Fermat's Last theorem short proof
in [Math]
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From: bassam king karzeddin on 25 May 2007 14:27 > On May 24, 9:19 am, bassam king karzeddin > <bas...(a)ahu.edu.jo> wrote: > > <snip> > > Ah, James, you admit to drinking while mathematic-ing > (Is that a > word? Well, now it is.). No wonder your posts are > full of drunken > tirades that you vomit from your distended belly. > > I'm sure you just drink to wash down the "solution" > of Quinn's > "mathematics" from your mouth? > > > I couldn't delete your ugly post, but to balance that, and I never expected that down level of reaction against facts with out a defense from THOSE who are fact searchers, so I have to make it alone I may ask any one the following questions Is it possible for a drunk to make those perfect equations? I guess also you generally don't like that, since it is not imposed from journals, but you should believe that is much larger than the journals capacity! Is it possible for a drunk to make those undefeatable conjectures? Of course, if prizes are offered, then there would be a very big deference! Is it ... ... How many proofs should I provide... I think you know the answer better than me, but... I think I came to adopt and rescue you poor mathematicians! B.Karzeddin
From: neilist on 26 May 2007 02:04 On May 25, 6:27 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > > On May 24, 9:19 am, bassam king karzeddin > > <bas...(a)ahu.edu.jo> wrote: > > > <snip> > > > Ah, James, you admit to drinking while mathematic-ing > > (Is that a > > word? Well, now it is.). No wonder your posts are > > full of drunken > > tirades that you vomit from your distended belly. > > > I'm sure you just drink to wash down the "solution" > > of Quinn's > > "mathematics" from your mouth? > > I couldn't delete your ugly post, but to balance that, > > and I never expected that down level of reaction against facts with out a defense from THOSE who are fact searchers, so I have to make it alone > > I may ask any one the following questions > > Is it possible for a drunk to make those perfect equations? > > I guess also you generally don't like that, since it is not imposed from journals, but you should believe that is much larger than the journals capacity! > > Is it possible for a drunk to make those undefeatable conjectures? > > Of course, if prizes are offered, then there would be a very big deference! > > Is it ... > .. > > How many proofs should I provide... > > I think you know the answer better than me, but... > > I think I came to adopt and rescue you poor mathematicians! > > B.Karzeddin- Hide quoted text - > > - Show quoted text - Harris, you're a troll. FEED THE TROLL! FEED THE TROLL!
From: bassam king karzeddin on 30 May 2007 07:46 Dear All Now, It is the time to tell you a message from the QUEEN And there are few things you should realize that some of the Queen's SECRETS have been revealed here, unfortunately some DEVILS began to realize them SECRETLY before you, that is why they already started making stories as one of their representative NEILIST is already here as their shadow only to divert you from the shining facts You should also realize that your DESIGN CHART has been OPENED recently where so many puzzles are awaiting you, and the need for you to be REPLACED is becoming URGENT as you will not be any more suitable for the coming centuries, and naturally there will be the need of only few of you Still the Queen is there in every time and where waiting your last digit to be infinite, B.Karzeddin
From: bassam king karzeddin on 14 Jun 2007 10:20 > Fermat's Last theorem short proof > > We have the following general equation (using the > general binomial theorem) > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + > x^p+y^p+z^p > > Where > N (x, y, z) is integer function in terms of (x, y, z) > > P is odd prime number > (x, y, z) are three (none zero) co prime integers? > > Assuming a counter example (x, y, z) exists such that > (x^p+y^p+z^p=0) > > > (x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z) > > CASE-1 > If (p=3) implies N (x, y, z) = 1, so we have > > (x+y+z)^3 =3* (x+y)*(x+z)*(y+z) > > Assuming (3) does not divide (x*y*z), then it does > not divide (x+y)*(x+z)*(y+z), > So the above equation does not have solution > (That is by dividing both sides by 3, you get 9 times > an integer equal to an integer which is not divisible > by 3, which of course is impossible > I think proof is completed for (p=3, and 3 is not a > factor of (x*y*z) > > My question to the specialist, is my proof a new one, > more over I will not feel strange if this was known > few centuries back > > Thanking you a lot > > Bassam King Karzeddin > Al-Hussein Bin Talal University > JORDAN May be I should have asked my question the way mathematicians in POWER like, like this for example: Dear Professors I had proved FLT, for the above mention case, but I think something went wrong that I didn't realize Can you please help me and point out that error PLEASE Thanking YOU very very very much SIRS OR should I ask DR.MATH please ADVICE ME Bassam Karzeddin
From: neilist on 14 Jun 2007 17:07
On Jun 14, 2:20 pm, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: > I had proved FLT Hahahahahahahahahahahahahahahahahahahahahaha (catch breath) Hahahahahahahahahahahahahahahahahahahahahaha <more snipping> > please ADVICE ME I _advise_ you with your own quote: "... GET OUT OF SCI.MATH ... B.Karzeddin" Hahahahahahahahahahahahahahahahahahahahahaha |