MHB Prove $x=-y$: A Math Challenge

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The discussion centers on proving that if \( (\sqrt{y^{2} - x} - x)(\sqrt{x^{2} + y} - y) = y \), then \( x = -y \). Participants suggest substituting \( -y \) for \( x \) in the equation to simplify and verify the equality. The identity \( (a+b)(a-b) = a^2 - b^2 \) is highlighted as a useful tool for the proof. Some participants clarify that demonstrating \( x = -y \) is not the same as proving the original statement. The conversation emphasizes the need for careful interpretation of the mathematical challenge.
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Let $x, y$ be real numbers such that
$$(\sqrt{y^{2} - x\,\,}\, - x)(\sqrt{x^{2} + y\,\,}\, - y)=y.$$
Prove $x=-y$.

Any suggestion would be appreciated.
 
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Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y

Writing it is this way it is more obvious that all you have to do is substitute (-y) for each of the x's in the left hand side and then simplify to show that the left hand side is equal to y.

To do this it will be helpful to remember the identity:
(a+b)(a-b)=(a^2-b^2)
 
Kiwi said:
Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y
No, this is the converse of what the original question is asking.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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