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Algebra Difficulty 4.8 AIME Prove it Estonia

Do there exist numbers aa, bb, cc that satisfy the equation
2a(c−a)−b(2a+b)+c(2b−c)=2020? 2a(c-a) - b(2a+b) + c(2b-c) = 2020?

Solution

Transforming the l.h.s. of the equation gives
2a(c−a)−b(2a+b)+c(2b−c)=2ac−2a2−2ab−b2+2bc−c2=−a2−(a+b−c)2. \begin{aligned} 2a(c-a) - b(2a+b) + c(2b-c) &= 2ac - 2a^2 - 2ab - b^2 + 2bc - c^2 \\ &= -a^2 - (a+b-c)^2. \end{aligned}
The equality −a2−(a+b−c)2=2020-a^2 - (a+b-c)^2 = 2020 cannot be valid since all terms of its l.h.s. are non-positive whereas the r.h.s. is positive.

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