math 135 - 0. Since ab < 0, either a...

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Math 135, Winter 2009, Bonus Question Solutions 1 Page 1 Question 1: Show that gcd( ab, c ) =gcd( b, c ) if gcd( a, c ) = 1. Is it true in general that gcd( ab, c ) = gcd( a, c ) · gcd( b, c )? Solution: Let d =gcd( b, c ). Then d | b, d | c implying d | ab and d | c . Hence d is a common divisor of ab and c . Since gcd( a, c ) = 1, there are integers x, y such that ax + cy = 1. Hence abx + bcy = b . Let e | ab and e | c . Then e | ( abx + bcy ) = b . Thus e | b and e | c . Since d =gcd( b, c ), we obtain e d . Therefore e | ab, e | c implies that e d . Since d is a common divisor of ab and c , we get gcd( ab, c ) = d =gcd( b, c ). In general, gcd( ab, c ) =gcd( a, c ) · gcd( b, c ) is not true. Let a = 4 , b = 6 , c = 2. Then gcd( ab, c ) = 2 but gcd( a, c ) = 2 and gcd( b, c ) = 2. ± Question 2: For what values of a and b , does the equation ax + by = c have an infinite number of positive solutions x, y ? In the board, I wrote ’ what values of c ’ in place of ’ what values of a and b ’. Solution: Let c be given. Let a, b Z be such that d =gcd( a, b ) | c and ab < 0. Then ax + by = c has a solution x = x 0 , y = y 0 and all the other solutions are given by x = x 0 + n b d , y = y 0 - n a d , n Z . Want infinitely n for which x > 0 and y >
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Unformatted text preview: 0. Since ab &lt; 0, either a &gt; , b &lt; 0 or a &lt; , b &gt; 0. Recall that d &gt; 0. Suppose a &lt; , b &gt; 0. Then-a d &gt; , b d &gt; 0 and hence x = n b d + x &gt; 0 and y = n (-a d ) + y &gt; 0 if n &gt;-x b d and n &gt;-y-a d = y a d . This is true for all n &gt; maximum(-x b d , y a d ) giving innite number of positive solutions. Suppose a &gt; , b &lt; 0. Then a d &gt; ,-b d &gt; 0 and hence x = (-n )(-b d )+ x &gt; 0 and y = (-n )( a d )+ y &gt; 0 if-n &gt;-x-b d = x b d and-n &gt;-y a d . or if n &lt;-x b d and n &lt;-y a d . This is true for all n &lt; minimum(-x b d ,-y a d ) giving innite number of positive solutions. Therefore whenever a, b Z , ab &lt; 0 and gcd( a, b ) | c , we get innite number of positive solutions. Remark: Whenever ab &gt; , we cannot get an innite number of positive solutions for any c ....
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