If n is an odd positive integer,show that n^2-1 is divisible by 8.
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We know that any positive integers is of the form 4q+1,4q+3.
So, when n= 4q+1,
n^2-1=(4q+1)^2-1
=16q^2+8q +1-1=16q^2+8q
=8q(2q+1)
=>n^2-1 is divisible by 8
When n=4q+3
n^2-1=(4q+3)^2-1
=16q^2+24q+9-1
=16q^2+24q+8
=8(2q^2+3q+1)
=>n^2-1 is divisible by 8.
Hence proved.
Hope this helps you.
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So, when n= 4q+1,
n^2-1=(4q+1)^2-1
=16q^2+8q +1-1=16q^2+8q
=8q(2q+1)
=>n^2-1 is divisible by 8
When n=4q+3
n^2-1=(4q+3)^2-1
=16q^2+24q+9-1
=16q^2+24q+8
=8(2q^2+3q+1)
=>n^2-1 is divisible by 8.
Hence proved.
Hope this helps you.
if u like it make it the brainiest.
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