# Elementary number theory: Instructor's notes by Clark W.E. PDF By Clark W.E.

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It follows as a consequence √ of our next result that the proportion of positive integers n having a prime factor > n is log 2. √ • Theorem 38. The number of positive integers n ≤ x having a prime factor > n x is (log 2)x + O . log x 47 • Proof. The desired quantity is 1= √ n≤x n

In each case, give a proof or a counterexample. (a) f (x) ∼ g(x) (b) f (x) = g(x) + O(1) (c) f (x) − g(x) ≍ 1 (d) f (x) = g(x) + o(g(x)) 1 (2) (a) Prove that ≤ 1 + log x for all x ≥ 1. k k≤x 33 1 ∼ log x. k k≤x 1 (c) Prove that = log x + O(1). k (b) Prove that k≤x (3) (a) How many positive integers ≤ 210 are not divisible by each of the primes 2, 3, 5, and 7? For example, 11 would be such an integer but 39 would not be. (b) Let A(x) = |{n ≤ x : each of 2, 3, 5, and 7 does not divide n}|. Prove that A(x) ∼ cx for some constant c and determine the value of c.

Since f (x) ≡ (x − 3)(x2 + 2) (mod 5) and = −1, the only 5 solutions of f (x) ≡ 0 (mod 5) are 3 modulo 5. Since f ′(3) ≡ 41 ≡ 1 ≡ 0 (mod 5) and f (3) = 45, we obtain from (iii) that the all solutions to f (x) ≡ 0 (mod 25) are congruent to 3 + 5(−9) ≡ 8 modulo 25. One checks directly that the incongruent solutions modulo 7 to f (x) ≡ 0 (mod 7) are 2, 4, and 6. It follows that there are exactly three incongruent solutions modulo 175, say x1 , x2 , and x3 , satisfying x1 ≡ 8 (mod 25), x1 ≡ 2 (mod 7), x2 ≡ 8 (mod 25), x2 ≡ 4 (mod 7), x3 ≡ 8 (mod 25) x3 ≡ 6 (mod 7).