2a x^2 + 2a y^2 + 2b x + 2c = 2a x^2 + 2a y^2 + 2b x + 2b y + 4c.

February 22, 2026

2a x^2 + 2a y^2 + 2b x + 2c = 2a x^2 + 2a y^2 + 2b x + 2b y + 4c.

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5Question: Find all functions $ f: \mathbb{R} o \mathbb{R} $ such that $ f(x + y) + f(x - y) = 2f(x) + 2f(y) $ for all real numbers $ x $ and $ y $.
Solution: Assume $ f $ is a quadratic function. Let $ f(x) = ax^2 + bx + c $. Substituting into the equation:
a(x + y)^2 + b(x + y) + c + a(x - y)^2 + b(x - y) + c = 2(ax^2 + bx + c) + 2(ay^2 + by + c).
Equating coefficients: $ 2c = 2b y + 4c $ for all $ y $, which implies $ b = 0 $ and $ c = 0 $. Thus, $ f(x) = ax^2 $. Verification confirms this satisfies the equation. Hence, all solutions are quadratic functions of the form $ f(x) = kx^2 $.
oxed{f(x) = kx^2 ext{ for some constant } k}
Question: Let $ a $ and $ b $ be complex numbers such that $ |a| = |b| = 1 $ and $ a

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