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By Michael Rosen, Kenneth Ireland

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This well-developed, obtainable textual content information the old improvement of the topic all through. It additionally presents wide-ranging insurance of vital effects with relatively ordinary proofs, a few of them new. This moment variation includes new chapters that supply an entire evidence of the Mordel-Weil theorem for elliptic curves over the rational numbers and an outline of contemporary growth at the mathematics of elliptic curves.

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Extra info for A Classical Introduction to Modern Number Theory (2nd Edition) (Graduate Texts in Mathematics, Volume 84)

Example text

B) a b (m) implies that b a (m). (c) If a = b (m) and b = C (m), then a = = = c (m). PROOF. (a) a - a = 0 and miO. (b) If mlb - a, then mla - b. (c) If mlb - a and mle - b, then mle - a = (c - b) + (b - a). 1 shows that congruence modulo m is an equivalence relation on the set of integers . If a E 7L, let adenote the set of integers congruent to a modulo m. a = {n E 7L In a (m) }. In other words a is the set of integers of the form a + km. If m = 2, then 0 is the set of even integers and T is the set of odd integers.

If m = 2, then 0 is the set of even integers and T is the set of odd integers. = Definition. A set of the form a is called a congruence class modulo m. 2. (a) (b) a = 5 iff a == b (m). a#-5 iffa n 5 is empty. (c) There are precisely m distinct congruence classes modulo m. PROOF. 5 = a, then a E a = 5. Thus a == b (m). Conversely, if a == b (m), then a E 5. If c == a (m), then c == b (m), which shows a ~ 5. Since a == b (m) implies that b == a (m), we also have 5 ~ a. Therefore a = 5. (a) If (b) Clearly, if an 5 is empty, then a#- 5.

Therefore a = 5. (a) If (b) Clearly, if an 5 is empty, then a#- 5. We shall show that an 5 not empty implies that a = 5. Let c E an 5. Then c == a (m) and c == b (m). It follows that a == b (m) and so by part (a) we have a = 5. (c) We shall show that 0, T, 2, ... , m - 1 are all distinct and are a complete set of congruence classes modulo m. Suppose that 0 ~ k < I < m. J( = T implies that k == I (m) or that m divides I - k. Since 0 < I - k < m this is a contradiction. Therefore J( #- T. Now let a E 71..

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