Exercise 1.2.23 (The square of any integer is not of the form $3k+2$ )

Question

Prove that the square of any integer is of the form $3k$ or $3k + 1$, but not of the form $3k+2$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} By Problem 20, any integer $n$ is of the form $3k, 3k + 1$ or $3k + 2$, and these three cases are mutually exclusive.
\begin{itemize}
\item If $n = 3k$, $n^3 = 9k^2 = 3(3k^2) = 3 K$, where $K = 3 k^2 \in \Z$, so $n^2$ is of the form $3k$.
\item If $n = 3k + 1$, $n ^3 = 9 k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 = 3 L + 1$, where $L = 3k^2 + 2k \in \Z$, so $n^2$ is of the form $3k+1$.
\item If $n = 3k + 2$, $n^3 = 9k^2 - 6k + 1 = 3(3k^2 - 2k) + 1 = 3 M + 1$, where $M = 3k^2 - 2k \in \Z$, so $n^2$ is of the form $3k+1$.
\end{itemize}
Hence $n^2$ is never of the form $3k+2$.
\end{proof}

Exercise 1.2.23 (The square of any integer is not of the form $3k+2$ )

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Exercise 1.2.23 (The square of any integer is not of the form $3k+2$ )

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