一名学生被要求证明：对 $p\in\mathbb{N}$，

“如果 $p^3$ 是 3 的倍数，那么 $p$ 必须是 3 的倍数”

该学生用反证法证明的开头已在下框中给出。

(a) 写出完成该证明所需要的计算和论述。

(b) 进而用反证法证明 $\sqrt[3]{3}$ 是无理数。

题框已经处理了 $p=3k+1$ 的情况。若自然数 $p$ 不是 $3$ 的倍数，它还可能写成 $3k+2$；展开其立方并证明仍不是 $3$ 的倍数，即可让两个可能情况都与原假设矛盾。

The case $p=3k+1$ has already been considered in the question.

For the other possible case, let

Then

Let

Since $H$ is an integer,

Therefore, $p^3$ is not a multiple of $3$.

If $p$ is not a multiple of $3$, it must be of the form $3k+1$ or $3k+2$. Both cases show that $p^3$ is not a multiple of $3$, contradicting the assumption that $p^3$ is a multiple of $3$.

Hence, if $p^3$ is a multiple of $3$, then

另一种情况也可写成 $p=3k-1$；这与 $3k+2$ 只相差 $k$ 的选取。此时

Let

Then $J$ is an integer and

which is not a multiple of $3$. Together with the $3k+1$ case in the question, this gives the same contradiction and conclusion.

原来的 $3k+2$ 展开也可直接说明：$27k^3$、$54k^2$、$36k$ 都能被 $3$ 整除， 但 $8$ 不能，所以总和不能被 $3$ 整除。这是同一余数判断，不是另一套解法。

承接 (a)，假设 $\sqrt[3]{3}$ 是有理数，并把它写成最简分数 $\frac{p}{q}$。立方后先由 (a) 推出 $p$ 是 $3$ 的倍数，再代回推出 $q$ 也是 $3$ 的倍数；这与 $p,q$ 互质矛盾。

Assume, for a contradiction, that $\sqrt[3]{3}$ is rational. Then it can be written in lowest terms as

where $p,q\in\mathbb{N}$, $q\ne0$, and $p$ and $q$ have no common factor greater than $1$.

Cubing both sides gives

so

Thus $p^3$ is a multiple of $3$. By part (a), $p$ is a multiple of $3$, so let

for some $k\in\mathbb{N}$. Substituting this into $p^3=3q^3$,

Hence $q^3$ is also a multiple of $3$. By part (a), $q$ is a multiple of $3$.

Therefore, both $p$ and $q$ have a common factor of $3$. This contradicts the assumption that $\frac{p}{q}$ is in lowest terms.

Hence