Question

Show that n square - 1 is divisible by 8 , if n is any odd positive integer

Answer 1

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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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Answer 2

$\huge{\bold{\underline{Solution-:}}}$

$\bold{\underline{According\ to\ Euclid's\ Division\ Lemma}}$

a = bq + r

Let a be n

Here,

b = 4

r = 0,1,2 & 3

n = 4q + 0

n = 4q + 1

n = 4q + 2

n = 4q + 3

$\bold{In\ the\ 1st\ case\ we\ have}$

n = 4q

${n}^{2}$ - 1

${4q}^{2}$ - 1

${16q}^{2}$ - 1

which is divisible by 8

$\bold{In\ the\ 2nd\ case\ we\ have}$

n = 4q + 1

${n}^{2}$ - 1

${4q + 1}^{2}$ - 1

${16q}^{2}$+ 8q +1 - 1

${16q}^{2}$+ 8q

which is divisible by 8

$\bold{In\ the\ 3rd\ case\ we\ have}$

n = 4q + 2

${n}^{2}$ - 1

${4q + 2}^{2}$ - 1

${16q}^{2}$+ 16q + 4 - 1

${16q}^{2}$+ 16q + 4

which is divisible by 8

$\bold{In\ the\ 4th\ case\ we\ have}$

n = 4q + 3

${n}^{2}$ - 1

${4q + 3}^{2}$ - 1

${16q}^{2}$+ 24q + 9 - 1

${16q}^{2}$+ 24q + 8

which is divisible by 8

$\bold{\underline{Hence\ proved}}$