By Edward Burger
2 DVD set with 24 lectures half-hour every 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 learn -- lecture 2. common 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 conception of best numbers -- lecture eight. Euler's product formulation and divisibility -- lecture nine. The leading 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 outline of the high-level ideas. subsequent, he offers a step by step clarification of the formulation and calculations that lay on the center of every conundrum. via transparent factors, exciting anecdotes, and enlightening demonstrations, Professor Burger makes this interesting box of analysis obtainable for an individual who appreciates the attention-grabbing nature of numbers. -- writer.
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Additional resources for An Introduction to Number Theory (Guidebook, parts 1,2)
6), we obtain β∈Z∗n β . 6) β∈Z∗n β ∈ Z∗n from the left- and right-hand αφ(n) = n . That proves the ﬁrst statement of the theorem. The second follows from the observation made above that αi = n if and only if the multiplicative order of α divides i. 16 (Fermat’s little theorem). For any prime p, and any integer a ≡ 0 (mod p), we have ap−1 ≡ 1 (mod p). Moreover, for any integer a, we have ap ≡ a (mod p). Proof. 15, and the fact that φ(p) = p − 1. The second statement is clearly true if a ≡ 0 (mod p), and if a ≡ 0 (mod p), we simply multiply both sides of the congruence ap−1 ≡ 1 (mod p) by a.
Therefore, it suﬃces to consider the problem of determining the solutions z to congruences of the form az ≡ b (mod n), for given integers a, b, n. 6. Let a, b, n ∈ Z with n > 0. If a is relatively prime to n, then the congruence az ≡ b (mod n) has a solution z; moreover, any integer z is a solution if and only if z ≡ z (mod n). Proof. The integer z := ba , where a is a multiplicative inverse of a modulo n, is clearly a solution. 5 holds if and only if z ≡ z (mod n). ✷ Suppose that a, b, n ∈ Z with n > 0, a = 0, and gcd(a, n) = 1.
Show that a0 + a1 x + · · · + ak xk ≡ a0 + a1 y + · · · + ak y k (mod n). 2. Let a, b, n, n ∈ Z with n > 0 and n | n. Show that if a ≡ b (mod n), then a ≡ b (mod n ). 3. Let a, b, n, n ∈ Z with n > 0, n > 0, and gcd(n, n ) = 1. Show that if a ≡ b (mod n) and a ≡ b (mod n ), then a ≡ b (mod nn ). 4. Let a, b, n ∈ Z such that n > 0 and a ≡ b (mod n). Show that gcd(a, n) = gcd(b, n). 5. Prove that for any prime p and integer x, if x2 ≡ 1 (mod p) then x ≡ 1 (mod p) or x ≡ −1 (mod p). 6. Let a be a positive integer whose base-10 representation is a = (ak−1 · · · a1 a0 )10 .