Maths Olympiad Prep

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, 2002

Algebra Difficulty 4.9 AIME Prove it Baltic Way

Problem:
Find all pairs (a,b)(a, b) of positive rational numbers such that
a+b=2+3. \sqrt{a} + \sqrt{b} = \sqrt{2 + \sqrt{3}}.

Solution

Solution:
Squaring both sides of the equation gives
a+b+2ab=2+3 a + b + 2 \sqrt{a b} = 2 + \sqrt{3}
so 2ab=r+32 \sqrt{a b} = r + \sqrt{3} for some rational number rr. Squaring both sides of this gives 4ab=r2+3+2r34 a b = r^2 + 3 + 2 r \sqrt{3}, so 2r32 r \sqrt{3} is rational, which implies r=0r = 0. Hence ab=3/4a b = 3 / 4 and substituting this into (8) gives a+b=2a + b = 2. Solving for aa and bb gives (a,b)=(12,32)(a, b) = \left(\frac{1}{2}, \frac{3}{2}\right) or (a,b)=(32,12)(a, b) = \left(\frac{3}{2}, \frac{1}{2}\right).

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