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March 18th, 2019, 11:45 AM   #11
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Originally Posted by Micrm@ss View Post
Trust me, Maschke is very well-versed in mathematics. He definitely knows what he's talking about.
We're not worthy! We're not worthy!

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March 24th, 2019, 08:45 AM   #12
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Originally Posted by Maschke View Post
Can you explain the meaning of $\alpha$?

I'm walking through your proof using the standard Pythagorean triple $3^2 + 4^2 = 5^2$. As far as I can tell, your proof shows this triple to be impossible, since you never used the fact that $n > 2$. You need to explain this else your proof is cooked.

So in this example, $x = 3$, $y = 4$, $z = 5$, and $n = 2$.

Now you say that $W = x^{\frac{n}{2}} - \alpha$. In this case we have $W = x - \alpha$ or $W = 3 - \alpha$.

So what is the meaning of $\alpha$? You write that $0 < \alpha < 1$ but I don't see what $\alpha$ is supposed to be. Do you mean to simply take $W = x$ in this case?

I hope you will take the time to respond to these concerns. To sum up, my questions are:

1) Where do you use that $n > 2$? Without that fact you have disproved the well-known existence of Pythagorean triples.

2) What is $\alpha$? In the case of $x = 3$, $y = 4$, $z = 5$, and $n = 2$, when you write that $W = 3 - \alpha$, what is $\alpha$?

ps -- Fermat's right triangle theorem asserts that not all of the labelled sides can be integers. But in the case at hand, if $0 < \alpha < 1$ then $W = 3 - \alpha$ is already not an integer, so of course you haven't proved anything.

On the other hand if $\alpha = 0$, then all the $\alpha$ terms in your squaring operations are zero and you have no contradiction.

I would say at this point that this basic example of $9 + 16 = 25$ cooks your proof. I do await your response.
I have thought about your question about a. I made a video to respond.


Last edited by skipjack; March 24th, 2019 at 02:58 PM.
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March 24th, 2019, 09:33 AM   #13
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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.
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Last edited by Maschke; March 24th, 2019 at 09:55 AM.
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March 25th, 2019, 11:38 AM   #14
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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)..
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March 25th, 2019, 04:13 PM   #15
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Originally Posted by michaelcweir View Post

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)..
My sense is that in order to prove this, you would have to assume FLT. In other words how do you know that for n > 2, you must necessarily need to subtract some epsilon off the hypotenuse?

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 04:17 PM.
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March 26th, 2019, 01:25 PM   #16
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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 re-written as
(Z(b))(2) = (Y(b))(2) + (X(b)) (2)

So what you have is a non-primitive 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 NON-PRIMITIVE TRIPLES WHICH ARE A ANOTHER VERSION OF N=2) FLT holds.
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March 26th, 2019, 05:23 PM   #17
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Originally Posted by michaelcweir View Post
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.
1) Suppose there are integers $x$, $y$, and $z$ such that $x^{10} + y^{10} = z^{10}$. Then there is a Pythagorean triangle with sides $x^5$, $y^5$, and $z^5$. In this case your $\alpha = 0$.

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?
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Last edited by Maschke; March 26th, 2019 at 05:46 PM.
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March 26th, 2019, 09:24 PM   #18
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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 re-reading 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?
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March 26th, 2019, 11:27 PM   #19
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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 26th, 2019 at 11:31 PM.
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March 27th, 2019, 11:56 AM   #20
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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 non-primitive 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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