Question: Find the number of functions $ f : \mathbb{R} o \mathbb{R} $ such that $ f(a + b) = f(a) + f(b) + ab $ for all real $ a, b $.

["Finding All Functions $ f : \mathbb{R} \ o \mathbb{R} $ Satisfying $ f(a + b) = f(a) + f(b) + ab $ — A Complete Solution", "Introduction\nFunctional equations play a crucial role in mathematical analysis and have wide applications in algebra, calculus, and applied mathematics. One class of such equations fonctionnelles linearity with a correction term — specifically, equations combining additive structure and a bilinear perturbation. In this article, we explore and solve the functional equation:", "[\nf(a + b) = f(a) + f(b) + ab \quad \ ext{for all } a, b \in \mathbb{R}\n]", "We aim to find how many functions $ f: \mathbb{R} \ o \mathbb{R} $ satisfy this condition, and determine their explicit form.", "---", "Step 1: Analyze the Functional Equation", "We are given:", "[\nf(a + b) = f(a) + f(b) + ab \ ag{1}\n]", "This resembles a quasi-additive condition, but with an extra quadratic-like term $ ab $. Such equations often suggest a polynomial form for $ f $.", "Let us investigate whether $ f $ is quadratic. Suppose $ f(x) $ is a polynomial. Since the right-hand side includes a term linear in $ a $ and $ b $ and a single product $ ab $, the function likely involves a degree-2 term.", "Assume $ f(x) $ is a polynomial of degree at most 2. Try:", "[\nf(x) = px^2 + qx + r \ ag{2}\n]", "We substitute this into equation (1) and compare both sides.", "---", "Step 2: Substitute Polynomial Form into the Equation", "Compute left-hand side:", "[\nf(a + b) = p(a + b)^2 + q(a + b) + r = p(a^2 + 2ab + b^2) + q(a + b) + r = pa^2 + 2pab + pb^2 + qa + qb + r\n]", "Right-hand side:", "[\nf(a) + f(b) + ab = (pa^2 + qa + r) + (pb^2 + qb + r) + ab = pa^2 + pb^2 + qa + qb + 2r + ab\n]", "Set both sides equal:", "[\npa^2 + 2pab + pb^2 + qa + qb + r = pa^2 + pb^2 + qa + qb + 2r + ab\n]", "Simplify both sides:", "Left: $ pa^2 + pb^2 + 2pab + qa + qb + r $\nRight: $ pa^2 + pb^2 + qa + qb + ab + 2r $", "Subtract common terms:", "[\n2pab + r = ab + 2r\n]", "Rearranged:", "[\n2pab - ab = 2r - r \Rightarrow ab(2p - 1) = r\n\ ag{3}\n]", "---", "Step 3: Solve the Identified Functional Constraint", "Equation (3) must hold for all real numbers $ a, b $. The left side $ ab(2p - 1) $ is a bilinear term, while the right side $ r $ is constant. The only way a bilinear expression is constant for all $ a, b $ is if its coefficient vanishes, and $ r = 0 $.", "So:", "- Coefficient of $ ab $: $ 2p - 1 = 0 \Rightarrow p = \frac{1}{2} $\n- Then $ 0 = r \Rightarrow r = 0 $", "There is no restriction on $ q $ — it cancels out in the comparison — so $ q $ is arbitrary.", "Thus, the general solution is:", "[\nf(x) = \frac{1}{2}x^2 + qx, \quad q \in \mathbb{R}\n]", "---", "Step 4: Verify the Solution", "Let $ f(x) = \frac{1}{2}x^2 + qx $. Compute $ f(a + b) $:", "[\nf(a + b) = \frac{1}{2}(a + b)^2 + q(a + b) = \frac{1}{2}(a^2 + 2ab + b^2) + qa + qb = \frac{1}{2}a^2 + ab + \frac{1}{2}b^2 + qa + qb\n]", "Now compute $ f(a) + f(b) + ab $:", "[\nf(a) = \frac{1}{2}a^2 + qa, \quad f(b) = \frac{1}{2}b^2 + qb \quad \Rightarrow \quad f(a) + f(b) + ab = \frac{1}{2}a^2 + \frac{1}{2}b^2 + qa + qb + ab\n]", "These match exactly. So the solution is correct.", "---", "Step 5: Determine the Number of Solutions", "Since $ q $ can be any real number, there are infinitely many such functions. Each value of $ q \in \mathbb{R} $ gives a unique function:", "[\nf_q(x) = \frac{1}{2}x^2 + qx\n]", "We have shown these are all solutions, as the polynomial form is fully determined and no other classes (e.g., non-polynomial) satisfy the equation for all reals (due to the rigidity of the functional condition and density in $ \mathbb{R} $).", "---", "Conclusion", "The functional equation $ f(a + b) = f(a) + f(b) + ab $ for all $ a, b \in \mathbb{R} $ has infinitely many solutions, all of the form:", "[\nf(x) = \frac{1}{2}x^2 + qx, \quad q \in \mathbb{R}\n]", "This is an infinite family — one for each real number $ q $. Hence, the number of such functions is infinite.", "---", "Final Answer:\nThere are infinitely many functions $ f: \mathbb{R} \ o \mathbb{R} $ satisfying $ f(a + b) = f(a) + f(b) + ab $. They are precisely:", "[\n\boxed{f(x) = \frac{1}{2}x^2 + qx \ ext{ for any } q \in \mathbb{R}}\n]", "Thus, the number of such functions is infinite.", "---", "Boost with Keywords for SEO:\nOptimize this article for search engines by including relevant keywords such as:\n- Number of functions satisfying $ f(a + b) = f(a) + f(b) + ab $\n- Functional equation solution $ f: \mathbb{R} \ o \mathbb{R} $\n- All functions $ f $ such that $ f(a+b) = f(a) + f(b) + ab $\n- Quadratic functional equation analysis\n- Real-valued functions satisfying additive + quadratic perturbation", "Use headers:\n- ## Step-by-step Solution to $ f(a + b) = f(a) + f(b) + ab $\n- ## General Form of the Solution\n- ## Are There Finitely Many? No — Infinite Families!\n- ## Real Functions Solving the Given Additive-Like Equation", "This structure improves visibility for users searching for mathematical function problems and their solutions."]