Exercise 1.2.14 (If $n$ is odd, $n^2 - 1$ is divisible by $8$)

Question

Prove that if $n$ is odd, $n^2 - 1$ is divisible by $8$.

Richard Ganaye · Accepted Answer

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

ewcommand{\N}{\mathbb{N}}

\begin{proof} If $n$ is odd, $n$ is of the form $n = 2k+1$ for some integer $k$. Then 
\begin{align*}
n^2 - 1 &= (2k+1)^2 - 1\
&= 4k^2 + 4k\
&= 8\,  \frac{k(k+1)}{2}.
\end{align*}
Moreover, as seen in Exercise 6 or 13, $k(k+1)$ is even, thus $\frac{k(k+1)}{2} \in \Z$. This shows that $8 \mid n^2 - 1$.
\end{proof}

Exercise 1.2.14 (If $n$ is odd, $n^2 - 1$ is divisible by $8$)

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Exercise 1.2.14 (If $n$ is odd, $n^2 - 1$ is divisible by $8$)

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