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So I think this can be done, in two steps: (1) understand sums of the 1/Q(k, m) for certain integral quadratic forms Q; and (2) express this sum as a linear combination of such sums. The first part …

A possible reformulation? If the equation of the circle is rational, then one rational point implies infinitely many. One sees that via lines of rational slope through the point, or the existence of a …

The papers by Roquette http://www.rzuser.uni-heidelberg.de/~ci3/rv.pdf, http://www.rzuser.uni-heidelberg.de/~ci3/rv2.pdf, http://www.rzuser.uni-heidelberg.de/~ci3/rv3.pdf and http://www.rzuser.uni- …

There is kind of an easy proof given that Berlekamp's algorithm works? In the notation of https://en.wikipedia.org/wiki/Berlekamp%27s_algorithm, the point is that the space of polynomials g congruen …

I'd lay odds on a counterexample. This is going to be a bit vague, but I assume the reason you're looking at a coset is down to Pontryagin duality at the local field level. That it is the coset for a …

You will end up taking m as around log n/loglog n, asymptotically, by the Prime Number Theorem? So again, your motivation. For small n something less easy to describe but computable will occur. For la …

I have had this question on my mind for two decades. We know, after Heath-Brown, that one out (say) of 3, 5, 7 is a primitive root mod p for infinitely many primes p. We just don't know which one. (We …

You have a typical recursively enumerable set S of integers, and a set X of lattice points cut out by a multivariate polynomial. We are talking about S being the projection (onto one axis) of X. Given …

A short answer: the kind of "structure" recognised in the analytic number theory of the period 1900 to 1930, successful as that theory was, doesn't go far enough. You need at least functions of severa …

Hecke started on this area a century ago https://en.wikipedia.org/wiki/Hilbert's_twelfth_problem. As far as I know his ideas have not been taken much further.

From the point of view of analytic number theory, the point usually would be asymptotic behaviour. This is typically well understood, and is in the massive book of Watson. Apart from that, yes, numero …

It looks to me like the intended method is to use the loglog divergence of the sum of the reciprocals of the primes on the first sum, and then presumably the Prime Number Theorem with error term on th …

Yes. The determination of Gauss sums with characters of prime order at least 5 is an unsolved problem.

This type of question can be answered (typically) by applying (first) summation by parts, and then estimating sums of fractional parts. The sums of fractional parts can be handled by the Fourier serie …

Edited: One point is that Hodge's original version of the conjecture was wrong, and in a couple of ways. You do need rational coefficients (integral is too much to ask for, see ref below). Also a mor …