Let $n$ be an integer. If $n^2$ is divisible by $3$, then $n$ is divisible by $3$.

Question

I am trying to prove the statement "Let $n$ be an integer. If $n^2$ is divisible by $3$, then $n$ is divisible by $3$." by using the direct proof technique.

The following is my proof.

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Consider the following definition.

Definition 5.1: A nonzero integer $a$ divides an integer $b$ if there is an integer $j$ such that $b = aj.$

Because $3$ divides $n^2$, by Definition 5.1 it follows that

\begin{align}
n^2 = 3j \text{ for some $j \in \mathbb{Z}$}
\end{align}

By taking the square root on both sides of the above equation, one obtains

\begin{align}
n = \pm \sqrt{3j} \text{ for some $j \in \mathbb{Z}$}
\end{align}

Since $n$ is an integer, for the $RHS$ of the above equation to be an integer, $j = 3k^2$ for each $k \in \mathbb{Z^*}$.

Hence,

\begin{align}
n = \pm \sqrt{9k^2} = \pm 3k \text{ for each $k \in \mathbb{Z^*}$}
\end{align}

By Definition 5.1, one can conclude that $3$ divides $n$.

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Is the proof correct?

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Reference:

Reading, Writing, and Proving: a Closer Look at Mathematics, by Ulrich Daepp and Pamela Gorkin, 2nd ed., Springer, 2011, pp. 52,55.

Answer 1

Your proof has a problem here

Since $n$ is an integer, for the $RHS$ of the above equation to be an integer, $j = 3k^2$ for each $k \in \mathbb{Z^*}$.

How do you know this?