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Unformatted text preview: 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 be the least solution such that 0 x < m . If x m ( a,m ) , then we claim x 00 = x m ( a,m ) is also a solution. Since x 00 < m , this would contradict the fact that x 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 m ( a,m ) ) (mod m ) ax am ( a,m ) (mod m ) b 0 (mod m ) b (mod m ) Therefore, we have a contradiction, and hence, x < m ( a,m ) . By prop. 5 (3C) the other solutions are x = x + n m ( a,m ) for all integers n . As argued above, the least solution in our range is x when n = 0....
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This note was uploaded on 06/21/2008 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 BILLERA
 Math, Algebra

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