Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Romania

Prove that the equation
$\frac{1}{\sqrt{x} + \sqrt{1006}} + \frac{1}{\sqrt{2012 - x} + \sqrt{1006}} = \frac{2}{\sqrt{x} + \sqrt{2012 - x}}$
has 2013 integer solutions.

Solution

One can easily check that the given relation holds for any admissible value of $x$ . Since the number $x$ is subject to the conditions $0 \le x \le 2012$ , the conclusion is easily reached.

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