By Titu Andreescu

This tough challenge booklet by means of well known US Olympiad coaches, arithmetic academics, and researchers develops a mess of problem-solving talents had to excel in mathematical contests and in mathematical examine in quantity concept. delivering suggestion and highbrow satisfaction, the issues through the booklet inspire scholars to precise their rules in writing to provide an explanation for how they conceive difficulties, what conjectures they make, and what conclusions they succeed in. employing particular thoughts and techniques, readers will collect a high-quality knowing of the elemental strategies and concepts of quantity theory.

**Read or Download 104 number theory problems : from the training of the USA IMO team PDF**

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**Additional info for 104 number theory problems : from the training of the USA IMO team**

**Example text**

A1 a2 aϕ(m) is also a reduced complete set of residue classes modulo m. One might attempt to generalize Wilson’s theorem by pairing residue classes that are inverses of each other. This approach fails, since there are residue classes other than 1 and −1 (or m − 1) that are inverses of themselves. ) For example, 62 ≡ 1 (mod 35) for m = 35. Let m be a positive integer, and let a be an integer relatively prime to m. Assume that b = na is a multiple of a; that is, n = ab is an integer. From a −1 a ≡ 1 (mod p), we have n ≡ a −1 an ≡ a −1 b (mod m).

A + m − 1} is a complete set of residue classes modulo m for any integer a. In particular, for a = 0, {0, 1, . . , m − 1} is the minimal nonnegative complete set of residue classes. Also, it is common to consider the complete set of residue classes {0, ±1, ±2, . . , ±k} for m = 2k + 1 and {0, ±1, ±2, . . , ±(k − 1), k} for m = 2k. 24. Let n be an integer. Then (1) n 2 ≡ 0 or 1 (mod 3); (2) n 2 ≡ 0 or ± 1 (mod 5); (3) n 2 ≡ 0 or 1 or 4 (mod 8); (4) n 3 ≡ 0 or ± 1 (mod 9); (5) n 4 ≡ 0 or 1 (mod 16); All the proofs can be done by checking complete sets of residue classes.

Let p be a prime with p > 5. Prove that p 8 ≡ 1 (mod 240). Proof: Note that 240 = 24 · 3 · 5. By Fermat’s little theorem, we have p 2 ≡ 1 (mod 3) and p 4 ≡ 1 (mod 5). Because a positive integer is relatively prime to 24 if and only if it is odd, ϕ(24 ) = 23 . By Euler’s theorem, we have p 8 ≡ 1 (mod 16). Therefore, p 8 ≡ 1 (mod m) for m = 3, 5, and 16, implying that p 8 ≡ 1 (mod 240). Note that this solution indicates that we can establish Euler’s theorem by Fermat’s little theorem. 24 (5)). Hence we can improve the result to p 4 ≡ 1 (mod 240) for all primes p > 5.