August 28th, 2015, 09:06 AM |
#51 |

Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 |
I just quoted the relevant parts of your text. Anyone interested in this thread can easily go back to see exactly what was written. You wrote: "The v^0 = w^0 = 1 does not mean v = w = 1." We are evidently not using the same rules of arithmetic and logic - and that seems to be the basic problem. I wrote: "It is certainly possible that at least one of a, b in a^n + b^n = c^n could be prime if n > 2." You replied: "It's absolutely unfounded statement." ... "If you can deduce your 'certainly possible' this way I would accept it." Consider this: How exactly would you _prove_ neither a nor b could be prime _without_ supposing FLT is true? I merely pointed out what is possible without that supposition. The burden of proof is not mine. You wrote: "But otherwise v and w (not v^n and w^n) can easily be primes but > 1." So now that's clear. I don't think it's necessary to go further. You obviously disagree with my and al-mahed's analyses of your argument. Why not take his suggestion to post on another forum? If you do, however, I believe you will get similar objections. Or, request Mr. Greathouse's opinion - he may be willing to comment. |

August 28th, 2015, 10:36 AM |
#52 | ||

Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 | Quote:
you using? The ones I am used to say that any number to the 0 power is equal to 1.Quote:
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August 28th, 2015, 11:06 AM |
#53 |

Global Moderator Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms | |

August 28th, 2015, 12:56 PM |
#54 | |||

Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 | Quote:
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required to satisfy this equation are deduced. Their correctness is proved for n=2 (Eq.26). Then it's proved that polynomials don't satisfy equation when n>2.This was opposed by pledge that there are other numbers that don't satisfy the equation. Certainly they are. FLT is true for all integers but only these corresponding to deduced polynomials should be tried. It's what I meant opposing suggestions about possible other numbers. | |||

September 1st, 2015, 08:50 AM |
#55 |

Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 |
What I am trying to understand is: Which part of my reply to Mr. Pogorsky did Country Boy not understand? Of course if w^0 = v^0 = 1 it then follows w = v = 1. Mr. Pogorsky seems to believe it does _not_ follow (see post #50). The only real issue here seems to be reading and comprehending what was written. Also, it follows from equation (10) if u = 0 when n = 1 then p = q = 1 is necessary; then from (7a) and (7b), a = vp and b = wq with all of p, q, v, w = 1, it follows that a = b = 1. The equations should of course reduce to a + b = c if n = 1 and they do not - that is what I missed when I first checked the algebra. Again, Mr. Pogorsky said my statement that at least one of a, b can be prime if n > 2 is "absolutely unfounded". No - it is certainly not at all unfounded because it _is_ possible unless he can _prove_ it is not. He also stated that v and w can be prime, so I do not see any problem with my assertion that if a is prime then v must equal a if v divides a. [This is what I meant when I had stated in a previous post that not all possibilities had been considered: One can not just _assume_ a^n and/or b^n have two _unique_ factors, that is, a and b are not necessarily composite. This all comes back to the improper statement, proof, and application of Lemma-3. I said near the beginning of this thread that the only thing this lemma seems to prove is 1 = 1, and that is hardly a basis for attempting to prove FLT.] Perhaps the mathematician whom he claimed reviewed his paper can clarify any objections that have been given on this forum - clearly, no one here has succeeded doing that to Mr. Pogorsky's satisfaction. I might soon be posting my own totally revised (and completely different than my last alleged) proof of FLT (everyone should have a backup proof of FLT just in case!); so I must refrain from any further comment on Mr. Pogorsky's paper. If he still believes his argument is a proof he should submit it to a professional mathematics journal. |

September 1st, 2015, 09:19 PM |
#56 | ||||

Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 | Quote:
(p-v^0)/w=(q-w^0)/v=u; (p-1)/w=(q-1)/v=u; v(p-1)=w(q-1)=uwv Did you concluded from this v=w=1? Now we bring here u=0 that follows from uwv+v+uwv+w=uwv+v+w. Since neither v nor w equal to zero the only conclusion - hurrah! -p=q=1. But when n=1 we have p=uw+1 and q=uv+1 and we have pretty zeros in parentheses None of the above steps impose any restriction on value of v and w. So I don't know which of them your conclusions are based on. Maybe you would explain, maybe not. Because Quote:
The presented polynomial expressions for a, b, c were obtained through succession of strict transformations (no errors still disclosed). Now somebody (in general) declares that he/she believes that there are kinds of a, b, different from deduced and demands from me to prove that the belief is wrong. By the way what does it mean at all? FLT is true for all numbers. My polynomials would satisfy the a^n+b^n=c^n if the theorem were not true. Is it asserted that prime a or b would satisfy equation too? Quote:
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Last edited by McPogor; September 1st, 2015 at 09:46 PM.
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September 4th, 2015, 01:29 AM |
#57 | |

Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 | Quote:
have noticed I _did_ refer to equation (10). Nevertheless, one more time - just to clarify: When n = 1 then w = v = 1 by equation (10). Then p = q = 1 is _required_ by (10) if u is to be zero. Now we have all of w, v, p, q = 1. Then by (7a) and (7b), a = vp = 1 and b = wq = 1. You can _not_ change the values of w, v, p, q once they have been determined by (10) for a particular case, and apparently that is what you have done because now you write: "But when n = 1 we have p = uw + 1 and q = uv + 1 and we have pretty zeros in parentheses None of the above steps impose any restriction on value of v and w." Exactly how are those statements to be interpreted? As for equations (4a) and (4b), you seem to believe that the right hand sides of each are _always_ the product of two _distinct_ factors, and that is not true if either a or b is prime. Again, it is not _my_ responsibility to prove neither a nor b can be prime. That at least one of a, b could be prime is certainly true if n = 1 or n = 2. To _assume_ it is _not_ true if n > 2 is totally unwarranted, and you either assumed it or you did not even consider the possibility - it really doesn't matter which. | |

September 4th, 2015, 09:22 AM |
#58 | |||||

Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 | Quote:
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Would FLT be untrue in this case? | |||||

September 6th, 2015, 04:42 AM |
#59 |

Senior Member Joined: Apr 2015 From: Barto PA Posts: 170 Thanks: 18 |
"I transformed this unfortunate (but true) Eq.(10) step by step trying to find where this conclusion w=v=1 may come from. There's not any clue. If you see one show at what step it is concluded. Or am I required to prove it's false according to your distorted logic?" (...) "And nothing except your unfounded statements proves that v and w cannot be of any value." SERIOUSLY??? You can't make sense of a set of equations YOU derived??? Incredible. How many times is it necessary for something to be explained before you finally understand it? Or is that even possible? AGAIN!: In equation (10) you have the terms w^(n-1) and v^(n-1) in the numerators of the fractions. The last time I bothered checking, when n = 1 then n - 1 = 1 - 1 = 0. And then (again!) w^0 = v^0 = 1. And that is _exactly_ where w = v = 1 comes from! Then (p - 1)/1 = (q - 1)/1 = u, and if u = 0 when n = 1 then p = q = 1 is _necessary_! And don't forget YOU are the one who said in post #39 that u = 0 when n = 1. Now I imagine the _next_ thing you will try telling me is w^0 and v^0 does NOT mean w = 1 and v = 1. Don't even go there. "I would not stop on it but I'm curious again what does it mean: a or b -prime?" Suggestion: Check the definition of 'prime number'. Hint: Those numbers have no factors. "Would FLT be untrue in this case?" You figure it out. After all, you're the one who 'proved' it. So: My logic is "distorted" and my statements are "unfounded". Hmm... if anyone's logic is distorted it is yours, but you are correct in one respect: You are definitely clueless. Enough is enough already. End of discussion. |

September 6th, 2015, 10:13 AM |
#60 | |

Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 | Quote:
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