This is known as the **quadratic functional equation**. The general solution over $ \mathbb{R} $, assuming regularity (e.g., continuity or boundedness), is $ f(x) = cx^2 $. We verify this form.

["The Quadratic Functional Equation: Understanding Its General Solution", "The quadratic functional equation stands as a foundational result in functional analysis and number theory, describing functions satisfying a specific relationship involving squares. Known formally as:", "[\nf(x + y) = f(x) + f(y) + 2xy \quad \ ext{for all } x, y \in \mathbb{R},\n]", "this elegant equation captures the structure of functions quadratic in nature. Assuming standard regularity conditions—such as continuity, boundedness, or differentiability—this equation uniquely determines the solution:", "[\nf(x) = cx^2 \quad \ ext{for some constant } c \in \mathbb{R}.\n]", "In this article, we explore the derivation and verification of this general solution, highlighting its significance and applicability.", "---", "### Understanding the Quadratic Functional Equation", "At first glance, the equation combines additive structure (the left-hand side) with a symmetric bilinear term (the (2xy) on the right). This form arises naturally when analyzing functional relationships tied to quadratic growth or symmetric transformations in one variable. It serves as a cornerstone in models involving area forms, quadratic mappings, and harmonic analysis.", "The presence of the degree-2 symmetric term (2xy) strongly suggests a quadratic dependence on (x), motivating the guess (f(x) = c x^2).", "---", "### Deriving the General Solution", "To verify that (f(x) = cx^2) satisfies the equation, substitute into both sides:", "Left side:\n[\nf(x + y) = c(x + y)^2 = c(x^2 + 2xy + y^2) = cx^2 + 2cxy + cy^2\n]", "Right side:\n[\nf(x) + f(y) + 2xy = cx^2 + cy^2 + 2xy\n]", "Both sides match exactly when (f(x) = cx^2), confirming it is indeed a solution.", "But is this the only solution under reasonable regularity assumptions?", "---", "### Uniqueness under Regularity Conditions", "To prove uniqueness, suppose (f : \mathbb{R} \ o \mathbb{R}) is continuous (or bounded on an interval, or differentiable). Consider a change of variable: define (g(x) = f(x) - c x^2). Choosing (c = f(1)), then (g(1) = 0).", "Substitute into the original equation:", "[\ng(x+y) + c(x+y)^2 = (g(x) + c x^2) + (g(y) + c y^2) + 2xy\n]", "Expand and simplify:", "[\ng(x+y) + c(x^2 + 2xy + y^2) = g(x) + g(y) + c(x^2 + y^2) + 2xy\n]", "[\ng(x+y) = g(x) + g(y) - 2cxy + 2xy\n]", "[\ng(x+y) = g(x) + g(y) + 2(c - 1)xy\n]", "Now, suppose (c = 1), which holds if (f(1) = 1); otherwise, the term (2(c - 1)xy) introduces a non-quadratic correction unless (c = 1). But under continuity, the only way to eliminate the uncontrolled bilinear term is (c = 1). Therefore, (g(x+y) = g(x) + g(y)), and by Cauchy’s functional equation with continuity, (g(x) = kx) for some constant (k).", "However, substituting back into the original structure forces (k = 0): suppose (f(x) = kx + x^2). Then:", "[\nf(x+y) = k(x+y) + (x+y)^2 = kx + ky + x^2 + 2xy + y^2\n]", "[\nf(x) + f(y) + 2xy = (kx + x^2) + (ky + y^2) + 2xy = kx + ky + x^2 + y^2 + 2xy\n]", "These match, so linear terms seem allowed. Wait — this contradicts uniqueness?", "Not quite: in the above verification, the added term (kx) survives, yet the functional equation still holds. This suggests that if no regularity is imposed, solutions include all functions of the form:", "[\nf(x) = x^2 + kx\n]", "But the problem specifies regularity assumptions such as continuity or boundedness — conditions that eliminate pathologies.", "With continuity, applications of Cauchy’s theorem force (g) to be linear, but the cross term (2(c - 1)xy) must vanish globally, which only happens if (c = 1). Therefore, under continuity, the only consistent solutions are those with (g(x) = 0), so:", "[\nf(x) = x^2 + kx \quad \ ext{only if } c = 1\n]", "But wait — re-examining the derivation: if (c <br/>\ne 1), the error term (2(c - 1)xy) is not constant and cannot vanish unless (c = 1). Hence, only when (c = 1) does the functional equation reduce to pure quadratic behavior.", "Thus, under continuity (or boundedness, or differentiability), the only solutions are:", "[\nf(x) = x^2 + kx \quad \ ext{is not valid unless } k = 0 \ ext{ and } c = 1\n]", "Wait — correction: this suggests non-quadratic solutions violate the equation unless controlled.", "Actually, more carefully: assume (f(x) = x^2 + kx). Then insert into:", "[\nf(x+y) - f(x) - f(y) - 2xy = (x+y)^2 + k(x+y) - x^2 - kx - y^2 - ky - 2xy = x^2 + 2xy + y^2 + kx + ky - x^2 - y^2 - kx - ky - 2xy = 0\n]", "It works with any (k)! So (f(x) = x^2 + kx) satisfies:", "[\nf(x+y) = f(x) + f(y) + 2xy\n]", "But this contradicts prior reasoning? Where did we go wrong?", "Ah — critical realization: the original equation is:", "[\nf(x+y) = f(x) + f(y) + 2xy\n]", "Then defining (g(x) = f(x) - c x^2), we get:", "[\ng(x+y) = g(x) + g(y)\n]", "So linear functions do satisfy this equation — without needing (c = 1). In fact, any function of the form:", "[\nf(x) = x^2 + g(x), \quad \ ext{where } g: \mathbb{R} \ o \mathbb{R} \ ext{ is additive}\n]", "satisfies the equation.", "But under regularity conditions such as continuity or boundedness on an interval, additive functions are linear: (g(x) = kx).", "Hence, under continuity:", "[\nf(x) = x^2 + kx\n]", "Thus, the general regular solution is (f(x) = x^2 + kx), not (f(x) = c x^2).", "But in classical functional equation literature, the symmetric quadratic equation:", "[\nf(x+y) - f(x) - f(y) = 2xy\n]", "has general solution (f(x) = x^2 + g(x)), with (g) additive. Under continuity, (g(x) = kx), so:", "[\nf(x) = x^2 + kx\n]", "But the problem states “the general solution is (f(x) = c x^2)”, implying no linear term.", "This suggests a special case: when we force symmetry or set (c = 1), or when additional constraints remove the linear part.", "However, standard references confirm: without regularity, solutions include all (f(x) = x^2 + kx). With continuity, still (f(x) = x^2 + kx); but uniqueness up to linear terms holds.", "Therefore, the statement that “the general solution is (f(x) = c x^2)" is inaccurate unless interpreted as only the quadratic homogeneous solution, ignoring additive perturbations.", "But reconsider: suppose we assume (f) is a quadratic function, meaning continuous and of the form (f(x) = ax^2 + bx + c). Plug into:", "[\nf(x+y) = a(x+y)^2 + b(x+y) + c = a x^2 + 2a xy + a y^2 + b x + b y + c\n]", "[\nf(x) + f(y) + 2xy = a x^2 + b x + c + a y^2 + b y + c + 2xy = a x^2 + a y^2 + b x + b y + 2c + 2xy\n]", "Equate:", "[\na x^2 + 2a xy + a y^2 + b x + b y + c = a x^2 + a y^2 + b x + b y + 2c + 2xy\n]", "Cancel terms:", "[\n2a xy + c = 2xy + 2c\n]", "Thus:", "[\n2a = 2 \implies a = 1, \quad c = 2c \implies c = 0\n]", "So under the assumption that (f) is quadratic and continuous, we must have (c = 0) and (b) arbitrary — but wait, the (b x + b y) matches, and no constraint on (b) from this equation?", "No: coefficient of (xy): left: (2a = 2"]