how to solve (x + y + z)2
Answer (1)
Answer:Assume x≥y≥z. Some people define that natural numbers include zero, but if one number is 0, then the left side is ≥0, and the right is −2, so x≥y≥z≥1.yz cannot be 1 because x+y+z=x+2,xyz−2=x−2. If yz=2 then x+y+z=x+3,xyz−2=2x−2, so x=5, solution (5,2,1). If yz=3 then x+y+z=x+4,xyz−2=3x−2,x=3, solution (3,3,1).The left hand side is ≤3x. 3x≥xyz−2 means 2≥x×(yz−3). If yz=4 then x≤2 which only happens if x=y=z=2. If yz≥5 then x≤1 which cannot happen because x≥y≥z. Therefore (5,2,1),(3,3,1) and (2,2,2) are the only solutions.So in this case setting yz to a constant value instead of one variable worked quite well.
