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mat336_hw3_solns

# mat336_hw3_solns - Partial Solutions to Homework 3 for MATH...

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Partial Solutions to Homework 3 for MATH 336 5E # 8 By definition, there is a solution to ax b (mod m ) if and only if there is an integer solution ( x, y ) to the equation ax + my = b . By prop 4. (3C), this solution exists if and only if ( a, m ) | b . Therefore if ( a, m ) - b there are 0 solutions. Otherwise, there is at least one solution. Let x 0 be the least solution such that 0 x 0 < m . If x 0 m ( a,m ) , then we claim x 00 = x 0 - m ( a,m ) is also a solution. Since 0 x 00 < m , this would contradict the fact that x 0 is the least solution in that range. We now show x 00 is a solution (remember that m ( a,m ) is an integer): ax 00 a ( x 0 - m ( a, m ) ) (mod m ) ax 0 - am ( a, m ) (mod m ) b - 0 (mod m ) b (mod m ) Therefore, we have a contradiction, and hence, x 0 < m ( a,m ) . By prop. 5 (3C) the other solutions are x = x 0 + n m ( a,m ) for all integers n . As argued above, the least solution in our range is x 0 when n = 0. If n = ( a, m ), then x = x 0 + ( a,m ) m ( a,m ) = x 0 + m 0 + m = m , which is not in the desired range. If 0 < n < ( a, m ), then x = x 0 + n m ( a,m ) < m ( a,m ) + (( a, m ) - 1) m ( a,m ) = ( a, m ) m ( a,m ) = m and hence is in the desired range. Therefore, n
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