# An Introduction to Number Theory (Guidebook, parts 1,2) by Edward Burger

By Edward Burger

2 DVD set with 24 lectures half-hour each one for a complete of 720 minutes...Performers: Taught through: Professor Edward B. Burger, Williams College.Annotation Lectures 1-12 of 24."Course No. 1495"Lecture 1. quantity conception and mathematical study -- lecture 2. normal numbers and their personalities -- lecture three. Triangular numbers and their progressions -- lecture four. Geometric progressions, exponential progress -- lecture five. Recurrence sequences -- lecture 6. The Binet formulation and towers of Hanoi -- lecture 7. The classical thought of major numbers -- lecture eight. Euler's product formulation and divisibility -- lecture nine. The best quantity theorem and Riemann -- lecture 10. department set of rules and modular mathematics -- lecture eleven. Cryptography and Fermat's little theorem -- lecture 12. The RSA encryption scheme.Summary Professor Burger starts off with an summary of the high-level innovations. subsequent, he presents a step by step clarification of the formulation and calculations that lay on the center of every conundrum. via transparent motives, interesting anecdotes, and enlightening demonstrations, Professor Burger makes this exciting box of analysis available for someone who appreciates the attention-grabbing nature of numbers. -- writer.

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**Extra info for An Introduction to Number Theory (Guidebook, parts 1,2)**

**Sample text**

This can easily be seen by considering the exponent a of any prime p which occurs in the prime-power factorization of (m*, (mi, m2)). Let the exponent of p in the factorization of my be Pj, for j = 1, 2 , , i. Then p occurs in (mi, m2) with exponent max (Pi, P2) r so tbat a = min (ft, max (ft, f t ) ) = max (min (Pi, Pi), min (ft, f t )). But our assumption is that pmSn<* ’*)| (ci - Ci) and Jf**<**)\ (C2 _ Ci)j and since p&l\(cx - f) and p^2\(c2 - f ) we see, by writing Cl Ci = (ci f) (f Ci), C2 Ci = (c2 f) (f Ci), that pBdnWb*>| f e - / ) and - /), so that also pa\(ci — / ) .

Problem s 1. In the notation of Problem 2, Section 2-3, show that

As a consequence of Theorem 3-12, we have the following im portant result. 3-13 (Chinese Remainder Theorem). Every system of linear congruences in which the moduli are relatively prime in pairs is solvable, the solution being unique modulo the product of the moduli. T heorem 36 CONGRUENCES [CHAP. 3 PROBLEM S 1. Solve the congruence 6s + Iby = 9 (mod 18). 2. Solve simultaneously: x = 1 (mod 2), 2 = 1 (mod 3), x = S (mod 4), x = 4 (mod 5). 3. Suppose that the system of congruences x = di (mod m*), i = 1, 2, .