Fermat's Last theorem short proof
in [Math]
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From: neilist on 22 May 2007 15:00 On May 22, 7:41 am, bassam king karzeddin <bas...(a)ahu.edu.jo> wrote: <snip> Hey Harris, I bet you picked your new pseudonym "bassam king ..." as a play on "assing king". Conscious choice, or subconscious Freudian slip? Or is that what Quinn affectionately calls you?
From: bassam king karzeddin on 24 May 2007 02:00 > In article > <24975251.1176279446561.JavaMail.jakarta(a)nitrogen.math > forum.org>, > bassam king karzeddin <bassam(a)ahu.edu.jo> writes: > >... > > > >(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) > >... > > The gap in the argument is in the marked lines. For > instance, > 3 does not divide 1*2*7, but it does divide > (1+2)*(1+7)*(2+7). > > > William C. Waterhouse > Penn State > I noticed your reply today only, and really speaking I didn't get the wisdom of your reply separately I wonder also if you realize that the issue is about FLT, where I have made it very clear in my original thread that for assumed counter example (x, y, z) must form a triangle, where also mentioned for a counter example to be considered the following inequality must hold true always: Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) Which also means that the angle opposing the larger side (z) must be between (60, 90) degrees Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: bassam king karzeddin on 24 May 2007 02:02 > In article > <24975251.1176279446561.JavaMail.jakarta(a)nitrogen.math > forum.org>, > bassam king karzeddin <bassam(a)ahu.edu.jo> writes: > >... > > > >(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) > >... > > The gap in the argument is in the marked lines. For > instance, > 3 does not divide 1*2*7, but it does divide > (1+2)*(1+7)*(2+7). > > > William C. Waterhouse > Penn State > I noticed your reply today only, and really speaking I didn't get the wisdom of your reply separately I wonder also if you realize that the issue is about FLT, where I have made it very clear in my original thread that for assumed counter example (x, y, z) must form a triangle, where also mentioned for a counter example to be considered the following inequality must hold true always: Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) Which also means that the angle opposing the larger side (z) must be between (60, 90) degrees Regards Bassam Karzeddin Al Hussein bin Talal University JORDAN
From: quasi on 24 May 2007 07:57 On Thu, 24 May 2007 06:00:36 EDT, bassam king karzeddin <bassam(a)ahu.edu.jo> wrote: >> In article >> <24975251.1176279446561.JavaMail.jakarta(a)nitrogen.math >> forum.org>, >> bassam king karzeddin <bassam(a)ahu.edu.jo> writes: >> >... >> > > >(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) >> >... >> >> The gap in the argument is in the marked lines. For >> instance, >> 3 does not divide 1*2*7, but it does divide >> (1+2)*(1+7)*(2+7). >> >> >> William C. Waterhouse >> Penn State >> >I noticed your reply today only, and really speaking I didn�t get the wisdom of your reply separately >I wonder also if you realize that the issue is about FLT, where I have made it very clear in my original thread that for assumed counter example (x, y, z) must form a triangle, where also mentioned for a counter example to be considered the following inequality must hold true always: > >Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) > >Which also means that the angle opposing the larger side (z) must be between (60, 90) degrees 3 does not divide 4*7*8, but it does divide (4+7)*(4+8)*(7+8). quasi
From: quasi on 24 May 2007 08:12
On Thu, 24 May 2007 06:57:20 -0500, quasi <quasi(a)null.set> wrote: >On Thu, 24 May 2007 06:00:36 EDT, bassam king karzeddin ><bassam(a)ahu.edu.jo> wrote: > >>> In article >>> <24975251.1176279446561.JavaMail.jakarta(a)nitrogen.math >>> forum.org>, >>> bassam king karzeddin <bassam(a)ahu.edu.jo> writes: >>> >... >>> > > >(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) >>> >... >>> >>> The gap in the argument is in the marked lines. For >>> instance, >>> 3 does not divide 1*2*7, but it does divide >>> (1+2)*(1+7)*(2+7). >>> >>> >>> William C. Waterhouse >>> Penn State >>> >>I noticed your reply today only, and really speaking I didn�t get the wisdom of your reply separately >>I wonder also if you realize that the issue is about FLT, where I have made it very clear in my original thread that for assumed counter example (x, y, z) must form a triangle, where also mentioned for a counter example to be considered the following inequality must hold true always: >> >>Sqrt (x^2+y^2) > z > Sqrt (x^2+y^2-x*y) >> >>Which also means that the angle opposing the larger side (z) must be between (60, 90) degrees > >3 does not divide 4*7*8, but it does divide (4+7)*(4+8)*(7+8). I withdraw the above example. It's not applicable in the context of the problem. Assuming the equation (x+y+z)^3 =3*(x+y)*(x+z)*(y+z) you are definitely correct in your statement that if 3 does not divide x*y*z then 3 does not divide (x+y)*(x+z)*(y+z). So yes, you can prove FLT for p=3, case 1, but that's the absolute easiest case. What about p=3, case 2? quasi |