n^{2} - 1 is divisible by 8 , if n is

Foundation

Mathematics Foundation

Real Number Properties

Question

n^{2} - 1

is divisible by

8

, if n is

an odd integer

none

an integer

a natural number

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Given,

Any odd positive integer

n

can be written in form of

4 q + 1

4 q + 3

n = 4 q + 1

, when

n^{2} - 1 = (4 q + 1)^{2} - 1 = 16 q^{2} + 8 q + 1 - 1 = 8 q (2 q + 1)

which is divisible by

8.

n = 4 q + 3

, when

n^{2} - 1 = (4 q + 3)^{2} - 1 = 16 q^{2} + 24 q + 9 - 1 = 8 (2 q^{2} + 3 q + 1)

which is divisible by

8

So, it is clear that

n^{2} - 1

is divisible by

8

, if

n

is an odd positive integer.