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DRMARJOHN
Posts:
116
From:
ROCHESTER NY
Registered:
4/28/10


fERMAT'S PROOF 2^n2 for prime n is mod n
Posted:
Apr 11, 2013 11:22 AM


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:
xx^3(1)(2) 28 117(81=7) 0016(71=6) 3 divides 6. The 7 is derived by 81. The 1 is derived by 10. These can be considered as the first derivatives. The 6 is derived by 71. This can be considered as the second derivative. In the 81 and 71 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: xx^3(1)(2) 327 2819(278=19) 11712(197=12) 0016 3 divides 12. The 279 and 81 are the first derivatives. The 197 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.



