Let $m > 1$ be a positive integer and let $f(x_1,\ldots,x_r)=0$ be a Diophantine equation. If integers $a_1,\ldots,a_r$ solving this equation^{1} exist, then the congruences $a_1(m),\ldots,a_r(m)$ solve also the Diophantine equation of congruences modulo $m$, i.e. we have $$(f(a_1,\ldots,a_r))(m)\equiv f(a_1(m),\ldots,a_r(m))\equiv 0(m).$$
Corollaries: 1
Proofs: 2 3 4
Propositions: 5 6
i.e. by setting $x_r=a_1,\ldots,x_r=a_r.$ Note that a solution does not have to exist, for instance $x^3+y^3=z^3$ has no integer solutions. ↩