Prove that $n^2 -1$ is divisible by $8$, for every odd integer $n$.

Question

Prove that $n^2 - 1$ is divisible by $8$, for every odd integer $n$.

I tried to factorize and find something but still have no idea.

What did you get by factorising? Have you managed to prove something, even if it is weaker than the thing you are asked to prove? — , Feb 01 '18 at 18:28

dssknj · Answer 1 · 2018-02-01T18:54:42.710

1

Since n is odd, n can be of form $8q+1, 8q+3, 8q+5$ or $8q+7 $. If $ n = 8q+1$ then $ n^2-1 = 8q(8q+2)$ If $n = 8q+3$ then $n^2-1 = 8(8q^2+6q+1)$ If $n = 8q+5$ then $n^2-1 = 8(8q^2+10q+3)$ and if $ n=8q+7$ then $n^2-1= 8(8q^2+14q+6)$. We see that all of them are divisible by 8. Hence $n^2-1$ is divisible by 8 where n is an odd integer.

edited Feb 01 '18 at 18:54

answered Feb 01 '18 at 18:31

dssknj

612

Yash Jain · Answer 2 · 2018-02-01T18:37:27.600

1

An odd integer must be in the form of $2n-1$, so $(2n-1)^2-1=4n^2-4n = 4n(n-1)$

$n-1$ must be even, since $n$ is odd. Thus, $n-1$ must be a multiple of $2$. $2*4=8$, proving $n^2-1$ is divisible by $8$ for every odd integer $n$.

edited Feb 01 '18 at 18:37

answered Feb 01 '18 at 18:33

Yash Jain

1

Your $n$ needn't be odd ... – Peter Smith Feb 01 '18 at 18:35

score 0 · Answer 3 · answered Feb 01 '18 at 18:29

0

Hint:

Since it is odd, we can write $\;n=2k+1\;$ , so

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

answered Feb 01 '18 at 18:29

DonAntonio

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Prove that $n^2 -1$ is divisible by $8$, for every odd integer $n$.

3 Answers3