# Proof problem: Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.

Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive Integer. Please help me to prove that whether this statement is true or false.

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Thanks for the answers... –  Alpha Oct 14 '12 at 16:11

If $n$ is an odd positive integer, then $n=2m+1$ for some non-negative integer $m$. Can you see how to finish? Hint- Do two cases , one where $m$ is odd and one where $m$ is even.

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Combine this with the fact that $n^2-1=(n+1)(n-1)$. –  copper.hat Sep 19 '12 at 15:52

Since $n$ is odd $n=4m+1$ or $n=4m+3$.

In the first case $n^2-1=(n-1)(n+1)=4m\cdot(4m+2)=8m(2m+1)$, while in the second case $n^2-1=(n-1)(n+1)=(4m+2)\cdot(4m+4)=8(2m+1)(m+1)$.

So $n^2-1$ is divisible by 8 if $n$ is odd.

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Nice try, you deserve vote up –  Alpha Sep 19 '12 at 16:21

Viewing the integer numbers modulo 8, write $a \equiv b$ for $8|(a-b)$.

(This structure, $\mathbb Z_8$ is generated by $8\equiv 0$ and is friendly with operations $+$ and $\cdot$.)

We have the following set of odd numbers: $\{ 1,3,5,7\}$. Or, rewriting by $5\equiv-3$ and $7\equiv -1$, this is only $$\{ 1,3,-3,-1\}$$ The squares of these: $$(\pm 1)^2 = 1\ \text{ and }\ (\pm 3)^2=9\equiv 1$$

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This is a nice example of a direct proof. You start with the facts that if $\phi$ is your positive odd integer, then it is in the form $\phi = 2n + 1$ where $n$ is an integer and $\phi^2 - 1 = 8p \quad( p\in\mathbb Z)$ is true.

Recall that if a number is divisible by 8, then 8 is one of its factors.

This is something like a “case-by-case” proof. You prove the hypothesis for $n \in2\mathbb Z$ and then $n \in 2\mathbb Z + 1$.

Let's start with assuming that $n = 2m + 1$ where $m$ is an integer (of course) to prove for odd numbers. It is clear that,\begin{align} \phi^2 - 1 =(2n + 1)^2 - 1 &= 4n^2 + 4n \\ &= 4n(n + 1) \\ &= (8mn + 4n)(2m + 2) \\ &=4n\cdot 2\cdot(2m + 1)\cdot (m + 1) \\ & = 8 \cdot (n(2m + 1)(m +1)) \\ & = 8k \end{align} $\rm 0.5\times Q.E.D$

Now $n = 2m$ which is gonna be easier. \begin{align}\phi^2 - 1 =(2n + 1)^2- 1 = 4n^2 + 4n &= 4n(n + 1) \\ &= 8mn(2m +1)\\&= 8(2m^2n + mn) \\ &=8k \end{align}

## Q.E.D

(yay)

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If $n$ is odd, then $n=8q+r$ with $r\in \{ 1,3,5,7 \}$. Then $n^2=64q^2+16qr+r^2=8(8q^2+2r)+r^2$. Thus, it suffices to prove that $r^2-1$ is divisible by $8$, which can be done by a simple calculation.

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Nice try @Ihf, you deserve vote up –  Alpha Sep 19 '12 at 16:20

Note that $n^2 - 1 = (n+1)(n-1)$. Because n is odd, both $n+1$ and $n-1$ are even. Let $m=n-1$. Noting that $2|m$, consider both $m$ and $m+2$ mod 4. After we see that either $m$ or $m+2$ is divisible by 4, we know that $m(m+2) = (2k)(4l) = 8kl$. (WLOG due the the commutativity of multiplication)

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$\rm n\ odd\, \Rightarrow\, n = 4k\pm 1\:$ so $4$ divides one of $\rm\:n\pm1,\:$ and $2$ divides other, so $\rm\,8\:|\:(n\!-\!1)(n\!+\!1) = n^2\!-1.$

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My favorite proof method is proof by induction. It consists of two steps:

1. you assume that it's true for $n$, and prove that it then has to be true for $n+2$ (next odd number) as well
2. prove that it's true for $n = 1$

Then, since it's true for 1 it's also true for 3 (from your first step), for 5, 7, etc.

$(n+2)^2 - 1 = n^2 + 4n + 4 - 1 = (n^2 - 1) + 4 (\overbrace{n+1}^{even} )$

We assumed the first term is divisible by 8, and the second is too. Now show that it's true for $n=1$:

$n^2-1=1-1=0$

which is divisible by 8.

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