```Date: Apr 11, 2013 11:22 AM
Author: DRMARJOHN
Subject: fERMAT'S PROOF 2^n-2 for prime n is mod n

One of Fermat's few proofs is that for prime n, n divides [(2^n) -2], or n is mod n. Can this be expanded for an appropriate formula for (3^n)? Is it possible to draw some parallel to Fermat's thinking and thence understand his mental process? [(2^n) -2] can be graphed, for x=3, as:------x------x^3----(1)----(2)------2-------8------1-------1------7--------------------(8-1=7)------0-------0------1------6-------------(7-1=6)3 divides 6. The 7 is derived by 8-1. The 1 is derived by 1-0. These can be considered as the first derivatives. The 6 is derived by 7-1. This can be considered as the second derivative.In the 8-1 and 7-1 there are two -1, i.e.,-2. The parallel here is Fermat's -2 is the derivatives -1 and -1.This chart can be expanded for x=3:------x------x^3----(1)----(2)------3------27------2-------8------19---------------------(27-8=19)------1-------1-------7------12-------------(19-7=12)------0-------0-------1-------63 divides 12.The 27-9 and 8-1 are the first derivatives. The 19-7 is the second derivatives. Please excuse my being pedantic, which I do because this approach is unfamiliar.  Can the logic of [2^n) -2] be applied to [(3^n)]? Can it be demonstrated that for prime n, n always  divides the second derivative of 3^n? Can anyone out there expand this further? If so, we may gain insight into the mind of Fermat.
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