If n is an odd integer then show that n^2-1 divisible by 8

1.	If n is an odd integer then show that n^2-1 divisible by 8
Answer» hence, we conclude that 4P² + 4P is divisible by 8 for all natural number .hence, n² -1 is divisible by 8 for all odd value of n . P =3 then,4P² + 4P = 4(3)² + 4(3) = 36 + 12 = 48 , divisible by 8 P =2 then,4P² + 4P = 4(2)² + 4(2) =16 + 8 = 24, it is also divisible by 8 . Now, when P = 1 then,4P² + 4P = 4(1)² + 4(1) = 4 + 4 = 8 , it is divisible by 8. We know ,odd number in the form of (2P +1) where P is a natural number ,so, n² -1 = (2P + 1)² -1= 4P² + 4P + 1 -1= 4P² + 4P If n is 4 n square + 1 here b is equals to 4the possible value of r is 0 ,1 ,2 ,3 but we take odd positive integer so we take 4q + 1 and 4q + 3 case first when n is equals to 4 Cube + 1 in this case we have any square - 14 Cube + 1 ka whole square minus 1 16 square + 8 + 1 - 1 16 square + 88 cube to cube + 1 and square minus 1 is divisible by 8 second case when was 24 Cube + 2 in this case we have

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