Consecutive Squares
The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum.
True or false?
What does this tell you about each side of the equality?
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 $$4n^2 + 28n + 70$$ $$(n+1)^2 + (n+2)^2 + (n+4)^2 +(n+7)^2 = n^2 + (n+3)^2 + (n+5)^2 + (n+6)^2$$
Students may square and add terms in an almost random way to start with and it is worth giving them time to play before disucssing what might be a more efiicient method.
Rather than squaring terms on a need-to-know basis - how about suggesting working them all out to save time re-calculating expansions as needed.