Question

Show that n2 - 1 is divisible by 8, if n is an odd positive integer.

NEET/Medical Exams

Maths

Solution

233

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We know that any odd positive integer is of the form ( 4 q+1 ) or, ( 4 q+3 ) for some integer ( q ) So, we have the following cases: Case I : When ( n=4 q+1 ) In this case, we have [ begin{aligned} n^{2}-1 &=(4 q+1)^{2}-1=16 q^{2}+8 q+1-1 &=16 q^{2}+8 q=8 q(2 q+1) Rightarrow quad n^{2}-1 text { is divisible by } 8 &[because 8 q(2 q+1) text { is divisible by } 8] end{aligned} ] Case II : When ( n=4 q+3 ) In the case, we have ( begin{aligned} n^{2}-1 &=(4 q+3)^{2}-1=16 q^{2}+24 q+9-1 &=16 q^{2}+24 q+8 Rightarrow quad & n^{2}-1=8left(2 q^{2}+3 q+1right)=8(2 q+1)(q+1) end{aligned} ) ( Rightarrow n^{2}-1 ) is divisible by 8 ( [because 8(2 q+1)(q+1) text { is divisible by } 8] ) Hence, ( n^{2}-1 ) is divisible by 8