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5.2 : Linear Congruences Revisted

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    The following theorem we stated in Chapter 4:

    Theorem \(\PageIndex{1})

    Let  \(m \in \mathbb{Z}_+\) and  \(a,b \in \mathbb{Z}\), with \(a\) be non-zero. Then

    \(ax \equiv b(mod\, m)\) has a solution if and only if \(\gcd(a,m) |b\). Moreover, if \(x=x_0\) is a particular solution to this congruence, then the set of all solutions is \(\{x \in \mathbb{Z} | x \equiv x_0(mod\, m)\}=\{x_0, x_0+\dfrac{m}{d}, x_0+\dfrac{2m}{d},\ldots, x_0+\dfrac{(d-1)m}{d}\}\), where \(d=\gcd(a,m).\)

    Let us learn how to use the theorem to solve linear congruence. Let \(m \in \mathbb{Z}_+\) and  \(a,b \in \mathbb{Z}\), with \(a\) be non-zero. Consider the  linear congruence \(ax \equiv b(mod\, m)\). Which is equivalent to \( m \mid ax- b \implies ax-b=my,\mbox{ for some} y \in \mathbb{Z} .\) Thus solving linear congruence is equivalent to solving the linear Diophantine equation \(ax+my=b\), for \(x, y \in \mathbb{Z}.\)

    Example \(\PageIndex{1}\)

    If possible,  solve \(2x \equiv 2 ( mod \,4 ) \).


    Since \(\gcd(2,4)=2\) and \(2\mid 2\), \(2x \equiv 2 ( mod \,4 ) \)  has a solution. Moreover, the set of all solutions is given by \(\{x\in \mathbb{Z}|x\equiv x_0 ( mod \,4 )\}\) . To find a particular solution we consider the corresponding Diophantine equation \(2x+4y=2.\) Since \(2(-1)+4(1)=2,\) one particular solution is given by \(x_0=-1.\)  Then all the solutions to the Diophantine equation \(2x+4y=2\) are of the form \(x=-1+2k, y=1-4k, k\in \mathbb{Z}.\)

    Thus \(x ( mod\,4)=-1,1\).

    Since \( (-1)(mod \,4)\equiv 3\) and \(1 (mod \,4)\equiv 1\), the set of all solutions are all integers  \(x\) such that \( x \equiv 3 (mod \,4)\) or \( x \equiv 1 (mod \,4)\).


    This page titled 5.2 : Linear Congruences Revisted is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

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