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 = 4n 2 + 1 + 4n + 4n 2 + 9 + 12n = 8n 2 + 16n + 10 Adding 6 to the sum of squares we get = 8n 2 + 16n + 10 + 6 = 8n 2 + 16n + 16 = 8(n 2 + 2n + 2) On dividing resultant sum by 8, we get : = n 2 + 2n + 2. Hence, proved that the new number is divisible by 8.