Show $8\mid n^2-1$ if $n$ is an odd positive integer.

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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: $$(\pm 1)^2 = 1\ \text{ and }\ (\pm 3)^2=9\equiv 1$$

Answer 4

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.

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

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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 $

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

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Q.E.D

(yay)

Answer 5

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)

Answer 6

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.

Answer 7

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.

Answer 8

We know that any odd positive integer is of the form $4q+1$ or, $4q+3$ for some integer $q$.

So, we have the following cases:

Case 1 when $n=4q+1$: In this case, we have
$n^2 -1=(4q+1)^2 -1=16q^2 +8q+1-1=16q^2 +8q=8q(2q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8q(2q+1)$ is divisible by 8]

Case 2 when $n=4q+3$: In this case, we have
$n^2 -1=(4q+3)^2 -1= 16q^2 +24q+9-1=16q^2 +24q+8$
⇒ $n^2 -1=8(2q^2 +3q+1)=8(2q+1)(q+1)$
$∴ n^2 -1$ is divisible by 8 [$∵ 8(2q+1)(q+1)$ is divisible by 8]

Hence, $n^2 -1$ is divisible by 8.

Answer 9

$\rm n\ odd\Rightarrow n \!=\! 4k\!\pm\! 1$ so $4$ divides one of $\rm n\!\pm\!1,\,$ & $\,2$ divides other, so $\rm\,4\cdot 2\:|\:\overbrace{(n\!-\!1)(n\!+\!1)}^{\textstyle n^2\!-1}$

Answer 10

Proof By Weak Mathematical Induction (weak induction simply means we consider only one base case)

RTP: 8|$n^2-1$

Step (1) [Base Case]: Test true for n = 1

$1^2 -1$ = 8(0)

Therefore true for n = 1

Step (2) [Inductive Step]:

Assume true for n = 2k + 1 where 2k + 1 is an odd +ve integer equal or greater than 1

$(2k+1)^2-1$ = 8m say, where m is an odd +ve integer

$4k^2+4k$ = 8m

Step (3): Test true for n = 2k + 3 (the next odd after 2k + 1)

WTS: 8|$(2k+3)^2-1$

$(2k+3)^2-1$ = $4k^2+12k+9-1$

= $4k^2+4k+8k+8$

= 8m + 8k +8 [From the assumption]

= 8 (m+k+1)

therefore 8|$(2k+3)^2-1$

Step(4) [Conclusion]: The result is true for n = 1. If it is true for n = 2k+1, where 2k+1 is an odd +ve integer equal or greater than 1, then it follows that it is also true for n = 2k+3. Thus by principle of weak mathematical induction is true for all n that is a +ve intger.

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