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 2017-02-22, 08:37 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×4,591 Posts 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.