inductive proof of Fermat’s little theorem proof
We will show
with prime. The equivalent statement
when does not divide follows by cancelling both sides (which can be done since then are coprime).
When , we have
Let’s examine . By the binomial theorem, we have
However, note that the entire bracketed term is divisible by , since each element of it is divsible by . Hence
It is easy to show that it also holds for whenever it holds for , so the statement works for all integers .
|Title||inductive proof of Fermat’s little theorem proof|
|Date of creation||2013-03-22 11:47:46|
|Last modified on||2013-03-22 11:47:46|
|Last modified by||mathcam (2727)|