20171221, 10:29  #12  
May 2004
2^{2}×79 Posts 
Quote:
Note that sqrt(m)^(p1) = m^(p1)/2. Now (p1)/2= p'(prime)example: p=7,p'=3 or it can be (p'1)example:p=13,p'=7 or (p^21)/2 = p' 1 example p = 19, p'= 181. All these subcases and variations thereof reduce case where a =0, b= 1 and m is a squarefree composite number to just cases of Fermat;s theorem. In other words modified Fermat's theorem has been further partly proved. We now come to the the tough part i.e. when (a,b,c) = 1 and m, the discriminant is (i) prime and (ii) m is a squarefree composite number ( to be continued). 

20171225, 05:43  #13  
May 2004
2^{2}·79 Posts 
Quote:
Now let M represent the binomial expansion of (a + bsqrt(m))^(p^21) excluding the first and last terms. Then, by the above, M = = 0 (mod (p^21)). This implies that M = = 0 (mod (p1)). This implies that M = = 1(mod p). Hence proved. Note i) m, the discriminant, is an integer; in other words both real and complex fields are covered ii) m can be prime or a squarefree composite number. 

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