Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8
(2n+3)^2-(2n+1)^2 4n^2+12n+9-4n^2-4n-1 8n+8 8(n+1), which is a multiple of 8
(2n+3)^2-(2n+1)^2 4n^2+12n+9-4n^2-4n-1 8n+8 8(n+1), which is a multiple of 8
