Question

Comment upon n, if ${n^2} - 1$ is divisible by 8.

Answer 1

Hint: In this question use the concept that any odd positive integer can be written as (4k + 1) or (4k + 3) and substitute these values in place of n for the given equation, the apply the concept that if a number p is divisible by 8 then eventually 8p will also be divisible by 8, to check whether it’s coming out to be divisible by 8 or not.

Complete step-by-step answer:

Given equation is $\left( {{n^2} - 1} \right)$ ............................. (1)
As we know any positive odd integer is in the form of (4k + 1) or (4k + 3) for some integer k.
Let n = 4k + 1.
So substitute this value in equation (1) we have,
$ \Rightarrow \left( {{n^2} - 1} \right) = {\left( {4k + 1} \right)^2} - 1$
Now expand the square according to property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,
$ \Rightarrow \left( {{n^2} - 1} \right) = {\left( {4k + 1} \right)^2} - 1 = 16{k^2} + 1 + 8k - 1 = 16{k^2} + 8k = 8k\left( {2k + 1} \right)$ So as we see this is the multiple of 8 which is divisible by 8.
Now let n = 4k + 3.
So substitute this value in equation (1) we have,
$ \Rightarrow \left( {{n^2} - 1} \right) = {\left( {4k + 3} \right)^2} - 1$
Now expand the square according to property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,
$ \Rightarrow \left( {{n^2} - 1} \right) = {\left( {4k + 3} \right)^2} - 1 = 16{k^2} + 9 + 24k - 1 = 16{k^2} + 24k + 8 = 8\left( {2{k^2} + 3k + 1} \right)$ So as we see this is also the multiple of 8 which is divisible by 8.
Therefore $\left( {{n^2} - 1} \right)$ is divisible by 8 if n is an odd positive integer.
So this is the required answer.

Note: In this problem if we would have taken n as an positive even integer than any positive integer could be written in the form of 4k, 4k+2……………. If the same concept would have been applied as of above for this positive even integer n then it would not have shown up as being divisible by 8.