We have seen that the Diophantine equations of congruences have a finite number of solutions, if they are solvable. Now, we are able to provide a counting principle for the solutions of solvable Diophantine equation of congruences.
Let $m_1,\ldots,m_r$ be positive integers with $m_i\perp m_j$ co-prime, whenever $i\neq j.$ Furthermore, let $$f(x)(m_1\cdots m_r)\equiv 0(m_1\cdots m_r)\label{eq:N18599}\tag{1}$$ be a Diophantine equation of congruences. The number of solutions of the equation $(\ref{eq:N18599})$ is equal the product of the numbers of solutions of the equations
$$\begin{array}{rcl}f(x)(m_1)&\equiv&0(m_1)\\ &\vdots&\\ f(x)(m_r)&\equiv&0(m_r). \end{array}$$
In particular, if $m=\prod_{i=1}^rp_i^{e_i}$ is a factorization of some positive integer $m > 0,$ then the number of solutions of $f(x)(m)\equiv 0(m)$ equals the product of the $r$ numbers of solutions of $f(x)(p_i^{e_i})\equiv 0(p_i^{e_i})$, $i=1,\ldots,r.$
Proofs: 1
