Question

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

Answer 1

1st number = n
2nd number = n+1
3rd number = n+2

sum of the squares of 3 consecutive numbers is 116

n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116  Last option.