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Old 2017-02-22, 08:37   #2
Batalov's Avatar
Mar 2008

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You seem to be describing P(x) = ((x+1)p-xp-1)/p
(only you have "+1/p" still dangling at the end. I will subtract it for symmetry).

After binomial expansion: P(x) = \sum_{k=1}^{p-1} {{C^k_p \over p} x^k}
It is obvious that for p>=7, the middle terms will have a factor precprime(p); it follows from
{C^k_p \over p} = {(p-1)(p-2)...({{p+1} \over 2}) \over { k ! }},
where k = (p-1)/2. This is because precprime(p) will be present in the denominator, but not in the numerator.

Note that your own observation that "the two most inner terms are divisible by Q(p-1)" is not holding for p<7 (e.g. for p=5, "the two most inner terms" are not divisible by 3), and the reason for this is obvious. For p>=7, it will hold, - as indeed shown above.
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