
Re: From Fermat little theorem to Fermat Last Theorem
Posted:
Jan 7, 2013 1:22 AM


On Sunday, January 6, 2013 9:05:08 PM UTC+2, M_Klemm wrote: > You should give a reason why you assume a < p. > > For p = 2 you have 3^2 + 4^2 = 5^2, and then by little Fermat 3 + 4 > > congruent 5 (mod 2) > > but not 3 < 2. > > > > Regards > > Michael > > > > wrote in message > > news:696982e8322848d0b4839cc3acb97341@googlegroups.com... > > > On Sunday, January 6, 2013 10:56:35 AM UTC+2, M_Klemm wrote: > > >> Hello, > > >> > > >> > > >> > > >> consider the case p =3, proved by Euler. Then you see that the assumption > > >> a > > >> > > >> < p in line 4 is not at all justified. > > >> > > >> Regards > > >> Michael > > > > > > I'm sorry but I don't understand what are you trying to say.
The reason is to prove FLT . Let's split in three steps : Step 1> prove a^p + b^p != c^p with a < p ,a,b,c, naturals Step 2> extend to rationals , still a < p Step 3> pick A >= p, assume A^p + b^p = c^p and scaling down to A/k < p ,k rational >contradiction to step 2
If a + b ? c>0 because 0<a?b<c implies b ? c < 0 , 0 ? a + b ? c < a < p then a + b ? c ? 1 and because a + b ? c < a implies a ? 2 and because a < p implies p > 2
Still a + b ? c>0 ,a + b  c must be at least 1 and with p=2 we can't find natural a between a + b  c and p there's no Step 1 for p = 2
(If a + b ? c ? 0 we have a + b ? c
(a+b)^p?c^p and using binomial theorem
a^p+b^p<(a+b)^p=a^p+?+b^p?c^p)

