sol2 - Math 465 Solution Set 2 1. If (a, c) = 1 and (b, c)...

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Math 465 Solution Set 2 Ae Ja Yee 1. If ( a, c ) = 1 and ( b, c ) = 1, then ( ab, c ) = 1. Solution. Method 1: Suppose ( ab, c ) > 1, namely x | ab and x | c for some x > 1. Since ( a, c ) = 1, x cannot divide a , that is ( x, a ) = 1. However, x | ab , so x | b , which is a contradiction. Method 2: We can write 1 = ax 1 + cy 1 = bx 2 + cx 2 . Thus ax 1 bx 2 = (1 - cy 1 )(1 - cy 2 ) = 1 - c ( y 1 + y 2 - cy 1 y 2 ) , which is equivalent to 1 = abx + cy, where x = x 1 x 2 and y = y 1 + y 2 - cy 1 y 2 . Therefore, ( ab, c ) = 1. 2 2. Show that if ( a, b ) = 1, then ( a - b, a + b ) = 1 or 2. When is the value 2? Solution. Method 1: Let n = a - b , m = a + b , and ( n, m ) = d . Then 2 a = m + n, 2 b = m - n Since ( n, m ) = d , d | 2 a and d | 2 b , that is d is a common divisor of 2 a and 2 b . We now compute (2 a, 2 b ): (2 a, 2 b ) = 2( a, b ) = 2 since ( a, b ) = 1. Therefore, d | 2, that is d = 1 or 2. Method 2: Since gcd is the least positive linear combination, (
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sol2 - Math 465 Solution Set 2 1. If (a, c) = 1 and (b, c)...

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