My Math Forum one page proof of Fermat's Last Theorem

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March 18th, 2019, 11:45 AM   #11
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 Originally Posted by Micrm@ss 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!

March 24th, 2019, 08:45 AM   #12
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 Originally Posted by Maschke 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.

Last edited by skipjack; March 24th, 2019 at 02:58 PM.

 March 24th, 2019, 09:33 AM #13 Senior Member   Joined: Aug 2012 Posts: 2,329 Thanks: 720 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. Thanks from Denis and topsquark Last edited by Maschke; March 24th, 2019 at 09:55 AM.
 March 25th, 2019, 11:38 AM #14 Member   Joined: Mar 2019 From: california Posts: 57 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,Y0 , then Z(n) > X(n) + Y(n)..
March 25th, 2019, 04:13 PM   #15
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 Originally Posted by michaelcweir 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.

 March 26th, 2019, 01:25 PM #16 Member   Joined: Mar 2019 From: california Posts: 57 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 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.
March 26th, 2019, 05:23 PM   #17
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 Originally Posted by michaelcweir 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?

Last edited by Maschke; March 26th, 2019 at 05:46 PM.

 March 26th, 2019, 09:24 PM #18 Senior Member   Joined: Aug 2012 Posts: 2,329 Thanks: 720 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?
 March 26th, 2019, 11:27 PM #19 Senior Member   Joined: Aug 2012 Posts: 2,329 Thanks: 720 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.
 March 27th, 2019, 11:56 AM #20 Member   Joined: Mar 2019 From: california Posts: 57 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 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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