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By Evangelos Kranakis (auth.)

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2: How many primes of a given length k exist? 19. 7. Let Pn be the n-th prime and let 0 < a < 1 be any real number. Let 7r(n, a) denote the number of primes in the interval (p~, Pn]. (n+t)'(I- ~-a). aPn+t In particular, for any 2 :5 k :5 n there exist arbitrarily large integers t such that (19) 4: Prove that the right hand side of inequality (18) holds, for all x > 200. Hint: Reduce to a simpler inequality and use a hand calculator. 16 Continued Fractions For any two positive real numbers a, {3, let [a,{3] = a+l/{3.

For the given primes p, q, let P = {1,2, ... ,(p-l)/2} and Q = {1,2, ... ,(q-l)/2}. It is clear that for any cEQ there exists be Q such that either pc == b modq or pc == -b modq (a similar property holds for Pl. Define n(p, q) = the number of times that pc is congruent modulo q to an integer in -Q, as c runs through Q. 2 (Gauss's Lemma) (plq) = (-l)n(p,q). Proof of the Lemma: For each cEQ, one can find Be, be such that (10) where be E Q and Be = +1 or -1. The mapping c ---+ be(c E Q, be E Q) is 1 - 1 (and hence also onto).

A n-1 = B nO'n+1 + B n-1 An I - Bn = (BnO'n+! 1 + B n-1)Bn ' It follows that 1 \An - O'Bn\ = - B ~ \A - O'B\ < \An - O'Bn\, n+1 'since Bn ~ B, and AlB is a Diophantine approximation of 0', which is a contradiction. Case 2: n is even. This is omitted because it is similar to the proof of case 1. Hence the proof of (i) is complete. 36 1 NUMBER THEORY (ii) Assume on the contrary that AlB is not a Diophantine approximation of a. This means that there exist integers C, D with D ~ Band C I D =I AlB such that the following inequality holds: (20) IA-Bal~ IC-Dal· In the proof below it will be assumed that a < AlB.

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