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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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).

Ask a homework question - tutors are online