New solvability conditions for congruence$ $ax \equiv b \pmod{n}$

Abstract

In \cite{2} it is proved that congruence $ax \equiv b \pmod{n}$ has a solution with $ t = gcd(x_0, n)$ if and only if $gcd (a; \frac{n}{t} )= gcd (\frac{b}{t},  \frac{n}{t} )$ thereby generalizing the result for $t = 1$ proved in \cite{1}, \cite{5}. We show that this generalized result follows from that given in \cite{1}, \cite{5}. Then we shall analyze this result from the point of view of a weaker condition that $gcd (a, \frac{n}{t}) |   gcd (\frac{b}{t}, \frac{n}{t} )$. We prove that given integers $a, b, n \geq 1$ and  $t \geq 1$, congruence $ax \equiv b \pmod{n}$ has a solution $x_0$ with $t$ dividing $gcd(x_0, n)$ if and only if $gcd (a,\frac{n}{t} | gcd (\frac{b}{t}, \frac{n}{t} )$.