Question: A function $ f: \mathbb{R} \to \mathbb{R} $ satisfies $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $ for all real $ a, b $. If $ f(1) = 1 $, find $ f(2) $.

["Title: Solving the Functional Equation: Finding $ f(2) $ Given $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $ and $ f(1) = 1 $", "Meta Description:\nExplore the functional equation $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $, solve for $ f(2) $ given $ f(1) = 1 $, and uncover the quadratic nature of $ f $.", "---", "### Introduction", "Functional equations often appear in advanced algebra, olympiad problems, and mathematical modeling. One particularly elegant equation is:", "[\nf(a + b) + f(a - b) = 2f(a) + 2f(b) \quad \ ext{for all } a, b \in \mathbb{R}\n]", "Given this condition and the value $ f(1) = 1 $, we are to determine $ f(2) $ — all while uncovering the general form of $ f $. This problem reveals deep structure behind seemingly simple equations.", "---", "### Step 1: Recognizing the Functional Form", "This equation resembles identities satisfied by quadratic functions. Suppose $ f(x) = cx^2 $. Let's test this hypothesis.", "Compute the left-hand side (LHS):", "[\nf(a + b) + f(a - b) = c(a + b)^2 + c(a - b)^2 = c(a^2 + 2ab + b^2 + a^2 - 2ab + b^2) = c(2a^2 + 2b^2) = 2c a^2 + 2c b^2\n]", "Right-hand side (RHS):", "[\n2f(a) + 2f(b) = 2(c a^2) + 2(c b^2) = 2c a^2 + 2c b^2\n]", "Since LHS = RHS, $ f(x) = cx^2 $ satisfies the equation for any constant $ c $. Thus, all quadratic functions $ f(x) = cx^2 $ are solutions to this functional equation (assuming continuity or mild regularity, which is standard in olympiad contexts).", "---", "### Step 2: Using the Given Value $ f(1) = 1 $", "We are told $ f(1) = 1 $. Plug into $ f(x) = cx^2 $:", "[\nf(1) = c(1)^2 = c = 1 \Rightarrow c = 1\n]", "Therefore, the unique function satisfying both the equation and the condition is:", "[\nf(x) = x^2\n]", "---", "### Step 3: Compute $ f(2) $", "Now, substitute $ x = 2 $:", "[\nf(2) = (2)^2 = 4\n]", "---", "### Additional Insight: Why Only Quadratic?", "Suppose $ f $ is twice differentiable. Differentiate both sides of the equation with respect to $ b $, treating $ a $ fixed:", "Differentiate once:", "[\nf'(a + b) - f'(a - b) = 2f'(b)\n]", "Differentiate again with respect to $ b $:", "[\nf''(a + b) + f''(a - b) = 2f''(b)\n]", "Now set $ b = 0 $:", "[\nf''(a) + f''(a) = 2f''(0) \Rightarrow 2f''(a) = 2f''(0) \Rightarrow f''(a) = f''(0)\n]", "So $ f''(a) $ is constant, meaning $ f'(a) $ is linear, and $ f(a) $ is quadratic. Hence, $ f(x) = cx^2 + dx + e $.", "But plugging $ f(x) = cx^2 + dx + e $ into the original equation shows $ d = 0 $, because linear terms do not cancel properly unless eliminated. Thus, only $ f(x) = cx^2 $ satisfies the equation.", "With $ f(1) = 1 $, $ c = 1 $, so $ f(x) = x^2 $, and $ f(2) = 4 $.", "---", "### Conclusion", "The functional equation $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $ has quadratic solutions, and with $ f(1) = 1 $, the unique solution is $ f(x) = x^2 $. Thus,", "[\nf(2) = 2^2 = 4\n]", "This functional equation appears in contexts ranging from algebraic number theory to physics and computer science — making $ f(2) = 4 $ not just a number, but a gateway to deeper mathematical insight.", "---", "Keywords: functional equation, $ f(a + b) + f(a - b) = 2f(a) + 2f(b) $, solution $ f(x) = cx^2 $, $ f(1) = 1 $, find $ f(2) $, quadratic functions, olympiad math", "ANC Tags: #FunctionalEquations #MathOlympiad #Algebra #QuadraticFunctions #FunctionalAnalysis #OlympiadProblem", "---", "> "Math reveals patterns — even in symmetry. This equation echoes the law of quadratic invariance.""]