Question

Hello friends !

Answer this ,

= > n^2 - 1 is divisible by 8 , if n is :

A)an integer
B)a natural number
C)an odd integer
D)an even integer

I need solution as well !

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All the best ! :)

Answer 1

( n^2 -1)

( n-1) ( n+1)

If n is odd

then n-1 and n+1 both even

Now if n-1 = 2k

then n+1 = 2k +2 = 2( k+1)

if k is odd then k+1 is even So 4 becomes factor of n+1

And 2 as factor of n-1 So 4×2= 8

if k is even then 2k contains 4 as factor

and 2(k+1) contains 2 as factor So 8

If n is even

( n-1) and ( n+1) both odd

As both are odd So both don't have 2 as factor So it can't be 8

So only odd integer

Answer 2

Answer:

Option(C)

Step-by-step explanation:

(i) When n is even:

We know that any even positive integer is of the form 2m for some integer m.

When n = 2m, then

n² - 1 = (2m)² - 1

         = 4m² - 1

∴ Not divisible by 8.

(ii) When n is odd:

We know that any odd positive integer is of the form 4m + 1 (or) 4m + 3 for some integer m.

When n = 4m + 1, then

n² - 1 = (4m + 1)² - 1

        = 16m² + 1 + 8m - 1

        = 16m² + 8m

        = 8m(2m + 1).   {∴ Divisible by 8}

n² - 1 = (4m + 3)² - 1

         = 16m² + 9 + 24m - 1

         = 16m² + 24m + 8

         = 8(2m² + 3m + 1) {∴Divisible by 8}

∴ Hence, n² - 1 is divisible by 8, n is an odd integer.

Hope it helps!