i have no idea about this?????????????

Question

If n=4k+3 does 8 divide n^2-1?

Answer 1

If n = 4k + 3 does 8 divide n^2-1 [evenly; i.e., no remainder]?

 

(n^2 - 1) / 8 = (n + 1)(n - 1) / 8

 

= (4k + 3 + 1)(4k + 3 - 1) / 8

 

= (4k + 4)(4k + 2) / 8

 

= 4(k + 1)(2)(2k + 1) / 8

 

= 8(k + 1)(2k + 1) / 8

 

= (k + 1)(2k + 1) with no remainder.

 

So the answer is YES.

Answer 2

Hi Daniel;

n=4k+3

I think the second equation is...

n²-1

(4k+3)²-1

Let's FOIL the (4k+3)(4K+3)

FIRST...(4K)(4K)=16K²

OUTER...(4k)(3)=12k

INNER...(3)(4k)=12k

LAST...(3)(3)=9

16k²+12k+12k+9

16k²+24k+9

The original equation includes -1...

16k²+24k+9-1

16k²+24k+8

Can we factor out 8?

8(2k²+3k+1)

YES!!!!