The reason for the sequence is constant when y m = sup ( A ) for some m .

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Chapter 7.5, Problem 3E

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To find: The reason for the sequence is constant when ym=sup(A) for some m .

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Use the Euclidean algorithm to find two sets of integers (a, b, c) such that 55a65b+143c: Solution = 1. By the Euclidean algorithm, we have: 143 = 2.65 + 13 and 65 = 5.13, so 13 = 143 – 2.65. - Also, 55 = 4.13+3, 13 = 4.3 + 1 and 3 = 3.1, so 1 = 13 — 4.3 = 13 — 4(55 – 4.13) = 17.13 – 4.55. Combining these, we have: 1 = 17(143 – 2.65) - 4.55 = −4.55 - 34.65 + 17.143, so we can take a = − −4, b = −34, c = 17. By carrying out the division algorithm in other ways, we obtain different solutions, such as 19.55 23.65 +7.143, so a = = 9, b -23, c = 7. = = how ? come [Note that 13.55 + 11.65 - 10.143 0, so we can obtain new solutions by adding multiples of this equation, or similar equations.]

- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?