Note that it is enough to prove
|
We will assume a to be relatively prime to p. This proof will make use of our base a multiplied by all the numbers from 1 to p-1. It turns out that if p is prime, the values 1a through (p-1)a (modulo p) are just the numbers from 1 through p-1 rearranged, a consequence of the following lemma. We then multiply all those numbers together, resulting in a formula from which the theorem follows.
Lemma: If a is relatively prime to p and x and y are integers such that xa = ya (mod p), then x = y (mod p).
Proof of lemma: xa = ya (mod p) means that p divides xa - ya = a (x - y). We know that a does not contain the prime factor p, so (x - y) must contain it, since the prime factorization is unique by the fundamental theorem of arithmetic. So p divides (x - y), which means x = y (mod p), which completes the proof of the lemma.
Proof of theorem: Consider the set P = {1a, 2a, 3a, ... (p-1)a}. These numbers are different modulo p by the lemma, and none of them is zero modulo p (again by the lemma: 0a = ka (mod p) would imply 0 = k modulo p, but k is too small for that). So modulo p, the set P is the same as the set N = {1, 2, 3, ... (p-1)}. So if we multiply the elements of these two sets together, we will get the same result modulo p:
Regrouping the left side:
Now we would like to cancel the common term (p-1)! from both sides. This is allowed by the lemma, since p and (p-1)! can have no factor in common, again by the fundamental theorem of arithmetic. Dividing out (p-1)!, we get:
Inductive proof with the binomial theorem
Here we use mathematical induction. First, the theorem is true for a=1, then one proves that that if it is true for a = k, it is also true for a = k + 1, concluding that the theorem is true for all a.
Before the main argument the following lemma is needed
<math> ({}_{i}^{p}) </math> is, by the fundamental theorem of arithmetic, a multiple of p, so the whole term <math> ({}_{i}^{p}) a^{i} b^{p-i} </math> is a multiple of p if 0 < i < p. This means the whole sum from i = 1 to i = p - 1 equals 0 mod p. So, (a + b)p mod p indeed equals ap + bp mod p when p is prime.
Back to the proof of the theorem. We proceed now with the two induction steps.
Here is an interesting proof which involves very little symbolic mumbo-jumbo.
Let us say that I make closed bracelets out of open chains that consist of p coloured links, with a choice of a different colours; and that I can use the links in any combination. Now, since these are closed bracelets, if I give you one, but you will be able to rotate it at will. So the 9-link bicolor bracelet ABAABBBBB is the same as BBABAABBB (you can rotate the bracelet), but it is different from BBBBBAABA (you cannot reverse it or recolour it).
Some 9-link bracelets can be made from only one "directional" open chain, such as AAAAAAAAA; however, some can be made from more than one such chain (ABBABBABB, BABBABBAB, BBABBABBA all make the same bracelet).
If you tell me how many links a bracelet is to have (call this number p), how many different bracelets of that size can I make, and out of how many distinct open chains can I make each one? Since it is Fermat's Little Theorem we are trying to prove, let us restrict ourselves to cases where p is prime. Let us find the answer thus:
So what do we have? We have a unicolor bracelets, each from one unicolor open chain; and the other ap-a (multicolor) open chains make (ap-a)/p bracelets, each from p open chains. This shows that p divides (ap-a). Q. E. D.
A proof using dynamical systems
Here we prove the special case a = 2 using the fixed points of a 1D-map which is commonly encountered in dynamical systems.
Define the "tent" map:
and consider the dynamical system xn+1 = f(xn), n = 0, 1, 2, ... If x0 is chosen in the interval [-1,1], then all xn will remain in that same interval.
The fixed points of f are -1 and 1/3.
If p is a prime number, then the p-iterated map fp has 2p fixed points which are solutions of:
But two of these fixed points are –1 and 1/3. All the others must have period p, since any period would have to be divisor of p but the prime number p doesn't have any non-trivial divisors.
So, we have 2p – 2 fixed points, forming disjoint orbits of period p. Then (2p – 2)/p is a natural number, the number of orbits of period p. So p' divides 2p - 2. QED
Take this into account: numerical calculations must fail for great p due to rounding errors of the calculator or the computer.
A second proof using Dynamical systems
For any n we define in the interval (0,1) ->(0,1) ( ) closed
Tn(x) = FP(nx), x<1
Tn(x) = 1, x=1
FP: Fractional Part
Lemma 1: Let n be an integer greater than 1. The function Tn(x) has n fixed points in (0,1).
Lemma 2: Let a and b be positive integers. Then for all x belognig to (0,1):
Ta(Tb(x)) = Tab(x)
Using values a and p, we consider the p-periodic point of Ta. These p-periodic points are fixed points of Ta iterated p times, which is Ta^p. This has a^p fixed points. Of these, exactly a are fixed points of Ta.
Since p is prime the rest of them have minimal period p under Ta. This means that there are a^p-a points that have minimal period p. Since each point with minimal period p lies in an orbit p, there are (a^p-a)/p orbits of size p. Since this is an integer, we see that p divides (a^p-a).
This is similar to the direct proof. Trivially, the integers 1, ..., p-1 form a group under multiplication mod p. This group is finite, so clearly the subgroup generated by any a in 1, ..., p-1 must have size q where q divides the size of the original group, p-1. That is, <math> a^{q} = 1 \mod p </math>. But since p-1 = rq for some integer r, <math>a^{p-1} = 1 \mod p</math>. Add the special case where a = p and we get the full proof.
Search Encyclopedia
|