[proofplan] Start with an integer solution satisfying the original side conditions and reduce its coordinates modulo an arbitrary modulus $m \geq 2$. The reduced vector automatically satisfies the reduced side conditions because those conditions were defined as the image of the original side-condition set under reduction. It remains only to check that polynomial evaluation commutes with coefficientwise and coordinatewise reduction modulo $m$, which turns the equality $F(a)=0$ in $\mathbb{Z}$ into the congruence $\bar F_m(\rho_m(a))=0$ in $\mathbb{Z}/m\mathbb{Z}$. [/proofplan] [step:Reduce the given integer solution modulo an arbitrary modulus] Assume there exists $a=(a_1,\dots,a_n)\in C$ such that $F(a)=0$. Fix an integer $m \geq 2$. Define \begin{align*} \bar a := \rho_m(a) = (\bar a_1,\dots,\bar a_n) \in (\mathbb{Z}/m\mathbb{Z})^n. \end{align*} Since $C_m := \rho_m(C)$ and $a \in C$, we have $\bar a \in C_m$. Thus the reduced vector satisfies the reduced side conditions by definition. [/step] [step:Show that polynomial evaluation commutes with reduction modulo $m$] Write $F$ as a finite sum \begin{align*} F(x_1,\dots,x_n)=\sum_{\alpha \in A} c_\alpha x_1^{\alpha_1}\cdots x_n^{\alpha_n}, \end{align*} where $A \subset \mathbb{N}_0^n$ is a finite set of multi-indices and each coefficient $c_\alpha$ lies in $\mathbb{Z}$. The reduced polynomial is \begin{align*} \bar F_m(x_1,\dots,x_n)=\sum_{\alpha \in A} \bar c_\alpha x_1^{\alpha_1}\cdots x_n^{\alpha_n} \end{align*} in $(\mathbb{Z}/m\mathbb{Z})[x_1,\dots,x_n]$. Because addition and multiplication in $\mathbb{Z}/m\mathbb{Z}$ are defined by reducing the corresponding integer operations modulo $m$, evaluating $\bar F_m$ at $\bar a$ gives \begin{align*} \bar F_m(\bar a) &= \sum_{\alpha \in A} \bar c_\alpha \bar a_1^{\alpha_1}\cdots \bar a_n^{\alpha_n} \\ &= \overline{\sum_{\alpha \in A} c_\alpha a_1^{\alpha_1}\cdots a_n^{\alpha_n}} \\ &= \overline{F(a)}. \end{align*} Since $F(a)=0$ in $\mathbb{Z}$, its residue class modulo $m$ is $\overline{F(a)}=\bar 0$. Therefore \begin{align*} \bar F_m(\bar a)=0 \end{align*} in $\mathbb{Z}/m\mathbb{Z}$. [/step] [step:Conclude that every local congruence test is passed] The integer $m \geq 2$ was arbitrary. For each such $m$, the vector $\bar a=\rho_m(a)$ lies in $C_m$ and satisfies \begin{align*} \bar F_m(\bar a)=0 \end{align*} in $\mathbb{Z}/m\mathbb{Z}$. Hence the congruence \begin{align*} F(x_1,\dots,x_n)\equiv 0 \pmod m \end{align*} has a solution satisfying the reduced side conditions for every modulus $m \geq 2$. [/step]