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QUESTION :

IF n IS AN ODD INTEGER THEN SHOW THAT
${n}^{2} - 1 \: is \: divisible \: by \: 8$
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Answer 1

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We know that,odd number is in the form of (2P +1) where P is a natural number.

So,

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

Now, Checking-

$p = 1 \\ {4p}^{2} + 4p = 4 \times {2}^{2} + 4(2) = 24$

It is also divisible by 8.

Now,p=3

${4p}^{2} + 4p = 4 {(3)}^{2} + 4(3) = 48$

Which is also divisible by 8.

Hence,we conclude that

${4p}^{2} + 4p$

is divisible by 8 for all natural number.Hence, n² -1 is divisible by 8 for all odd value of n.

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Answer 2

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If 'n' is an odd positive integer, show that (n2-1) Is divisible by 8??

Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.

Let n = 4p+ 1,

(n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1)

⇒ (n2 – 1) is divisible by 8.

(n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1)

⇒ n2– 1 is divisible by 8.

Therefore, n2– 1 is divisible by 8 if n is an odd positive integer.

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