题目

Problem

Given that $n$ is an integer, use algebra, to prove by contradiction, that if $n^3$ is even then $n$ is even.

(4)

题目中文翻译

已知 $n$ 是整数，使用代数，通过反证法证明：如果 $n^3$ 是偶数，那么 $n$ 是偶数。

解答

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要用反证法证明“ $n^3$ 为偶数蕴含 $n$ 为偶数”，先假设前件成立而结论不成立，即 $n^3$ 为偶数但 $n$ 为奇数。把奇数写成 $2r+1$ 并展开其立方，证明 $n^3$ 必为奇数，从而产生矛盾。

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Suppose, for a contradiction, that $n^3$ is even but $n$ is not even.

Since $n$ is an integer, $n$ must then be odd. Hence, for some integer $r$ ,

n=2r+1.

Cubing,

\begin{align*} n^3 =&\,(2r+1)^3\\ =&\,8r^3+12r^2+6r+1\\ =&\,2\big(4r^3+6r^2+3r\big)+1. \end{align*}

Since $4r^3+6r^2+3r$ is an integer, this shows that $n^3$ is odd. This contradicts the assumption that $n^3$ is even.

Therefore, if $n^3$ is even, then

\boxed{n\text{ is even}}.