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

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 April 2nd, 2019, 06:22 PM #31 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 What you are describing as a Pythagorean Triangle meets the conditions for the right triangle proof. Where one of X,Y,Z cannot be integer. In practice, there is a smaller triangle where (Y+1)(T)- a = X(T) + Y(T) where X,Y,T are integer. Since Z cannot be smaller than Y+1 Last edited by skipjack; April 2nd, 2019 at 11:32 PM.
April 2nd, 2019, 06:29 PM   #32
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Quote:
 Originally Posted by michaelcweir What you are describing as a Pythagorean Triangle meets the conditions for the right triangle proof. Where one of X,Y,Z cannot be integer. In practice, there is a smaller triangle where (Y+1)(T)- a = X(T) + Y(T) where X,Y,T are integer. Since Z cannot be smaller than Y+1
You are now denying that $(x^5)^2 + (y^5)^2 = (z^5)^2$ is a right triangle where $\alpha = 0$? You are denying this? What's $\alpha$ then?

Last edited by skipjack; April 2nd, 2019 at 11:32 PM.

 April 3rd, 2019, 10:58 AM #33 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 what I am saying is that you can apply Fermat's right triangle theorem and say that there exists a smaller triangle where Z is smaller than Y+1.I see now that calling it a non primitive is incorrect, but I DON'T THINK THERE IS A TERM THAT DESCRIBES THIS SITUATION. what THE RIGHT TRIANGLE PROVES IS THAT THERE EXIST SOME SMALLER TRIANGLE WHERE AT LEAST ONE OF Z,X,Y IS NOT INTEGER. For example 3(3) + 4(3) is not equal to 5(3) but to approx. 4.3(3) a in this case 0.7
April 3rd, 2019, 11:02 AM   #34
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 Originally Posted by michaelcweir what I am saying is that you can apply Fermat's right triangle theorem and say that there exists a smaller triangle where Z is smaller than Y+1.I see now that calling it a non primitive is incorrect, but I DON'T THINK THERE IS A TERM THAT DESCRIBES THIS SITUATION. what THE RIGHT TRIANGLE PROVES IS THAT THERE EXIST SOME SMALLER TRIANGLE WHERE AT LEAST ONE OF Z,X,Y IS NOT INTEGER. For example 3(3) + 4(3) is not equal to 5(3) but to approx. 4.3(3) a in this case 0.7
You're pointedly ignoring the direct counterexample I showed to your false "proof." I'm done here. All the best.

 April 4th, 2019, 02:38 PM #35 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 I have a hard time understanding why that is a counter example when your question made me think of something that I think inspired Fermat when he was thinking about Pythagorean Triplets. WE know that FLT only needs to be proven for power n, when n is odd. It is possible to construct a right triangle X(n), Y(n), with hypotenuse (x(2n) + Y(2n)) (1/2) We know the square of the hypotenuse is a whole number. The question is the triplet formed a triplet with integer lengths.- ie. is (x(2n) + Y(2n)) (1/2) an integer? if not , then (x(2n) + Y(2n)) (1/2) must be some irrational? Or is (x(2n) + Y(2n)) (1/2) a integer (example 2(1/2) when squared is 2) Also if (x(2n) + Y(2n)) (1/2) is integer, is it possible that the triplet formed a Pythagorean triplet?
 April 4th, 2019, 03:30 PM #36 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timey-wimey stuff. So.. Would you be willing to show us how to prove $\displaystyle a^3 + b^3 \neq c^3$. Is there a reason you won't do this? -Dan
April 4th, 2019, 04:17 PM   #37
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 Originally Posted by michaelcweir I have a hard time understanding why that is a counter example when your question made me think of something that I think inspired Fermat when he was thinking about Pythagorean Triplets. WE know that FLT only needs to be proven for power n, when n is odd.
Yes that is true. So why did I use $n = 10$ in my counterexample?

It's true that if there's a Fermat triple $x^{10} + y^{10} = z^{10}$ with $n = 10$ then we already have another triple $(x^2)^5 + (y^2)^5 = (z^2)^5$; showing that for purposes of FLT it's sufficient to prove it for $n = 5$; and in general only for odd primes. [$n = 4$ is a corner case that must be handled separately).

However, the $n = 10$ example is relevant to YOUR PROOF; not FLT proofs in general. You are talking about Pythagorean triples, which are Fermat triples with $n = 2$. So now if $x^{10} + y^{10} = z^{10}$ then we also have $(x^5)^2 + (y^5)^2 = (z^5)^2$.

Thus $x^5$, $y^5$, and $z^5$ form a RIGHT TRIANGLE whose SIDES ARE INTEGERS. Therefore when you draw your picture, alpha must be zero. And that breaks the rest of your proof.

Now of course we know from FLT that there can be no such integer triple for $n = 10$. But if we are trying to prove FLT, then we can't assume it. We do not know a priori that such a triple for $n = 10$ can not exist. It is incumbent on you to explain this point. You CLAIM that alpha must be strictly positive; but here is an example where it is zero.

Last edited by Maschke; April 4th, 2019 at 04:22 PM.

 April 8th, 2019, 12:14 PM #38 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 I do appreciate your ability to explain your statements. I would like to express myself better instead of stumbling around like I do..So yes, you are right under the conditions that I presented. There is a range that exists, and I will find it But I seem to have stumbled onto something interesting while trying to figure out the range. I would be very interested n your comments.. FLT for powers of n odd, n>2. X(n) + Y(n) = Z(n) We can construct a right triangle with legs X(n/2), Y(n/2), Z(n/2) We can label z(n/2) =C So C(2) = Z(n) By re-arranging the formula we get Z=C(2)/Z(n-1) where C(2) and Z(n-1) are integer. The claim is that Z must be irrational. The proof of that is to claim Z is rational and prove a contradiction. The proof is modeled after the proof that square root of 3 is irrational. First, Z=C(2)/Z(n-1) is in simplest form. Amuse Z(n-1), Z are even, then C(2) must also be even. This is a contradiction that Z. Z(n-1) , C(2) are simplest for, as the equation can be divided by 2 again. Now assume that Z(n-1), Z are odd, then C(2) must also be odd Let C(2) = 2h +1 and Z(n-1) = 2k +1 where h,k are integer (2h +1) = Z(2) ( 2k +1) Now square the the entire equation (2h +1) (2)= Z(4) ( 2k +1)(2) expanding equation becomes 4h(2) + 4 h +1 = Z(4)(4k(2) + 4k + 1) expanding rhs 4h(2) + 4 h +1 = Z(4)(4k(2) +Z(2) 4k + Z(2) Subtract 1 from both sides 4h(2) + 4 h = Z(4)(4k(2) +Z(4) 4k + Z(4) – 1 Both LHS and RHS are even so we divide both sides by 2 2h(2) + 2 h = Z(2)(2k(2) +Z(2) 2k + (Z(2) – 1)/2 (Z(2) – 1)/2 IS ODD , SINCE Z(4) – 1 IS EVEN But now it is apparent that LHS is even, but RHS is odd So the assumption that Z is integer is false. It should now be evident that the proof also works when the power n , when positive, also works. This proves FLT, just as Fermat was able to prove for himself.
April 8th, 2019, 07:54 PM   #39
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 Originally Posted by michaelcweir I do appreciate your ability to explain your statements. I would like to express myself better instead of stumbling around like I do..
Okay. To that end, I'd only like to respond to the first couple of paragraphs of your response and see if we can get agreement on specific points.

Quote:
 Originally Posted by michaelcweir So yes, you are right under the conditions that I presented.
Do you mean to acknowledge that your proof is refuted by my example that a Fermat triple for n = 10 would give rise to an integer right triangle with alpha = 0? It's important to me to have perfect clarity on this point.

Quote:
 Originally Posted by michaelcweir There is a range that exists, and I will find it
A range of what? What do you mean? It's clear that sometimes alpha is greater than zero and sometimes it's zero. But knowing that alpha must always be greater than zero amounts to proving FLT. Can you clarify your thoughts about what this "range" is? What is ranging, and what range is it ranging over?

Quote:
 Originally Posted by michaelcweir But I seem to have stumbled onto something interesting while trying to figure out the range. I would be very interested n your comments..
I prefer not to go down this path right away. Toward the goal of more clear exposition, can you respond directly to the points I've made please?

However I will say this much:

Quote:
 Originally Posted by michaelcweir FLT for powers of n odd, n>2. X(n) + Y(n) = Z(n) We can construct a right triangle with legs X(n/2), Y(n/2), Z(n/2)
Suppose n = 3. Then n/2 = 1.5 and this makes no sense. Earlier you used the greatest integer function in the exponent. But then you have int(n/2) = 1 and you just have X + Y = Z. I don't see your point at all.

Your idea depends on a real parameter alpha; and the greatest integer function is to coarse to provide meaning here IMO.

But frankly to focus the conversation it's better to talk about your alpha and your "range" and see if you can explain those, rather than go down another path.

 April 17th, 2019, 06:13 PM #40 Member   Joined: Mar 2019 From: california Posts: 74 Thanks: 0 Thank you for your good suggestions. Sometimes you can't really interfere with the flow of the mind.So i have been working on the proof using the proof for the square root of 3 being irrational. I think that Fermat 's proof is something like that. Anyway I did go back to the proof i was working on, bu tit turned out differently than I thought. The range of numbers for the proof shows that Z must be between Y and Y+1. I have not really investigated how to make the proof for n =3. Of more concern to me is why the case n =2 is able to be excluded. Perhaps you can shed some light on that.If it can't be excluded there may be something wrong with the proof. I did load up a video on YouTube and there is talking in the video.If you just look at the equations and diagrams you will the idea easily. I would like to extend an invitation to help me with the notation. i would like to include you as one of the authors of the proof. Thank you.

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