## Number Theory: Structures, Examples, and Problems by Titu Andreescu, Dorin Andrica

By Titu Andreescu, Dorin Andrica

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

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For positive integers s and t, the minimal element of the set Ms ∩ Mt is called the least common multiple of s and t and is denoted by lcm(s, t). The following properties are easily obtained from the definition above: (1 ) If m = lcm(s, t), m = ss = tt , then gcd(s , t ) = 1. (2 ) If m is a common multiple of s and t and m = ss = tt , gcd(s , t ) = 1, then m = m. (3 ) If m is a common multiple of s and t, then m | m . β β (4 ) If s = p1α1 · · · pkαk and t = p1 1 · · · pk k , αi , bi ≥ 0, i = 1, .

9. Find all positive integers n, n ≥ 1, such that n 2 + 3n is a perfect square. Solution. Let m be a positive integer such that m 2 = n 2 + 3n . Since (m −n)(m +n) = 3n , there is k ≥ 0 such that m −n = 3k and m +n = 3n−k . From m − n < m + n follows k < n − k, and so n − 2k ≥ 1. If n − 2k = 1, then 2n = (m + n) − (m − n) = 3n−k − 3k = 3k (3n−2k − 1) = 3k (31 − 1) = 2 · 3k , so n = 3k = 2k + 1. We have 3m = (1 + 2)m = 1 + 2m + 22 m2 + · · · > 2m + 1. Therefore k = 0 or k = 1, and consequently n = 1 or n = 3.

Pn are distinct primes, and α1 , . . , αn , β1 , . . , βn , γ1 , . . , γn are nonnegative integers. Then n gcd(a, b) gcd(b, c) gcd(c, a) = gcd(a, b, c)2 min{αi ,βi } n min{βi ,γi } pi i=1 n pi i=1 n min{γi ,αi } pi i=1 2 min{αi ,βi ,γi } pi i=1 n min{αi ,βi }+min{βi ,γi }+min{γi ,αi }−2 min{αi ,βi ,γi } = pi i=1 and n lcm(a, b) lcm(b, c) lcm(c, a) = lcm(a, b, c)2 max{αi ,βi } n i=1 n max{βi ,γi } pi pi i=1 n i=1 2 max{αi ,βi ,γi } pi i=1 n = max{αi ,βi }+max{βi ,γi }+max{γi ,αi }−2 max{αi ,βi ,γi } pi i=1 max{γi ,αi } pi .