Find three consecutive integers such that the sum of their squares of 77?

Answer 1

let first number be x. then other two are x+1 and x+2 respectively. now x^2 +(x+2)^2 +(x+1)^2 = 77 3x^2+6x+5=77 3x^2+6x-72=0 x^2+2x-24=0 (x+6)(x-4)=0 hence x=-6 or 4 hence numbers are -6,-5,-4 or 4,5,6 From Dhruv