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1.55 (i) Find d = gcd(12327, 2409), ﬁnd integers s and t with d = 12327s +
2409t , and put the fraction 2409/12327 in lowest terms.
Solution. One uses the Euclidean algorithm to get: (12327, 2409) =
3 and 3 = 12327 · 299 − 2409 · 1530; the fraction 2409/12327 =
803/4109 is in lowest terms. (ii) Find d = gcd(7563, 526), and express d as a linear combination
of 7563 and 526.
Solution. The Euclidean algorithm gives
(7563, 526) = 1 and 1 = 532 − 526 − 37 − 7563.
(iii) Find d = gcd(73122, 7404621) and express d as a linear combination of 73122 and 7404621.
Solution. Here are the equations of the Euclidean algorithm:
7404621 = 101 · 73122 + 19299
73122 = 3 · 19299 + 15225
19299 = 1 · 15225 + 4074
15225 = 3 · 4074 + 3003
4074 = 1 · 3003 + 1071
3003 = 2 · 1071 + 861
1071 = 1 · 861 + 210
861 = 4 · 210 + 21
210 = 10 · 21.
We conclude that the gcd is 21. Following the algorithm in the
text, we ﬁnd that
21 = 34531 · 73122 − 341 · 7404621.
1.56 Let a and b be integers, and let sa + tb = 1 for s , t in Z. Prove that a and
b are relatively prime.
Solution. If sa + tb = 1, then any common divisor of a and b must divide
1; hence, a and b are relatively prime.
1.57 If d = (a , b), prove that a /d and b/d are relatively prime.
Solution. Absent.
1.58 Prove that if (r, m ) = 1 = (r , m ), then (rr , m ) = 1.
Solution. Since (r, m ) = 1, we have ar + bm = 1; since (r , m ) = 1, we
have sr + tm = 1. Multiplying,
1 = (ar + bm )(sr + tm ) = (as )rr + (ar t + bsr + btm )m . ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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