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Here is what I know

if n is any odd integer then $n$ can be expressed as $n=2k+1 ~~~ where~k\in\mathbb{Z}$.So

$n^2-1=(2k+1)^2 -1=4k^2+4k=4k(k+1)$

but $k(k+1)~~ is~~even$. Thus $k(k+1)=2t, t\in \mathbb{Z}$

Now, $n^2-1=4[k(k+1)]=4(2t)=8t$ which implies $n^2 -1$ is a multiple of 8.

But, I don't know how to prove the above statement by using Division algorithm. Is there any one who could show me that proof?

Thanks!

Bart Michels
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albo
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1 Answers1

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We divide $n^2-1$ by $n-1$ and obtain $n+1$ by the division algorithm. Hence $$ (n+1)(n-1)=n^2-1. $$ This is of course also clear without using the division algorithm. Both $n-1$ and $n+1$ are divisible by $2$ by assumption, and one of them is even divisible by $4$, because their difference is $2$. Hence their product is divisible by $8$.

Dietrich Burde
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