Question

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

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

Solution