By Leo Moser

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**Extra resources for An Introduction to the Theory of Numbers**

**Example text**

P pep we find that n log n > log n! = ep (n) log p > n(log n − 1). The reader may justify that the error introduced in replacing ep (n) by course ep (n) = n pi ) is small enough that p≤n n log p = n log n + O(n) p or p≤n log p = log n + O(1). p We can now prove Theorem 4. R(x) = p≤x 1 = log log x + O(1). p In fact x R(x) = S(n) − S(n − 1) log n n=2 x = S(n) n=2 x 1 1 − log n log(n + 1) (log n + O(1)) = n=2 x = + O(1) log(1 + n1 ) + O(1) (log n) log(n + 1) 1 + O(1) n log n n=2 = log log x + O(1), as desired.

Ak and no others, we call this set compatible (mod n). Let the number of compatible sets (mod n) be denoted by C(n). Since the number of subsets of the set consisting of 0, 1, 2, . . , n − 1 is 2n , we call c(n) = C(n) 2n the coefficient of compatibility of n. If n = p is a prime then the congruence (x − a1 )(x − a2 ) · · · (x − ak ) ≡ 0 (mod n) has precisely the roots a1 , a2 , . . , an . Hence c(p) = 1. In a recent paper M. M. Chokjnsacka Pnieswaka has shown that c(4) = 1 while c(n) < 1 for n = 46 Chapter 5.

Hence c(p) = 1. In a recent paper M. M. Chokjnsacka Pnieswaka has shown that c(4) = 1 while c(n) < 1 for n = 46 Chapter 5. Congruences 6, 8, 9, 10. We shall prove that c(n) < 1 for every composite n = 4. , 1 (c(1) + c(2) + · · · + c(n)) = 0. n→∞ n Since c(n) = 1 for n = 1 and n = p we consider only the case where n is composite. Suppose then that the unique prime factorization of n is given by 1 α2 pα 1 p2 · · · with α2 1 pα 1 > p2 > · · · lim Consider separately the cases (1) n has more than one prime divisor, and (2) n = pα , α > 1.