20171117, 11:18  #1 
May 2004
2^{2}×79 Posts 
Conjecture pertaining to modified Fermat's theorem
Let a = x + ysqrt(m) be a quadratic integer. As stated before modified
Fermat's theorem is valid for m = 1 and 5 as practically proved by Hardy (An introduction to the theory of numbers). Conjecture: it is valid for all integer values of m subject to conditions: i) a and p are coprime ii) m not equal to p. Recall that modified Fermat's theorem is a^(p^21) = = 1 (mod p). Here x,y and m belong to Z. Last fiddled with by devarajkandadai on 20171117 at 11:21 Reason: important poimt omited 
20171117, 15:09  #2  
Aug 2006
5,987 Posts 
Quote:
Given any integer m which is not a square, any (rational) prime p, and any integers x and y, if
Is this correct? 

20171117, 17:12  #3  
Feb 2017
Nowhere
2^{2}·3·499 Posts 
Quote:
x^2  x  1 = 0 which by quadratic formula are . Second, you need to make it clear what you mean by "relatively prime to p." But, assuming you mean "a" can be any algebraic integer in the maximal order R (ring of algebraic integers) in Q(sqrt(m)), such that the ideal aR + pR is all of R, or alternatively, a is invertible in the residue ring R/pR, then assuming p does not divide the field discriminant, your "conjecture" is trivial and has been known since the precambrian era. OK, maybe not that long, but Lordy, it's been known a long time. Under this additional assumption, the residue ring R/pR is either the field of p^2 elements, of which the invertible elements form a cyclic group of order p^21; or the direct product of two copies of the field of p elements. The group of invertible elements in this ring is the direct product of two cyclic groups of order p1. In either case, the exponent of any invertible element divides p^2  1. If p does divide the field discriminant, your conjecture is in trouble. For example, with m equal to 1, p = 2 (which is not equal to m as per your condition), and a = i, p^2  1 is 3, but i^3 is not congruent to 1 (mod 2). In fact, i^2 = 1 is congruent to 1 (mod 2). Last fiddled with by Dr Sardonicus on 20171117 at 17:14 Reason: Fixing typos and editing mistakes 

20171118, 11:08  #4  
May 2004
100111100_{2} Posts 
Quote:


20171127, 10:08  #5  
May 2004
13C_{16} Posts 
Quote:
( to be continued). 

20171127, 20:45  #6 
Mar 2016
2·5·41 Posts 
Please consider that
(x+y)^p=x^p+y^p mod p if p is prime for (x+yI)^p = x^p+(yI)^p mod p if p=1 mod 4 => x^p+(y^p)I if p=3 mod 4 => x^p(y^p)I same calculation for A=sqrt (2) for example (x+yA)^p = x^p +(yA)^p = x^p+(y^p)(A^p) You get a criteria for an "extended Fermat" 
20171128, 12:26  #7  
May 2004
100111100_{2} Posts 
Quote:
2) If, with respect to a given base as given above Fermat's theorem is valid for a given p needless to say modified Fermat's theorem is valid; the converse is not necessarily true. 3) If for a given base and p modified Fermat's theorem is valid in the relevant real field it will also be valid in the corresponding complex field (example: if it is valid in the field Mod(x^2  5) it will be valid in the field Mod(x^2 + 5) Now let me state the conjecture in the case where the discriminant, m, is a prime number : Let a be an algebraic quadratic integer. Then a^(p^21)==1 (mod p) subject to the following conditions: i) p is coprime with Norm of a ii) p not equal to m, the discriminant. 

20171129, 10:46  #8  
May 2004
2^{2}·79 Posts 
Quote:


20171130, 05:16  #9  
May 2004
316_{10} Posts 
Quote:
Then a^(p^21) = = 1 subject to the following conditions: i) p is coprime with norm of a ii) p not equal to p_1, p_2,....p_r 

20171211, 05:04  #10  
May 2004
2^{2}×79 Posts 
Quote:


20171220, 05:19  #11 
May 2004
13C_{16} Posts 
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). 
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