Exercise 3.17

Proof. If $x$ is such that $f(x)\equiv 0(\mathit{mod}n)$, as $p_i^{{\alpha }_i}\mid n$, $f(x)\equiv 0(\mathit{mod}{p}_i^{a_i})$.

Conversely, let $x_1,x_2,\dots ,x_t$ be integers such that

As $p_i^{a_i}\wedge p_j^{a_j}=1$ if $i\ne j$, the Chinese Remainder Theorem gives an integer $x$ such that $x\equiv x_i(\mathit{mod}{p}_i^{a_i}),i=1,2,\dots ,t$. As $f(x)\in \mathbb{Z}[x]$, $f(x)\equiv f(x_i)\equiv 0(\mathit{mod}{p}_i^{a_i})$. Thus $p_i^{a_i}\mid f(x),i=1,2,\dots ,t$, where $p_i^{a_i}\wedge p_j^{a_j}=1$ if $i\ne j$, then $n=p_1^{a_1}\cdots p_t^{a_t}\mid f(x)$, so $x$ is a solution of $f(x)\equiv 0(\mathit{mod}n)$.

Conclusion: $f(x)\equiv 0(\mathit{mod}n)$ has a solution iff $f(x)\equiv 0(\mathit{mod}{p}_i^{a_i})$ has a solution for $i=1,\dots ,t$. □