Question

The sum of the squares of 3 consecutive positive integers is 116. What are the numbers?

Answer 1

Let's say, the first number is "n"

well, the consecutive number of "n" is "n + 1"and the consecutive number of "n+1" is "n + 1" +1 or n + 2

so, the numbers are (n), (n+1) and (n+2), whatever "n" is

now, the sum of their squares is 116

$\bf (n)^2+(n+1)^2+(n+2)^2=116$

expand the binomials by FOIL or binomial theorem, and then simplify