By Chaohua Jia, Kohji Matsumoto
Contains numerous survey articles on top numbers, divisor difficulties, and Diophantine equations, in addition to examine papers on a number of elements of analytic quantity thought difficulties.
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Extra resources for Analytic Number Theory
It is straightforward, for the most part, to handle the various integrals over the major arcs 331 that we encounter later. In order to dispose of such routines simultaneously, we prepare the scene with an exotic lemma. 1. Let s be either 1 or 2, and let k and kj (0 5 j s) be natural numbers less than 6. Suppose that w(P) is a function satisfying w(P) = Cuko(p) O(Xko(logN)-2) with o constant C, and that the function h ( a ) has the property + I(n) = + O(log L))I(n), < for 1 5 j s. 15 of Hua .
Mod n2), Theorem 3. If n is odd, then is Euler's quotient of n with base r . 4nl4 (-l)qn((&;/2) Corollary 1. If n is odd, then =4 m - dln 1 7 = 9 2 ( 4 - nq:(n)/2 i= 1 (mod n2) for 3t n (mod n2/3) for 3 1 n. Similarly as Theorem 1,we can generalize other congruences by Lehmer t o modulo arbitary positive integers. Among those, the most interesting one might be the following, Theorem 2. If n is odd, then (mod n3) for (mod n3/3) for 3 1n 3 ( n, (3) where p(n) is Mobius' function, and +(n) is Euler's function.
Consequently we must have 89 < Xk2 and integers x and y satisfying > Obviously we see ~ ( n 5) R1(n). 1. 34) valid for all even n E [N, (615)N], where for u 10. Note now that Cl (u) = 1 and C2(u) = log(u - 1) for u 2 2, by the definition. 35) that whence 2K (4)/8' < 5. 34) that R(n) - ~ ( n 2) R(n) - R1(n) > 0 for every even n E [N, (6/5) N], which establishes Theorem 4 (i) . 5. 34) that R(n) - ~ ( n > ) 0 again for every even n E [N, (615)N], which completes the proof of Theorem 4 (ii). References To examine K(k2), the following observation is useful.