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Topic: fERMAT'S PROOF 2^n-2 for prime n is mod n
Replies: 0

 DRMARJOHN Posts: 116 From: ROCHESTER NY Registered: 4/28/10
fERMAT'S PROOF 2^n-2 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:

------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-------6
3 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.