If n is an odd integer, then show that n²-1 is divisible by 8
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we know that every odd interger is of form 4q+1 or 4q+3 for some integer q.
So case 1
when n=4q+1
then n^2-1=(4q+1)^2-1
=16q^2+8q+1-1
=8 (q+2q^2) which is divisible by 8
Case 2
when n=4q+3
then n^2-1=(4q+3)^2-1
=16q^2+24q+9-1
=8 (2q^2 + 3q +1) which is again divisible by 8.
Hence, if n is an odd integer then n^2-1 is divisible by 8
So case 1
when n=4q+1
then n^2-1=(4q+1)^2-1
=16q^2+8q+1-1
=8 (q+2q^2) which is divisible by 8
Case 2
when n=4q+3
then n^2-1=(4q+3)^2-1
=16q^2+24q+9-1
=8 (2q^2 + 3q +1) which is again divisible by 8.
Hence, if n is an odd integer then n^2-1 is divisible by 8
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