Some examples:
Let's use p=7
- $2^{\small{7-1}}-1%3D2^6-1%3D64-1%3D63,$ a multiple of 7
- $3^{\small{7-1}}-1=3^6-1=729-1=728,$ a multiple of 7
- $4^{\small{7-1}}-1=4^6-1=4096-1=4095,$ a multiple of 7
Let's use p=5
- $2^{\small{5-1}}-1=2^4-1=16-1=15,$ a multiple of 5
- $3^{\small{5-1}}-1=3^4-1=81-1=80,$ a multiple of 5
- $4^{\small{5-1}}-1=4^4-1=256-1=255,$ a multiple of 5
|
|
Fermat's little theorem (not to be confused with
Fermat's last theorem)
states that if p is a
prime number, then for any
integer a, $a^p-a$ will
be evenly divisible by p. This can be expressed in the notation of
modulo arithmetic as follows:
$a^p\equiv{a}$ (mod p)
A variant of this theorem is stated in the following form:
- If p is a prime number
- and a is an integer with no factors in common with p
- then $a^{p-1}-{1}$ is a multiple of p.
Fermat's little theorem is the basis for the Fermat primality test.
Euler's extension of Fermat's little theorem is the basis of the RSA public-key cryptosystem.
Page Challenge
This theorem only works in one direction. Every prime
satisfies the equation, but there are non-primes that
also satisfy it.
Can you find one?
|
|
- If n is a positive integer
- and $\phi(n)$ is how many numbers that are
- less than n and have no factors in common with n,
- then if a is one of those numbers,
- then $a^{\small\phi(n)}-1$ is a multiple of n.
Proof
This proof can be adapted to prove Euler's result by taking the product of all the numbers
co-prime to n, rather than using the numbers from 1 to p-1. |
|
Let
p be a prime. What we're going to do is write down one product, and then evaluate it
in two different ways. One way will give some number $F,$ the other will give $a^{p-1}F,$
and so these are equal.
So suppose we have a number $a$ which is between 1 and $p-1$ inclusive.
Everything is being done modulo p.
We start by looking at the product $(p-1)!$ which is $F=1\text{x}{2}\text{x}\ldots\text{x}(p-1).$
This is some number which, because we are working modulo a prime, has a multiplicative inverse, $F^{-1}.$
In the integers modulo a prime, every number
from 1 to p-1 has a multiplicative inverse.
We'll use that several times. |
|
Now let's look at the product
- $P{\equiv}(1.a)\text{x}(2a)\text{x}(3a)\text{x}\ldots\text{x}((p-1)a).$
This is the magic formula that we'll evaluate in two different ways.
If $ma{\equiv}na,$ multiplying by $a^{-1}$ shows that $m=n.$
That means that each of the terms $a,~2a,~3a,~\ldots~(p-1)a$
must be different. |
|
Firstly, each of the elements here are different. The element
a has a multiplicative
inverse, so multiplying all the number from 1 up to
p-1 can be undone. That means we
can't get the same answer twice, so they're all different. That means that the product
is just the same elements multiplied in a different order, and so the product is again
F.
But let's rearrange it to bring all the a's to the front. Then
- $P{\equiv}(1.a)\text{x}(2a)\text{x}(3a)\text{x}\ldots\text{x}((p-1)a)\quad\quad$ (equation 1)
is the same as
- $P{\equiv}a^{p-1}(1)\text{x}(2)\text{x}(3)\text{x}\ldots\text{x}(p-1)\quad\quad$ (equation 2)
Equation (1) shows that $P{\equiv}F$
(because it's the same product in a different order)
Equation (2) shows that $P{\equiv}a^{p-1}F$
So $F{\equiv}a^{p-1}F$
Now we can multiply by $F^{-1}$ and we can see that $1{\equiv}a^{p-1}$
which is the result we wanted!
Local neighbourhood - D3