Take two consecutive odd numbers. Find the sum of their squares, and then add 6 to the result. Prove that the new number is always divisible by 8.

Let two consecutive odd numbers be (2n + 1) and (2n + 3) for some integer n.

Sum of squares = (2n + 1)² + (2n + 3)²

= 4n² + 1 + 4n + 4n² + 9 + 12n

= 8n² + 16n + 10

Adding 6 to the sum of squares we get

= 8n² + 16n + 10 + 6

= 8n² + 16n + 16

= 8(n² + 2n + 2)

On dividing resultant sum by 8, we get :

$\Rightarrow \dfrac{8(n^2 + 2n + 2)}{8}$ = n² + 2n + 2.

Hence, proved that the new number is divisible by 8.