By Titu Andreescu, Dorin Andrica

This introductory textbook takes a problem-solving method of quantity concept, situating every one notion in the framework of an instance or an issue for fixing. beginning with the necessities, the textual content covers divisibility, distinct factorization, modular mathematics and the chinese language the rest Theorem, Diophantine equations, binomial coefficients, Fermat and Mersenne primes and different certain numbers, and certain sequences. integrated are sections on mathematical induction and the pigeonhole precept, in addition to a dialogue of alternative quantity platforms. through emphasizing examples and purposes the authors encourage and interact readers.

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**Extra resources for Number Theory: Structures, Examples, and Problems**

**Example text**

Note that 2s divides 2n = ab + 1 and a + 1, so that ab ≡ a ≡ −1 (mod 2s ). Hence, b ≡ 1 (mod 2s ), or 2s | b − 1, so that s ≤ t. Similarly, ab ≡ −b ≡ −1 (mod 2t ), so a ≡ −1 (mod 2t ), and 2t | a + 1. Thus, t ≤ s. Therefore, s = t, the highest power of 2 dividing (a + 1)(b − 1) is 2s, and ab − (a − b) − 1 = k · 22s for some odd k. 3. Find all nonnegative integers m such that (22m+1 )2 +1 is divisible by at most two different primes. (2002 Baltic Mathematics Competition) Solution. We claim m = 0, 1, 2 are the only such integers.

Find all integers n > 1 such that any prime divisor of n 6 − 1 is a divisor of (n 3 − 1)(n 2 − 1). 8. Let f (n) be the number of permutations a1 , . . , an of the integers 1, . . , n such that (i) a1 = 1; (ii) |ai − ai+1 | ≤ 2, i = 1, . . , n − 1. Determine whether f (1996) is divisible by 3. 9. For natural numbers m, n, show that 2n −1 is divisible by (2m −1)2 if and only if n is divisible by m(2m − 1). 10. Suppose that n is a positive integer and let d1 < d2 < d3 < d4 be the four smallest positive integer divisors of n.

A positive integer n is called m-good if for every positive integer a relatively prime to n, one has n | a m − 1. Show that every m-good number is at most 4m(2m − 1). (2004 Romanian International Mathematical Olympiad Team Selection Test) Solution. If m is odd, then n | (n − 1)m − 1 implies n | 2; hence n ≤ 2. Take now m = 2t q, t ≥ 1, q odd. If n = 2u (2v + 1) is m-good, then (2v + 1) | (2v − 1)m − 1; hence (2v + 1) | 2m − 1. Also, if a = 8v + 5, then gcd(a, n) = 1, so t t−1 2u | (a q )2 − 1 = (a q − 1)(a q + 1)(a 2q + 1) · · · (a 2 q + 1).