But [x] = x for all integers x, so the integer solutions of [x][y] = x + y are the same as the integer solutions of xy = x + y, (x, y) in {(0, 0), (2, 2)}.
first, I must say that "Guessing" (and then checking your guess) is a perfectly "good and proper" method! For n an integer, [n]= n so your equation is simply xy= x+ y. You can write this as xy- y= (x-1)y= x or, if x is not 1, y= x/(x-1). If x is not 0, that says x-1 divides x. The only integer x such that x-1 is a factor of x, is x= 2. You can then check x= 1 or x= 0 separately: If x= 1, xy= x+ y becomes y= 1+ y which is never true.; If x= 0, then xy= x+ y becomes 0= 0+ y which is true for y= 0. The only only solutions are x= y= 0 and x= y= 2.
We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling We Value Civility
• Positive and compassionate attitudes
• Patience while debating We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving