Z(n/2) must be greater than Y(n/2) and a greater than 0. When you square Z(n/2) it is greater than the sum of the squares of the absolute values of X(n/2) and Y(n/2) I did not complete the proof for n =3 as I did want to use the the assumption that if case for true for number = n and then prove from that n+1 must be true.I did some of the first terms ie 3,4,5 when n=3. I also did for 4,5,6 and 5,6,7. that's where I found out a could be larger than 1, but the inequality still held. If you followed the proof so far,, the proof fails for n=2. So if you exclude the proof for the case n=2, then every case for n > 2 (AND NONPRIMITIVE TRIPLES WHICH ARE A ANOTHER VERSION OF N=2) FLT holds. 
March 26th, 2019, 06:23 PM  #17  
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760  Quote:
How do you know this can't be the case? It seems to me that the assertion that $\alpha > 0$ depends on already assuming FLT. 2) Suppose $x^3 + y^3 = z^3$. Then your resulting triangle has sides of $x$, $y$, and $z$, which might perfectly well be a right triangle. Why couldn't that be? Your use of the greatest integer function fails here. 3) You said (in your latest exposition) that $x^{\frac{n}{2}}$ might not be an integer, but $ x^{\frac{n}{2}} $ must be an integer. Now that is of course absurd. If a real number is not an integer its absolute value can't be an integer either. I'm only relying on your written page. I haven't watched the video, only took a screen capture of your exposition. It's murky in the extreme, proves nothing at all. If you explained anything verbally, can you please write it down? Last edited by Maschke; March 26th, 2019 at 06:46 PM.  
March 26th, 2019, 10:24 PM  #18 
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760 
ps  Never mind #3, I see that you were saying that $x^{\frac{n}{2}}$ PLUS $y^{\frac{n}{2}}$ might not be an integer. Perfectly correct. I mistook your plus sign for a handwritten "and" symbol. But rereading your proof it's perfectly clear that your assumption that $\alpha > 0$ is a consequence of FLT, not a proof. If you don't know beforehand that FLT is true, how do you know $\alpha$ must be greater than zero? 
March 27th, 2019, 12:27 AM  #19 
Senior Member Joined: Aug 2012 Posts: 2,426 Thanks: 760 
Wait on that #3 again ... You said now that $x^{\frac{n}{2}} + y^{\frac{n}{2}}$ may not be an integer. But now here's a mystery. In your earlier video you used the exponent $\lfloor \frac{n}{2} \rfloor$, the greatest integer or floor function. But in your latest video you are just using $\frac{n}{2}$ by itself. Surely you meant to include the floor function, otherwise your idea makes no sense at all. But then $x^{\lfloor \frac{n}{2} \rfloor } + y^{\lfloor \frac{n}{2} \rfloor}$ must be an integer. So this looks like a mystery with your exposition. But again, no matter, since you have a problem with claiming $\alpha > 0$ without justification. Last edited by Maschke; March 27th, 2019 at 12:31 AM. 
March 27th, 2019, 12:56 PM  #20 
Member Joined: Mar 2019 From: california Posts: 74 Thanks: 0  a > 0
Yes, you are right. I have a problem with notation, but notation can be fixed. You are also right that the idea behind the notation is what really matters. if a <0 then Z(n/2) is longer than Y(n/2) and by extension Z > Y. This violates the assumption that Z>Y. So a = 0 or a > 0. If a=0 then the proof shows the right triangle must be the representation of a nonprimitive Pythagorean triple. When a >0 then Z(n) is greater than X(n) + Y(n). I did want to thank you for bringing up so many issues. This has made me examine the theorem in a different way. It may be possible to construct a different proof. So thank you for that. 

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