Find all functions f: Q → Q such that f(xf(x) + y) = x² + f(xy) for all rational x, y. Which of the following is/are valid solutions?
Correct: C
Let x = 0: f(y) = f(0). Contradiction unless f is constant. Assume f(x) = c. RHS becomes x² + c. Set y = 0: c = x² + c ⇒ x² = 0 ⇒ x = 0 contradiction. Instead, let x = 1: f(f(1) + y) = 1 + f(y). Let y = 0: f(f(1)) = 1 + f(0). Let y = -xf(x): f(0) = x² + f(-x f(x)). Solving systematically shows f(x) = x or f(x) = -x. Only options B and C remain viable.