Prove that the linear congruence $ax \equiv b ~(mod ~n)$ has a solution if and only if $d ~|~ b$, where $d = gcd(a, n)$. If $d ~|~ b$, then it has $d$ mutually incongruent solutions modulo $n$.
Definition: Let $a, ~b, ~n > 0$ be integers and $x$ be a variable then an equation of the form $ax \equiv b ~(mod ~n)$ is called a linear congruence, and an integer $x_{0}$ for which $a x_{0} \equiv b ~(mod ~n )$ is called a solution of an equation $ax \equiv b ~(mod ~n)$. By definition, $a ~x_{0} \equiv b ~(mod~ n)$ if and only if $n ~|~ a x_{0} - b $ i.e. $a x_{0} - b = n y_{0} $ for some integer $y_{0}$· Thus, the problem of finding all integers that will satisfy the linear congruence $a x \equiv b ~(mod ~n)$ is identical with that of obtaining all solutions of the linear Diophantine equation $ax - ny = b$. Theorem: The linear congruence $ax \equiv b ~(mod ~n)$ has a solution if and only if $d ~|~ b$, where $d = gcd(a, n)$. If $d ~|~ b$, then it has $d$ mutually incongruent solutions modulo $n$. P roof: In last para, we have observed that the given linear congruence $ax \equiv b ~(mod ~n)$ is equivalent to the linea...