Show that (n^2) + (n+1)^2 + (n+2)^2 = 3n^2 + 6n + 5, Hence show that the sum of 3 consecutive square numbers is always 2 away from a multiple of 3.

Expanding out the Brackets: (n²)+ (n² + n + n + 1) + (n² + 2n + 2n +4) = (n²) + (n²+2n+1) + (n²+4n+4) =3n² + 6n +5 Using this result: Three consecutice numbers: n, n+1, n+2 So three consecutive squares:  n², (n+1)², (n+2)² Sum of those brackets = 3n² + 6n + 5 (from first part) Simplifying 3n² + 6n + 5 = 3n² + 6n + 3   +2 = 3(n² + 2n +1)  + 2 First part is always a multiple of 3, and the final result is always 2 away from that.