Conjecture pertaining to modified Fermat's theorem

[devarajkandadai]

Originally Posted by devarajkandadai

Let us now consider the simplest case i.e. case (i) in which a = 0, b and c =1.Since m is
prime, raising sqrt(m) or sqrt(-m) to an even power and recalling that m is not equal to p reduces the case to nothing but Fermat's theorem. Hence partly proved. (to be continued).

Note that sqrt(m)^(p-1) = m^(p-1)/2. Now (p-1)/2= p'(prime)-example: p=7,p'=3 or it can be (p'-1)-example:p=13,p'=7 or (p^2-1)/2 = p'- 1 -example p = 19, p'= 181. All these sub-cases 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).

[devarajkandadai]

Originally Posted by devarajkandadai

Note that sqrt(m)^(p-1) = m^(p-1)/2. Now (p-1)/2= p'(prime)-example: p=7,p'=3 or it can be (p'-1)-example:p=13,p'=7 or (p^2-1)/2 = p'- 1 -example p = 19, p'= 181. All these sub-cases 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).

It is well known that (x + y)^n = = x^n + y^n.
Now let M represent the binomial expansion of (a + bsqrt(m))^(p^2-1) excluding the first and last terms. Then, by the above, M = = 0 (mod (p^2-1)). This implies that M = = 0 (mod
(p-1)). 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 square-free composite number.