San José State University |
---|
applet-magic.com Thayer Watkins Silicon Valley & Tornado Alley USA |
---|
|
Pierre de Fermat is best known to the general public for what is called his Last Theorem which was really an unproved conjecture. Most likely he did not have α proof for his Last Theorem. However he may well have had α proof for what came to be called his Little Theorem even though he did not communicate it.
Consider the remainders of the set of numbers {α, 2α, 3α, …, (p-1)α} upon division by p. These remainders have to all be distinct and different from 0. If the remainder were 0 for some k it would mean that p divides α contrary to the hypothesis because p does not divide any integer in the set {1, 2, 3, … (p-1)}.
Suppose gα≡hα (mod p) for some g and h in the set {1, 2, 3, … (p-1)}. If gα≡hα (mod p) then ga−ha=kp for some k. This is equivalent to (g−h)α=kp. Without any loss of generality it can be assumed that g≥h. Since p does not divide α it must divide (g−h). The only integer in the possible range of (g−h) which p divides is 0. Thus g−h=0 and hence g=h.
NOte that
The remainders of {α, 2α, 3α, …, (p-1)α } upon division by p are just some permutation of {1, 2, 3, … (p-1)}.
Therefore the product of the remainders of the elements of {α, 2α, 3α, …, (p-1)α } is equal to the product of the elements of the set {1, 2, 3, … (p-1)}. This means
But if αp-1(p-1)! ≡ (p-1)! (mod p) then p divides [αp-1(p-1)! − (p-1)!] which is equivalent to p dividing (αp-1−1)(p-1)!. Since p divides none of the factors of (p-1)! it must divide (αp-1−1), which is equivalent to
If p divides (αp-1−1) it also divides a(αp-1−1), which is equivalent to (αp−α) and thus
Consider p=7 and α=4. Four to the sixth power is 4096. This number divided by 7 is 585 with a remainder of 1. Four to the seventh power is 16384, which has a remainder of 4 upon division by 7.
HOME PAGE OF Thayer Watkins |