| Consider the values of f(p) for p<2^u.
The number of primes below 2^u is k(u)*2^u/u, where k(u)->some k>0.
Let f(p) be of degree n; choose K so f(p) is bounded by K*p^n. Then
lg f(p) <= lg K + n lg p <= lg K + nu
for p<2^u (where lg is log base 2). So there are a number of values which is
exponential in u being spread among a number of values that is linear in u as u
increases, which means that eventually this will force there to be at least n+1
distinct primes p0,...,pn such that f(p) is the same power of 2 at each pi. But
this means that f(p) must be a constant.
So the answer is: f(p) = 2^n, for some n>=0.
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| This will not be rigorous; I'm just trying to indicate how to get a proof that
"uses less."
Assume f isn't a constant, and f(p) is a power of 2 for all primes p. Then the
leading coefficient of f must be positive, and for large enough n, f(n) is
monotone increasing. Therefore, for large enough (consecutive) primes p[i] and
p[i+1], f(p[i+1])/f[p] is some power of 2 greater than 1. So, for large enough
i (say i >= I), we have f(p[I+k]) >= 2^k*f(p[I]).
Now for large enough p and q, f(p)/f(q) is very nearly (p/q)^n where n is the
degree of f; so for large enough i, p[i+1]/p[i] >= 2^(-n) - epsilon for any
epsilon we find convenient. Choose an epsilon so that 2^(-n) - epsilon > 1, and
we have that for large enough i, p[i] is exponential in i.
This implies that sum(1/p[i]) converges, which we know isn't true, so f must be
a constant.
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