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December 18th, 2016, 09:37 AM  #1 
Newbie Joined: Dec 2016 From: Saudi Arabia Posts: 5 Thanks: 0  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: (ki)^p = (ki1+1)^p=(k(i+1))^p +1 +sum_of_binomial coefficients (that have p as divisor as stated above) we'll have then (ki)^p  (k(i+1))^p = 1 [p] by summing up this formula from i=0 to i =k1, 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. 
December 18th, 2016, 10:02 PM  #2 
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 
Nice proof.


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