Question is from Fibonacci\'s Liber Quadratorum Given the squares of three successive odd numbers, show that thelargest square exceeds the middle square by eight more than themiddle square exceeds the smallest. I think I have the answer (2n+5)2 = (2n+3)2 +[(2n+3)2 - (2n+1)2] + 8 but I am not sure howto show it algebraically. Solution three consecutive odd numbers: (2x+1), (2x+3), (2x +5) (2x+1)2 = 4x2 + 4x + 1 (2x+3)2 = 4x2 + 12x + 9 (2x+5)2 = 4x2 + 20x + 25 largest square - middle square = 4x2 + 20x + 25 -(4x2 + 12x + 9) = 8x + 16 middle square - smallest square = 4x2 + 12x + 9 -(4x2 + 4x + 1) = 8x + 8 so the difference between the above two is 8x+16 - (8x+8) = 8.