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A note on Fermat's theorem

Keywords: 
Fermat's theoem
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Any prime in k(1) with shape 4m+1 can be factorised in k(i). Examples : 5 = (2+i)(2-i), 13 = (2+3i)(2-3i) or (3 +2i)(3 - 2i). Now if we use any one of such factors and apply Fermat’s theorem with respect to a prime of the same shape ( excepting the prime in k(1) of which the base is a factor) we get quotients which are integers in k(i). Examples: a)((2+i)^12-1)/13 = 904 - 782i b) ((2+i)^16-1)/17 = 9696 + 20832i. However neither 13 nor 17 are primes in k(i) - in fact they are pseudo primes in k(i).

Some more examples pertaining to the previous message: a) ((6+i)^4-1)/5 = 216 + 168i b)((6 + i)^12-1)/13 = - 78849720 + 180928440i c) ((2 +3i)^4-1)/5 = -24 +24i

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341 is a pseudoprimePlanetmathPlanetmath in k(1). It is also a pseudoprime in k(i) i.e. ((1+i)^340 + 1)/341 = a Gaussian integer. Note it is +1 in parenthesis because -1 is also a unity in k(i). Also 1105, a Carmichael number in k(1) is a pseudoprime in k(i) i.e. ((6 + i)^1104 -1)/1105 = a Gaussian integer . However, it is too big to be copied here.

Primes of shape p’= 4m + 3 are prime in both k(1) and k(i). With reference to Frermat’s theorem in k(i) what is the nature of the bases? In fact they form a group isomorphic with z_n; the bases are given by (1 + kp’ + i). Let me illustrate this with examples p’ = 11 and 19. Here k belongs to W. The congruence in all these cases is to -i and not 1; recall that -i is also one of the unities in k(i). Hence ((1 + i)^10 + i)/11 = 3i and ((12 + i)^10 + i)/11 = 3926445133 + 4305498635i. ((1 + i)^18 + i)/19 and ((20 + i)^18 + i)/19 are also equal to Gaussian integers.

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