Question 362881
one of the classics..observe..
Let's say x,x+1,x+2 for the consecutive integers..

x*(x+2)-(x+1) = 3*(x+2)+1

x^2+2x-x-1 = 3x+6+1

x^2-2x-8 = 0

(x-4)*(x+2) = 0

x={4,-2}

You must choose as x = 4 because the integers are positive..
So that,4,5,6 are the integers..

RF.