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December 18th, 2016, 08:37 AM   #1
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Lightbulb 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 09:28 PM.
moussaid521 is offline  
December 18th, 2016, 09:02 PM   #2
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Nice proof.
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