Pairs of Linear Congruences

Since linear congruences of the form axb $\left(\vphantom{ \limfunc{mod}%
m}\right.$$\limfunc$modm$\left.\vphantom{ \limfunc{mod}%
m}\right)$ can be reduced to simple congruences of the form xc $\left(\vphantom{
\limfunc{mod}m}\right.$$\limfunc$modm$\left.\vphantom{
\limfunc{mod}m}\right)$, we consider systems of congruences in this latter form.



\begin{example}
Consider the system of two congruences
\begin{eqnarray*}
x &...
...king, $14265\limfunc{mod}237=45$\ and $14265\limfunc{mod}419=19$.
\end{example}


\begin{example}
The complete set of solutions is given by
\begin{displaymath}...
...equiv 14265\,\left( \limfunc{mod}99303\right)
\end{displaymath}
\end{example}

In general, if m and n are relatively prime, then a solution to the pair

x a $\displaystyle \left(\vphantom{ \limfunc{mod}m}\right.$$\displaystyle \limfunc$modm$\displaystyle \left.\vphantom{ \limfunc{mod}m}\right)$  
x b $\displaystyle \left(\vphantom{ \limfunc{mod}n}\right.$$\displaystyle \limfunc$modn$\displaystyle \left.\vphantom{ \limfunc{mod}n}\right)$  

is given by

x = a + m$\displaystyle \left[\vphantom{ (b-a)/m\limfunc{mod}n}\right.$(b - a)/m$\displaystyle \limfunc$modn$\displaystyle \left.\vphantom{ (b-a)/m\limfunc{mod}n}\right]$

A complete set of solutions is given by

x = a + m$\displaystyle \left[\vphantom{ (b-a)/m\limfunc{mod}n}\right.$(b - a)/m$\displaystyle \limfunc$modn$\displaystyle \left.\vphantom{ (b-a)/m\limfunc{mod}n}\right]$ + rmn

where r is an arbitrary integer.