Consecutive Squares

If we take any 8 consecutive numbers: $$n, n+1, n+2, n+3, n+4, n+5, n+6, n+7$$ then the sum of the squares of four of these numbers is equal to the sum of the squares of the other four.

This means that the terms in $x^2$, in $x$, and the constant term must be split equally.

If we sum the squares of each of the eight consecutive numbers, and then halve the result, this will equal the sum of each of the four terms needed. So adding all the squares we have: $$n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 (n+5)^2 + (n+6)^2 + (n+7)^2 = $$$$= 8n^2 + 56n + 140 $$ So the two sides of the equality must have the value $$4n^2 + 28n + 70$$ $$(n+1)^2 + (n+2)^2 + (n+4)^2 +(n+7)^2 = n^2 + (n+3)^2 + (n+5)^2 + (n+6)^2$$