Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to solve $ax \equiv b \mod n$ given a solution $x_0$. How can i prove that there are exactly $(a,n)$ solutions ?

share|improve this question
add comment

2 Answers

Suppose $x$ is a soloution, then $a(x-x_0)\equiv 0$ mod $n$. Hence $a(x-x_0)=nk$ for some integer $k$, that is $x=x_0+\frac{nk}{a}$. Suppose $d=(a,n)$ and $a=db,n=dm,(b,m)=1$. Then you have $$x=x_0+\frac{dmk}{db}=x_0+\frac{mk}{b}$$ This implies $b|mk$ but $(b,m)=1$, hence $b|k$. So one can write $$x=x_o+\frac{nk}{db}=x_0+\frac{n(k/b)}{d}$$ So every soloution is of the form $x_0+\frac{nt}{d}$ and you can easily check that $x_0+\frac{nt}{d}$ is indeed a soloution for all integer $t$. Now observe that for $t\geq d$ the same soloutions for $t=0,\ldots d-1 $ are repeated. Now to prove that these are the only soloutions assume $$x_0-\frac{nt_1}{d}=x_0+\frac{nt_2}{d}$$ where $0\leq t_1<t_2\leq d-1$, then we have$$\frac{nt_1}{d}\equiv \frac{nt_2}{d}\mbox { mod } n$$ Now $(n/d,n)=n/d$ and we have $$t_1\equiv t_2\mbox{ mod } d$$ but this is a contradiction to $0<t_2-t_1<d$. Hence there are exactly $d$ soloutions.

share|improve this answer
add comment

Hint $\ $ Let $\rm\:c = x_0.\:$ Working $\rm\:mod\ n,\:$ if $\rm\:ac \equiv b\:$ then $\rm\:ax\equiv b\iff ax\equiv ac\iff n\mid a(x - c).\:$ Denote the gcd $\rm\:d = (n,a) = d(\bar n,\bar a)\:$ for $\rm\:\bar n = n/d,\ \bar a = a/d,\:$ and $\rm\:\color{#0A0}{(\bar n,\bar a) = 1}.\:$ Then, canceling $\rm\,d,$

$$\rm n\mid a(x-c)\iff \color{#0A0}{\bar n\mid \bar a}\, (x-c)\iff \bar n\mid x-c\ \ \ by\ \ \color{#0A0}{ Euclid's\ \ Lemma}$$

This solution $\rm\:x \equiv c\pmod{\bar n}\:$ splits into precisely $\rm\:\color{#C00}d\:$ classes mod $\rm\: \color{#C00}d\bar n = n,\:$ namely

$$\rm x\equiv c\,\ (mod\ \bar n)\iff x\,\equiv\, c,\,\ c\! + \bar n,\,\ c\! + 2\bar n,\,\ c\! + 3\bar n,\, \ldots,\, c\! + (\color{#C00}{d}\!-\!1)\bar n\ \ (mod\ {\color{#C00}d\bar n})$$

since, by the $\rm\color{blue}{Division}$ Algorithm, $\rm\: c + \color{blue}{j}\,\bar n\, =\, c + (\color{blue}{r\! +\! qd})\,\bar n \,\equiv\, c + r\,\bar n\, \ (mod\ \color{#C00}d\bar n)\ $ for $\rm\: r\in [0,\color{#C00}{d}\!-\!1],\ $ and this interval contains precisely $\rm\:\color{#C00}d\:$ integers, namely $\rm\:0,1,2,\, \ldots,\, \color{#C00}{d}\!-\!1.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.