如何在证明环境中保持列表缩进？

Question

我有以下证明环境：

\begin{proof}
\leavevmode

\begin{enumerate}
    \item[Case 1:] $3\mid n$. Then $n=3k$ for some integer $k$. Thus $n^2 = 9k^2 = 3(3k^2)$, and so $3\mid n^2$.
    \item[Case 2:] $3\nmid n$. By the division algorithm, either $n=3k+1$ or $n = 3k+2$ for some integer k.
    \begin{enumerate}
        \item[Case 2a)] Suppose $n=3k+1$. Then $n^2 = (3k+1)^2 = 9k^2 + 6k + 1$ and so $n^2 = 3(3k^2 + 2k) + 1$. Therefore, $n^2 = 3m+1$ for some integer m, and so $n^2 - 1 = 3m$, thus $3\mid n^2 - 1$.
        \item[Case 2b)] Suppose $n = 3k + 2$. Then $n^2 = (3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2+4k + 1) + 1$. $n^2 - 1 = 3(3k^2  +4k + 1)$, thus $3\mid n^2 -1$.
    \end{enumerate}

\end{enumerate}

\end{proof}

其呈现效果如下：

我希望案例 1 和案例 2 与证明环境中的“证明”部分一致，但我所做的一切（包括使用 \indent 添加手动缩进）似乎都不起作用。我该怎么办？

Answer 1

如果你添加，showframe你就会看到出了什么问题。

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{amsthm}

\usepackage{showframe}

\begin{document}

\begin{proof}
\leavevmode

\begin{enumerate}
    \item[Case 1:] $3\mid n$. Then $n=3k$ for some integer $k$. Thus $n^2 = 9k^2 = 3(3k^2)$, and so $3\mid n^2$.
    \item[Case 2:] $3\nmid n$. By the division algorithm, either $n=3k+1$ or $n = 3k+2$ for some integer k.
    \begin{enumerate}
        \item[Case 2a)] Suppose $n=3k+1$. Then $n^2 = (3k+1)^2 = 9k^2 + 6k + 1$ and so $n^2 = 3(3k^2 + 2k) + 1$. Therefore, $n^2 = 3m+1$ for some integer m, and so $n^2 - 1 = 3m$, thus $3\mid n^2 - 1$.
        \item[Case 2b)] Suppose $n = 3k + 2$. Then $n^2 = (3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2+4k + 1) + 1$. $n^2 - 1 = 3(3k^2  +4k + 1)$, thus $3\mid n^2 -1$.
    \end{enumerate}

\end{enumerate}

\end{proof}

\end{document}

使用enumitem并记住\qedhere（虽然添加一些结束语会更好）。

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{amsthm}
\usepackage{enumitem}

\usepackage{showframe}

\begin{document}

\begin{proof}
We divide our proof into cases.
\begin{enumerate}[label=Case \arabic*:,ref=\arabic*,leftmargin=*]
  \item $3\mid n$. Then $n=3k$ for some integer $k$. Thus $n^2 = 9k^2 = 3(3k^2)$,
        and so $3\mid n^2$.

  \item \label{case2} $3\nmid n$. By the division algorithm, either $n=3k+1$ or
        $n = 3k+2$ for some integer k.
    \begin{enumerate}[label=Case \ref{case2}\alph*),leftmargin=2em]
      \item Suppose $n=3k+1$. Then $n^2 = (3k+1)^2 = 9k^2 + 6k + 1$ and so
            $n^2 = 3(3k^2 + 2k) + 1$. Therefore, $n^2 = 3m+1$ for some integer m,
            and so $n^2 - 1 = 3m$, thus $3\mid n^2 - 1$.
      \item Suppose $n = 3k + 2$. Then $n^2 = (3k+2)^2 = 9k^2 + 12k + 4 = 3(3k^2+4k + 1) + 1$.
            $n^2 - 1 = 3(3k^2  +4k + 1)$, thus $3\mid n^2 -1$.\qedhere
    \end{enumerate}
\end{enumerate}
\end{proof}

\end{document}