Question

show that n²-1 is divisible by 8 where n is the odd positive integer.​

Answer 1

✑Question :-

Show that $\rm n^2-1$ is divisible by $\rm 8$ where, n is the odd positive integer.

✑Solution :-

Any odd positive integer n can be written in the form of :-

• 4q+1
• 4q+3

✎If n = 4q+1, then $\rm n^2-1$ is :-

$\blue \implies\rm (4q+1)^2-1$

✢Using the identity $\rm (a+b)^2=a^2+b^2+2ab$

$\blue \implies \rm 16q^2+8q + 1 - 1$

$\blue \implies \rm 8( 2{q}^{2} +q)$

$\rm 8(2q^2+q)$ is divisible by 8.

✎If n = 4q+3, then $\rm n^2-1$ is :-

$\blue \implies\rm (4q+3)^2-1$

✢Using the identity $\rm (a+b)^2=a^2+b^2+2ab$

$\blue \implies \rm 16q^2+12q+9-1$

$\blue \implies \rm 8(2q^2+3 q+ 1)$

$\rm 8(2q^2+3q+1)$ is divisible by 8.

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❖ Therefore, it is clear that $\rm n^2-1$ is divisible by 8, when n is an odd positive integer.

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