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On elementary proofs of the Prime Number Theorem for arithmetic progressions, without characters.
, 1993
"... : We consider what one can prove about the distribution of prime numbers in arithmetic progressions, using only Selberg's formula. In particular, for any given positive integer q, we prove that either the Prime Number Theorem for arithmetic progressions, modulo q, does hold, or that there exists a s ..."
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: We consider what one can prove about the distribution of prime numbers in arithmetic progressions, using only Selberg's formula. In particular, for any given positive integer q, we prove that either the Prime Number Theorem for arithmetic progressions, modulo q, does hold, or that there exists a subgroup H of the reduced residue system, modulo q, which contains the squares, such that `(x; q; a) ¸ 2x=OE(q) for each a 62 H and `(x; q; a) = o(x=OE(q)), otherwise. From here, we deduce that if the second case holds at all, then it holds only for the multiples of some fixed integer q 0 ? 1. Actually, even if the Prime Number Theorem for arithmetic progressions, modulo q, does hold, these methods allow us to deduce the behaviour of a possible `Siegel zero' from Selberg's formula. We also propose a new method for determining explicit upper and lower bounds on `(x; q; a), which uses only elementary number theoretic computations. 1. Introduction. Define `(x) = P px log p, where p only denot...
The Möbius function and the residue theorem
 Journal of Number Theory
"... A classical conjecture of Bouniakowsky says that a nonconstant irreducible polynomial in Z[T] has infinitely many prime values unless there is a local obstruction. Replacing Z[T] with κ[u][T], where κ is a finite field, the obvious analogue of Bouniakowsky’s conjecture is false. All known counterex ..."
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A classical conjecture of Bouniakowsky says that a nonconstant irreducible polynomial in Z[T] has infinitely many prime values unless there is a local obstruction. Replacing Z[T] with κ[u][T], where κ is a finite field, the obvious analogue of Bouniakowsky’s conjecture is false. All known counterexamples can be explained by a new obstruction, and this obstruction can be used to fix the conjecture. The situation is more subtle in characteristic 2 than in odd characteristic. Here we illustrate the general theory for characteristic 2 in some examples.