Each of x and y is a real number that satisfy this equation:
• {x+√(1+x2)}{y+√(1+y2)}= 1
Determine the value of (x+y)2
The first thing that came to my mind was to multiply both sides by the conjugate of the x-term: -x+sqrt[1+x^2].
Then we get -x+sqrt[1+x^2] = y+sqrt[1+y^2]. From this it is obvious that -x=y is a solution.
Then (x+y)^2 = 0.
But what if we missed a second branch? So lets just brute force it, starting with where we left off:
-x+sqrt[1+x^2] = y+sqrt[1+y^2]
sqrt[1+x^2] - sqrt[1+y^2] = x+y
(sqrt[1+x^2] - sqrt[1+y^2])^2 = (x+y)^2
2 + x^2 + y^2 - 2sqrt[1+x^2+y^2+(xy)^2] = x^2 + 2xy + y^2
-sqrt[1+x^2+y^2+(xy)^2] = xy - 1
(sqrt[1+x^2+y^2+(xy)^2])^2 = (xy - 1)^2
1+x^2+y^2+(xy)^2 = 1 - 2xy + (xy)^2
x^2+y^2 = -2xy
(x+y)^2 = 0.
Solution
