Partial Solutions to Homework 3 for MATH 3365E # 8By definition, there is a solution toax≡b(modm) if and only if there is an integer solution (x, y)to the equationax+my=b. By prop 4. (3C), this solution exists if and only if (a, m)|b. Thereforeif (a, m)-bthere are 0 solutions.Otherwise, there is at least one solution.Letx0be the leastsolution such that 0≤x0< m. Ifx0≥m(a,m), then we claimx00=x0-m(a,m)is also a solution. Since0≤x00< m, this would contradict the fact thatx0is the least solution in that range. We now showx00is a solution (remember thatm(a,m)is an integer):ax00≡a(x0-m(a, m))(modm)≡ax0-am(a, m)(modm)≡b-0(modm)≡b(modm)Therefore, we have a contradiction, and hence,x0<m(a,m). By prop. 5 (3C) the other solutions arex=x0+nm(a,m)for all integersn. As argued above, the least solution in our range isx0whenn= 0.Ifn= (a, m), thenx=x0+(a,m)m(a,m)=x0+m≥0 +m=m, which is not in the desired range. If0< n <(a, m), thenx=x0+nm(a,m)<m(a,m)+ ((a, m)-1)m(a,m)= (a, m)m(a,m)=mand hence isin the desired range. Therefore,n
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