|December 18th, 2016, 09:37 AM||#1|
Joined: Dec 2016
From: Saudi Arabia
New proof for Fermat's little theorem
Again, I don't really know whether this proof is known or not but I'm trying the chance again hhh.
Ok so Fermat's little theorem says: a^p = a [p], for p prime and a any integer.
We will use the following property: if p is prime, p is a divisor of the C(p,j) for j>0 and j<p.
with this identity for every k and i integers:
(k-i)^p = (k-i-1+1)^p=(k-(i+1))^p +1 +sum_of_binomial coefficients (that have p as divisor as stated above)
we'll have then (k-i)^p - (k-(i+1))^p = 1 [p]
by summing up this formula from i=0 to i =k-1, we will have:
k^p = k [p]
I hope it's not rubbish writing.
Last edited by skipjack; December 18th, 2016 at 10:28 PM.
|fermat, proof, theorem|
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