quasi wrote: >John Jens wrote: > >>Finished !!!!!!!!! :) >> >>http://primemath.wordpress.com/ > >You're an idiot. > >Several responders have alerted you to the fact that your >proof is fatally flawed, and yet you obliviously post >proof after proof, all with the same blatant error. > >Your proof fails almost right away. I'll quote from your >latest version: > >>Fermat?s little theorem states that if p is a prime number, >>then for any integer a, the number a^p is an integer >>multiple of p.
Still another error above ...
Presumably you meant to say
"the number a^p - a is an integr multiple of p"
"the number a^p is an integr multiple of p".
Of course that error was probably just a typo, and easily corrected.
As has been pointed out to you several times, the irreparable error is your claimed inequality a < p.
>> a^p = a(mod p) > >However Fermat's little Theorem doesn't force a < p. > >For example, using a = 4, p = 3, we have > > a^p = a (mod p) > >but a is not less than p. > >>Assume that a,b,c naturals and p prime and >> >> 0 < a <= b < c > >Ok so far. > >> and a < p > >Stop right there -- you can't justify a < p. > >Moreover, there's no repair -- your proof is dead.