Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 6D / a / Solution

Let $d=\gcd (a,m)$. If the linear congruence $ar\equiv b(\mathrm{mod}m)$ holds, then $b=ar-ms$ for some integer $s$, so the greatest common divisor $d$ divides $b$. Conversely, Bézout's identity gives integers $u,v$ with $au+mv=d$. If $b=dj$, multiply this identity by $j$ to obtain $a(uj)\equiv b(\mathrm{mod}m)$. HenceFor the number of solutions, write $a=d{a}^{\prime }$, $m=d{m}^{\prime }$ and $b=d{b}^{\prime }$. The linear congruence becomes $a^{\prime }r\equiv b^{\prime }(\mathrm{mod}{m}^{\prime })$, where $a^{\prime },m^{\prime }$ are coprime integers. Multiplication by $a^{\prime }$ is invertible modulo $m^{\prime }$ by Bézout's identity, so the solutions form one residue class $r\equiv r_0(\mathrm{mod}{m}^{\prime })$. Modulo $m=d{m}^{\prime }$, these are exactlyThey are distinct: if two have the same residue class modulo $m$, then $d{m}^{\prime }$ divides $(j-\ell )m^{\prime }$, forcing $d\mid j-\ell$; with $0\le j,\ell <d$, this forces $j=\ell$. Every other solution differs from $r_0$ by a multiple of $m^{\prime }$, so these exhaust the $d$ solutions.

Finally,Combining this with the existence criterion gives the requested equivalent modular congruence.