Proof: By Induction
(related to Proposition: Existence and Number of Solutions of Congruence With One Variable)
By hypothesis, $a,b$ are integers, $m > 0$ is a positive integer. We first show that the solvability of the congruence is equivalent to $\gcd(a,m)\mid b.$
"$\Rightarrow$"
 Assume, the congruence with one variable $(axb)(m)\equiv 0(m)$ is solvable.
 Therefore, we have also $(axb)(\gcd(a,m))\equiv 0(\gcd(a,m)).$
 Since the greatest common divisor $\gcd(a,m)$ is a divisor of both, $a$ and $m$, we $(b)(\gcd(a,m)) \equiv 0(\gcd(a,m)).$
 It follows $\gcd(a,m)\mid b.$
"$\Leftarrow$"
 Assume, $\gcd(a,m)\mid b.$
 Consider the congruence with one variable $$\frac{a}{\gcd(a,m)}x\equiv\frac{b}{\gcd(a,m)}\mod \frac{m}{\gcd(a,m)}.\label{eq:E18453}\tag{1}$$
 Note that $\frac{a}{\gcd(a,m)}\perp \frac{m}{\gcd(a,m)}$ are coprime, by generating coprime numbers knowing the $\gcd$.
 According to congruences and division with quotient and remainder (see also explanation), the integers $0,1,\ldots,m1$ build a complete residue system modulo $m$.
 According to creation of complete residue systems from others, this holds also for the system $c\cdot 0,c\cdot 1,\ldots,c\cdot(m1)$ with $c:=\frac{a}{\gcd(a,m)}.$
 Therefore, the congruence with one variable $(\ref{eq:E18453})$ has exactly one solution.
 With multiplication of congruences with a positive factor, it follows that also the congruence $(ax)(m)\equiv b(m)$ is solvable.
It remains to be shown that the congruence has exactly $\gcd(a,m)$ different solutions.
* The congruence with one variable $(\ref{eq:E18453})$ has exactly one solution modulo $\frac{m}{\gcd(a,m)}.$
* Since $\gcd(a,m)\mid m$, there are $\gcd(a,m)$ as many solutions modulo $m.$
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:
