Solution: Assume $ f $ is quadratic: $ f(x) = ax^2 + bx + c $. Substitute into equation: $ a(x + y)^2 + b(x + y) + c = ax^2 + bx + c + ay^2 + by + c + 2xy $. Expand left: $ ax^2 + 2axy + ay^2 + bx + by + c $. Right: $ ax^2 + ay^2 + bx + by + 2c + 2xy $. Equate coefficients: $ 2a = 2 \Rightarrow a = 1 $, and $ 2c = c \Rightarrow c = 0 $. Thus, $ f(x) = x^2 + bx $. Check: $ (x + y)^2 + b(x + y) = x^2 + y^2 + 2xy + bx + by $, which matches. Any $ b $ works, so infinitely many solutions. \boxed\infty

📅 March 1, 2026 👤 scraface

Mar 01, 2026