Eldad Eshel Posted July 7, 2017 Share Posted July 7, 2017 I heard once of an equation that was proved in a simple way and then the solution was lost. This happened some hundreds of years ago, I think. Does anyone know this equation ? I forgot the exact context. Link to comment Share on other sites More sharing options...

studiot Posted July 7, 2017 Share Posted July 7, 2017 Fermat's Last Theorem? 1 Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 7, 2017 Author Share Posted July 7, 2017 Fermat's Last Theorem? Yes, thank you. x^n + y^n = z^n This equation only works with certain numbers with n = 1 or 2. x^n + y^n =/ z^n For n > 2 That last solution from long ago was simply wrong. Link to comment Share on other sites More sharing options...

John Cuthber Posted July 7, 2017 Share Posted July 7, 2017 That last solution from long ago was simply wrong. How do you know? It's unlikely, but possible that Fermat came up with some brilliant solution. Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 7, 2017 Author Share Posted July 7, 2017 How do you know? It's unlikely, but possible that Fermat came up with some brilliant solution. n = 3 x = 2 y = 3 z = 4 2^3 + 3^3 =/ 4^3 8 + 27 = 35 8 + 27 =/ 64 End of story. -2 Link to comment Share on other sites More sharing options...

Acme Posted July 8, 2017 Share Posted July 8, 2017 n = 3 x = 2 y = 3 z = 4 2^3 + 3^3 =/ 4^3 8 + 27 = 35 8 + 27 =/ 64 End of story. That's just a special case and not the same as a proof of the general equation. As was pointed out, there is no evidence Fermat had a proof; he simply claimed to have had one. If he did, it would most certainly be different than Wile's proof which runs some 200 pages and uses mathematics not extant in Fermat's time. Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 8, 2017 Author Share Posted July 8, 2017 That's just a special case and not the same as a proof of the general equation.As was pointed out, there is no evidence Fermat had a proof; he simply claimed to have had one. If he did, it would most certainly be different than Wile's proof which runs some 200 pages and uses mathematics not extant in Fermat's time. Special case ? Are we talking mathematics or rubjub ? There are an infinite number of these "Special case"s. It just doesn't work AT ALL, NEVER, EVER, EVER for n > 2. Try it and see. This Wile's "proof" is also some bullshit. Fermat came up with one proof for n=2 and thought he prooved it for the entire equation. This is the so called "Lost solution". He went out and talked about it, and created a big excitement. He later found out it doesn't work always like he thought and destroyed the so called "Lost solution". That is all there is to this story. It is one big Bullshit. This is more a psychological story than a mathematical one. Something is destroyed, erased, lost or forgotten, and it's value and merit get bloated to the heavens. The interesting part of this equation from the mathematical point of view is how it never ever works for n>2. Not that it does. -4 Link to comment Share on other sites More sharing options...

MigL Posted July 8, 2017 Share Posted July 8, 2017 Really ? So YOUR proof is to actually multiply it out for each number from 3 to infinity ? Indulge us, then. Show the proof that it doesn't work for n=99999999999999999999. We'll wait several months. Then you can start on n+1 ! Link to comment Share on other sites More sharing options...

Acme Posted July 8, 2017 Share Posted July 8, 2017 Special case ? Are we talking mathematics or rubjub ? There are an infinite number of these "Special case"s. It just doesn't work AT ALL, NEVER, EVER, EVER for n > 2. Try it and see. The point is that you, nor I, nor any person can try all the special cases, therefore in order to say with surety "It just doesn't work AT ALL, NEVER, EVER, EVER for n > 2", a mathematical proof of the general case is required. Do you have such a mathematical proof? This Wile's "proof" is also some bullshit. So you say. What proof do you have? Fermat came up with one proof for n=2 and thought he prooved it for the entire equation. This is the so called "Lost solution". He went out and talked about it, and created a big excitement. He later found out it doesn't work always like he thought and destroyed the so called "Lost solution". That is all there is to this story. Please direct me to your source for these assertions. It is one big Bullshit.Again, so you say. This is more a psychological story than a mathematical one. Something is destroyed, erased, lost or forgotten, and it's value and merit get bloated to the heavens. The interesting part of this equation from the mathematical point of view is how it never ever works for n>2. Not that it does. ? But you offer no mathematical proof, which makes your assertions uninteresting. PS Sorry MigL, you posted in between when I was voting my conscience. Hopefully some well meaning member will bind your wound and mend my error. Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 8, 2017 Author Share Posted July 8, 2017 (edited) Really ? So YOUR proof is to actually multiply it out for each number from 3 to infinity ? Indulge us, then. Show the proof that it doesn't work for n=99999999999999999999. We'll wait several months. Then you can start on n+1 ! Big fucking deal.My claims are proovable with a serious computer, for any number. Edited July 8, 2017 by hypervalent_iodine Inappropriate content removed -5 Link to comment Share on other sites More sharing options...

Acme Posted July 8, 2017 Share Posted July 8, 2017 Big fucking deal. My claims are proovable with a serious computer, for any number. ... That would be a proof by exhaustion, but since natural numbers are infinite, such a proof is impossible by any computer the same as for a people. The prevalence of digital computers has greatly increased the convenience of using the method of exhaustion. Computer expert systems can be used to arrive at answers to many of the questions posed to them. In theory, the proof by exhaustion method can be used whenever the number of cases is finite. However, because most mathematical sets are infinite, this method is rarely used to derive general mathematical results. ... Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted July 8, 2017 Share Posted July 8, 2017 ! Moderator Note I have hidden the last two posts. Quit with the personal comments / insults. MigL, that goes for you too. Eldad, you need to tone down the language. If you can't post respectfully, you will not be allowed to post, 3 Link to comment Share on other sites More sharing options...

John Cuthber Posted July 8, 2017 Share Posted July 8, 2017 Big fucking deal. My claims are proovable with a serious computer, for any number. With a computer you can test it for any number- but not for every number. Do you understand the difference? Your "proof" is like saying Pythagorean triples don't exist because 2^2 =3^2 is 13 and 13 isn't a square number. Just because it doesn't work for that choice of numbers doesn't mean there are not other numbers where it does work (for example 3 and 4) This Wile's "proof" is also some bullshit. Please summarise his proof- just to show that you understand it well enough to comment meaningfully on it. 3 Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 8, 2017 Author Share Posted July 8, 2017 x^n + y^n = z^n x = y x = z x =/ 0 x^n + x^n =/ x^n 2x^n =/ x^n 2 =/ 1 For this case an infinite number of numbers will not work Link to comment Share on other sites More sharing options...

John Cuthber Posted July 8, 2017 Share Posted July 8, 2017 This bit "x^n + y^n = z^n x = yx = z" is nonsense x is not the same as y and x is not the same as z. In the case for n=2 (where it does work) the bet known example is 3^2 +4^2 =5^2 x=3 y=4 and z=5 (and n =2) All the 4 numbers can be different. 1 Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 8, 2017 Author Share Posted July 8, 2017 This bit "x^n + y^n = z^n x = yx = z" is nonsense x is not the same as y and x is not the same as z. In the case for n=2 (where it does work) the bet known example is 3^2 +4^2 =5^2 x=3 y=4 and z=5 (and n =2) All the 4 numbers can be different. They CAN be the same. And then my proof stands firm. Link to comment Share on other sites More sharing options...

John Cuthber Posted July 8, 2017 Share Posted July 8, 2017 They CAN be the same. And then my proof stands firm. And they can be different, in which case the "proof" fails. A "proof" that doesn't always work isn't a proof. Did you not understand that? It's like saying that all pairs of numbers add up to 6 because 2 + 4 is 6. And when someone says but 5 and 5 make 10 you say "But they can be 2 and 4 and so my proof stands". It's meaningless. Also, please give us an explanation of Wiles' proof so that you can show you are in a position to criticise it. 1 Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 8, 2017 Author Share Posted July 8, 2017 And they can be different, in which case the "proof" fails. A "proof" that doesn't always work isn't a proof. Did you not understand that? It's like saying that all pairs of numbers add up to 6 because 2 + 4 is 6. And when someone says but 5 and 5 make 10 you say "But they can be 2 and 4 and so my proof stands". It's meaningless. Also, please give us an explanation of Wiles' proof so that you can show you are in a position to criticise it. This equation works for a few x,y,z when n=1,2. For n>=3 it simply does not work at all. Trying to prove it ALWAYS works is just ridiculous. One false answer, like x=2 y=3 z=4 n=3, debunks the entire case of both fermat and wile. Link to comment Share on other sites More sharing options...

DrKrettin Posted July 8, 2017 Share Posted July 8, 2017 This equation works for a few x,y,z when n=1,2. For n>=3 it simply does not work at all. Trying to prove it ALWAYS works is just ridiculous. One false answer, like x=2 y=3 z=4 n=3, debunks the entire case of both fermat and wile. Why do you think it ridiculous? That is exactly what Wiles did. It took him about 4 years, so why not try it yourself? 2 Link to comment Share on other sites More sharing options...

studiot Posted July 8, 2017 Share Posted July 8, 2017 Yes, thank you. x^n + y^n = z^n This equation only works with certain numbers with n = 1 or 2. x^n + y^n =/ z^n For n > 2 That last solution from long ago was simply wrong. This thread had become so vituperous that I had withdrawn. However I have noticed something in the above. Eldad Eschel, You don't seem to know what Fermat's Last Theorem actually says. Fermat There are no three positive integers, x, y and z which satisfy the equation x^{n} + y^{n} = z^{n} for any n>2 This is perfectly consistent with the first part of what you say, but perhaps your difficulty is contained in your last two lines where you seem to think that Fermat or someone offered three such numbers for some n>2 He did not. Have you perhaps misread the theorem? 2 Link to comment Share on other sites More sharing options...

John Cuthber Posted July 8, 2017 Share Posted July 8, 2017 This equation works for a few x,y,z when n=1,2. For n>=3 it simply does not work at all. Trying to prove it ALWAYS works is just ridiculous. One false answer, like x=2 y=3 z=4 n=3, debunks the entire case of both fermat and wile. Actually, the whole point was that he proved it NEVER works. 1 Link to comment Share on other sites More sharing options...

Acme Posted July 8, 2017 Share Posted July 8, 2017 Actually, the whole point was that he proved it NEVER works.Precisely. And to clarify 'he', Fermat proposed the theorem/conjecture & claimed he proved it never works (saying the proof is too long for this margin), while Wiles actually did prove it never works. Link to comment Share on other sites More sharing options...

John Cuthber Posted July 8, 2017 Share Posted July 8, 2017 It's also important to recognise that Wiles proved it it true, even for numbers that are too big to fit into any computer. There are only something like 10^80 particles in the known universe so no computer -even one which is described by the potty mouthed Mr Eidad as "serious" - could ever test the theorem for numbers like n= 10^81 because there isn't enough stuff in the universe to make it from Link to comment Share on other sites More sharing options...

swansont Posted July 9, 2017 Share Posted July 9, 2017 This equation works for a few x,y,z when n=1,2. For n>=3 it simply does not work at all. Trying to prove it ALWAYS works is just ridiculous. One false answer, like x=2 y=3 z=4 n=3, debunks the entire case of both fermat and wile. ! Moderator Note Indeed, since that's the opposite of what the conjecture states. Let's get this discussion back on track Link to comment Share on other sites More sharing options...

Eldad Eshel Posted July 9, 2017 Author Share Posted July 9, 2017 You guys spill out alot of contradictions. Some of you say it always works some of you say it never works, and then all sorts of bs in between. It never works for n>=3. Wiles "proof" is some big bs I assure you. -1 Link to comment Share on other sites More sharing options...

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