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Ukrainian National Mathematical Olympiad - Third Round

Algebra Difficulty 5.0 AIME Prove it Ukraine

**Can real numbers xx, yy, zz satisfy**
1(x−y)(x+y)+1(y−z)(y+z)+1(z−x)(z+x)=0? \frac{1}{(x-y)(x+y)} + \frac{1}{(y-z)(y+z)} + \frac{1}{(z-x)(z+x)} = 0?

Solution

Denote a=x2−y2a = x^2 - y^2, b=y2−z2b = y^2 - z^2, then −a−b=z2−x2-a - b = z^2 - x^2 and the equation can be rewritten as:
1a+1b=1a+b⇔(a+b)2=ab⇔a2−ab+b2=0⇔(a−b2)2+3b24=0. \frac{1}{a} + \frac{1}{b} = \frac{1}{a+b} \Leftrightarrow (a+b)^2 = ab \Leftrightarrow a^2 - ab + b^2 = 0 \Leftrightarrow \left(a - \frac{b}{2}\right)^2 + \frac{3b^2}{4} = 0.
The last equality holds only with a=b=0a = b = 0, which is impossible. Therefore, the equation has no solutions.

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