Exercise 1.27

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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

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

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

For all odd $n$ show that $8\mid n^2 - 1$. If $3
mid n$,
show that $6\mid n^2 - 1$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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

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

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
As $n$ is odd, write $n=2k+1, n \in \Z$. Then
$$n^2 - 1 = (2k+1)^2-1 = 4k^2+4k = 4k(k+1).$$
As $k$ or $k+1$ is even, $8 \mid n^2 - 1$.

$(n-1)n(n+1) = n (n^2-1)$, product of three consecutive numbers, is a multiple of 3.

As $3 
mid n$, and 3 is a prime, $3 \wedge n = 1$, so $3 \mid n^2-1$.
$$3 
mid n \Rightarrow 3 \mid n^2 -1.$$
(This is also a consequence of Fermat's Little Theorem.)

As $n$ is odd, $n^2-1$ is even.
$3 \mid n^2-1, 2 \mid n^2-1$ and $2 \wedge 3 = 1$, so $6 \mid n^2-1$.
\end{proof}

Exercise 1.27

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Exercise 1.27

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