Question

If n is an odd integer, prove that n^4+4n^2+11 is of the form 16k

Answer 1

Answer:

The answers is 8K (not 16K)

Step-by-step explanation:

we are given  

n^4 + 4n^2 + 11

taking common n² we get

n²(n² + 4) + 11

since given n is odd then it must be of form

                         2x + 1  where x in integer

so putting n = 2x+1 we get

  (2x+1)²{(2x+1)² +4} + 11 = (4x²+ 4x + 1)( 4x²+ 4x + 1 +4) + 11

= (4x²+ 4x + 1)( 4x²+ 4x + 5) + 11  

= (16x^4+ 16x^3 + 4x²) +( 4x²+ 4x + 5) +(16x^3 + 16x² + 20x) + 11  

= (16x^4+ 32x^3 + 24x² + 24x + 16

Taking 8 common

= 8(2x^4+ 4x^3 + 3x² + 3x + 2)

let   (2x^4+ 4x^3 + 3x² + 3x + 2) = K

where K is any integer

Then

8(2x^4+ 4x^3 + 3x² + 3x + 2) = 8K

So if n is odd we see that n^4 + 4n^2 + 11 will of form 8K