104 number theory problems : from the training of the USA by Titu Andreescu

By Titu Andreescu

This hard challenge e-book by means of popular US Olympiad coaches, arithmetic lecturers, and researchers develops a mess of problem-solving talents had to excel in mathematical contests and in mathematical examine in quantity idea. providing concept and highbrow pride, the issues in the course of the publication motivate scholars to specific their rules in writing to provide an explanation for how they conceive difficulties, what conjectures they make, and what conclusions they achieve. utilising particular ideas and methods, readers will collect a pretty good realizing of the basic recommendations and ideas of quantity theory.

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The set T = aS + b = {as + b | s ∈ S} is also a complete set of residue classes modulo n. 26 104 Number Theory Problems Now we are better equipped to discuss linear congruence equations a bit further. 25. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. There exist integers x such that ax ≡ b (mod m), and all these integers form exactly one residue class modulo m. Proof: Let {c1 , c2 , . . , cm } be a complete set of residue classes modulo m. 24, {ac1 − b, ac2 − b, .

18. (a) a ≡ a (mod m) (reflexivity). (b) If a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod m) (transitivity). (c) If a ≡ b (mod m), then b ≡ a (mod m). 20 104 Number Theory Problems (d) If a ≡ b (mod m) and c ≡ d (mod m), then a + c ≡ b + d (mod m) and a − c ≡ b − d (mod m). (e) If a ≡ b (mod m), then for any integer k, ka ≡ kb (mod m). (f) If a ≡ b (mod m) and c ≡ d (mod m), then ac ≡ bd (mod m). In general, if ai ≡ bi (mod m), i = 1, . . , k, then a1 · · · ak ≡ b1 · · · bk (mod m). In particular, if a ≡ b (mod m), then for any positive integer k, a k ≡ bk (mod m).

Proof: Let d = gcd(a1 , . . , an ). If b is not divisible by d, then (∗) is not solvable, since for any integers x1 , . . , xn the left-hand side of (∗) is divisible by d and the right-hand side is not. If d | b, then we obtain the equivalent equation a1 x 1 + · · · + an x n = b , 1. Foundations of Number Theory 39 where ai = ai /d for i = 1, . . , n and b = b/d. Clearly, we have gcd(a1 , . . , an ) = 1. We use induction on the number n of the variables. In the case n = 1 the equation has the form x1 = b or −x1 = b, and thus the unique solution does not depend on any parameter.

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