Show that if a is an odd integer, then \frac{({\ a}^4+{\ 4a}^2\ -11)}{16} is an integer, or 16| (a4 + 4a2−11) leaves remainder zero.

 

Solution;

 Let a = 2k + 1 , for any k ∈ ℕ

 

now, a⁴ + 4a² + 11 

= (2k +1)⁴ + 4(2k + 1)² + 11 

= (4k² + 4k + 1)² + 4(4k² + 4k + 1) + 11 

= (16k⁴ + 16k² + 1 + 32k³ + 8k + 8k²) + 16k² + 16k + 4 + 11 

= 16k⁴ + 32k³ + 40k² + 24k + 16 

= 16[k⁴ + 2k³ + 1] + 40k² + 24k 

= 16[k⁴ + 2k² + 1] + (80k² + 48k)/2 

= 16[k⁴ + 2k² + 1] + 16k(5k + 3)/2 

here,

k(5k + 3)/2 is also an integer for all k ∈ ℕ

so, assume k(5k + 3)/2 = m 

now, a⁴ + 4a² + 11 

= 16[k⁴ + 2k² + 1] + 16m 

= 16[k⁴ + 2k² + 1 + m] 

hence, it is clear that 16 divides a⁴ + 4a² + 11 when a is  any odd integers.