Question: For all real numbers $ x $ and $ y $, find the number of functions $ f : \mathbb{R} o \mathbb{R} $ such that $ f(x + y) = f(x) + f(y) + 2xy $.

["Question:\nFor all real numbers $ x $ and $ y $, find the number of functions $ f : \mathbb{R} \ o \mathbb{R} $ such that\n$$ f(x + y) = f(x) + f(y) + 2xy $$?", "---", "Understanding the Functional Equation", "We are given a functional equation for all real numbers $ x, y $:\n$$\nf(x + y) = f(x) + f(y) + 2xy\n$$\nThis is a Cauchy-type functional equation with an additional quadratic term. Our goal is to determine how many such functions $ f: \mathbb{R} \ o \mathbb{R} $ exist that satisfy this identity.", "---", "Step 1: Try a Polynomial Ansatz", "Since the right-hand side includes a quadratic term $ 2xy $, it's natural to suppose that $ f $ might be a quadratic function. Let us assume:\n$$\nf(x) = ax^2 + bx + c\n$$\nWe will substitute this into the functional equation and verify when equality holds.", "Compute left-hand side:\n$$\nf(x + y) = a(x + y)^2 + b(x + y) + c = a(x^2 + 2xy + y^2) + b(x + y) + c = ax^2 + ay^2 + 2axy + bx + by + c\n$$", "Now compute right-hand side:\n$$\nf(x) + f(y) + 2xy = (ax^2 + bx + c) + (ay^2 + by + c) + 2xy = ax^2 + ay^2 + bx + by + 2c + 2xy\n$$", "Set both sides equal:\n$$\nax^2 + ay^2 + 2axy + bx + by + c = ax^2 + ay^2 + bx + by + 2c + 2xy\n$$", "Cancel common terms from both sides:\n$$\n2axy + c = 2xy + 2c\n$$", "Rearranging:\n$$\n2axy - 2xy = 2c - c \Rightarrow 2xy(a - 1) = c\n$$", "This identity must hold for all real $ x $ and $ y $, which is only possible if both sides are identically zero. Otherwise, the left side varies with $ x, y $ while the right side is constant.", "So, for all $ x, y $:\n$$\n2xy(a - 1) = c\n$$", "- If $ a <br/>\ne 1 $, the left-hand side depends on $ x $ and $ y $, but the right-hand side is constant — contradiction unless $ c = 0 $ and $ a = 1 $, but even then expression becomes 0 = 0 only if $ c = 0 $.\n- If $ a = 1 $, then left-hand side becomes $ 0 $, so we get $ c = 0 $.", "Thus, necessary conditions:\n- $ a = 1 $\n- $ c = 0 $", "So the general quadratic candidate reduces to:\n$$\nf(x) = x^2 + bx\n$$\nwith $ b \in \mathbb{R} $ arbitrary so far.", "---", "Step 2: Check if All Such Functions Satisfy the Equation", "Let $ f(x) = x^2 + bx $. Compute both sides.", "Left-hand side:\n$$\nf(x + y) = (x + y)^2 + b(x + y) = x^2 + 2xy + y^2 + bx + by\n$$", "Right-hand side:\n$$\nf(x) + f(y) + 2xy = (x^2 + bx) + (y^2 + by) + 2xy = x^2 + y^2 + bx + by + 2xy\n$$", "These match exactly. Therefore, for every real number $ b $, $ f(x) = x^2 + bx $ satisfies the functional equation.", "---", "Step 3: Are There Any Other Solutions?", "We are working over $ \mathbb{R} $, and the equation is a functional equation involving real numbers. Under mild regularity conditions (like continuity, or measurability, or boundedness on an interval), the only solutions to\n$$\nf(x + y) = f(x) + f(y) + 2xy\n$$\nare the quadratic functions found above.", "Even without assuming continuity a priori, it is known (from the theory of Cauchy-type equations) that all solutions over $ \mathbb{R} $ to this equation are precisely the functions of the form\n$$\nf(x) = x^2 + bx\n$$\nfor some constant $ b \in \mathbb{R} $.", "Why? The equation defines $ f $ uniquely at all $ x $ via additive extensions, and $ x^2 $ is the unique quadratic solution. The linear terms absorb the rest.", "---", "Step 4: Count the Number of Such Functions", "Each choice of $ b \in \mathbb{R} $ gives a distinct function $ f(x) = x^2 + bx $. Since there are uncountably many real numbers $ b $, there are infinitely many such functions — specifically, a continuum many.", "But the question asks: "find the number" — and in math olympiad contexts, when infinitely many solutions exist, the answer is often phrased precisely.", "However, note: the functional equation is defined for all real $ x, y $, and we are to find how many functions satisfy it. Since $ b $ is arbitrary real, the solution set is in bijection with $ \mathbb{R} $, so the number is infinite.", "But more precisely: there are infinitely many (uncountably many) such functions.", "Wait — but sometimes “number” implies cardinality. However, in olympiad-style problems like this, if the solution set is a family parameterized by one real parameter, the expected answer is that there are infinitely many, but the question may expect a specific finite count if only discrete solutions exist.", "But from our analysis: $ f(x) = x^2 + bx $, $ b \in \mathbb{R} $, so solution set is $ { f_b(x) = x^2 + bx \mid b \in \mathbb{R} } $ — a one-parameter family.", "Hence, there are infinitely many such functions — in fact, continuum many.", "But let us double-check: is it possible only one? No. $ b = 0 $ gives $ f(x) = x^2 $, $ b = 1 $ gives $ f(x) = x^2 + x $, etc. All satisfy.", "But is there a constraint we missed?", "Wait: suppose we try $ x = 0 $. Set $ x = 0, y = 0 $:\n$$\nf(0) = f(0) + f(0) + 0 \Rightarrow f(0) = 2f(0) \Rightarrow f(0) = 0\n$$", "Our solutions give $ f(0) = 0^2 + b\cdot 0 = 0 $ — good.", "No restriction on $ b $. So $ b $ is arbitrary.", "Thus, the number of such functions is equal to the cardinality of $ \mathbb{R} $.", "But in olympiad problems, answers are usually boxed integers or radicals — so perhaps reconsider: is the solution truly one-parameter?", "Wait — could there be only one solution?", "No — $ b $ is arbitrary. But let us test with specific values.", "Let $ x = y = 1 $:\nLeft: $ f(2) = 4 + 2b $\nRight: $ f(1) + f(1) + 2(1)(1) = (1 + b) + (1 + b) + 2 = 2 + 2b + 2 = 4 + 2b $ — works.", "Any $ b \in \mathbb{R} $ works.", "Hence, the solution set is parameterized by $ \mathbb{R $.", "Therefore, the number of such functions is infinite — in fact, uncountably infinite.", "But perhaps the question expects the form of the solution and a count in terms of basis.", "However, standard interpretation: since $ b $ is arbitrary real, and each $ b $ gives a distinct function, the number is infinite.", "But in strict mathematical terms, the solution set has cardinality $ \mathfrak{c} $, the continuum.", "But in olympiad context, if the answer is “infinitely many”, and no finite number fits, but often such questions expect “uncountably many” or “infinite”.", "But let us recall: in functional equations over $ \mathbb{R} $, the general solution to\n$$\nf(x+y) = f(x) + f(y) + kxy\n$$\nis\n$$\nf(x) = \frac{k}{2}x^2 + bx\n$$\nExactly our case with $ k = 2 $, so $ f(x) = x^2 + bx $.\nThus, one quadratic parameter.", "Conclusion: There is a one-parameter family of solutions, so infinitely many functions satisfy the equation.", "But the question is: Find the number — so the answer is infinitely many.", "But can we write that in boxed form?", "In olympiads, sometimes “infinitely many” is acceptable, or we box $ \infty $, though not standard.", "But let’s reconsider: is there a possibility that only one function satisfies?", "Suppose $ f $ is differentiable — then we can derive the form via calculus.", "Let us take partial derivatives (formally). Set $ y $ small and consider increment.", "Differentiate both sides with respect to $ y $:\nLHS: $ \frac{d}{dy} f(x+y) = f'(x + y) $\nRHS: $ \frac{d}{dy} [f(x) + f(y) + 2xy] = f'(y) + 2x $", "Set $ y = 0 $:\n$ f'(x) = f'(0) + 2x $", "Integrate:\n$ f(x) = f'(0)x + x^2 + C $", "But $ f(0) = 0 $ (from earlier), plug $ x = 0 $:\n$ 0 = 0 + 0 + C \Rightarrow C = 0 $", "So $ f(x) = x^2 + bx $, $ b = f'(0) $", "Same result.", "So no restriction. $ b $ free.", "Thus, infinitely many functions.", "But wait — is that correct? Yes.", "However, let us check with $ f(x) = x^2 $. Then:\n$ f(x+y) = (x+y)^2 = x^2 + 2xy + y^2 $\n$ f(x) + f(y) + 2xy = x^2 + y^2 + 2xy $ — equal.", "Now $ f(x) = x^2 + 3x $:\n$ f(x+y) = (x+y)^2 + 3(x+y) = x^2 + 2xy + y^2 + 3x + 3y $\n$ f(x)+f(y)+2xy = x^2 + 3x + y^2 + 3y + 2xy $ — same.", "So yes, all work.", "Therefore, the number of such functions is infinite, specifically uncountably infinite.", "But how to express this in boxed form?", "In math olympiads, if the answer is “infinitely many”, and cannot be a finite number, it is often phrased as so.", "But perhaps the expected answer is the dimension of the solution space? That is 1 — but “number of functions” means cardinality.", "Alternatively, maybe only one if we consider symmetry?", "No — $ f(x) = x^2 + bx $ are distinct.", "Wait — could $ b = 0 $ be forced?", "No — no condition eliminates $ b $.", "Unless $ f $ must be bounded or continuous — but even then, under mild conditions, only quadratic solutions exist.", "But without additional constraints, uncountably many.", "But let us suppose that the only solution is $ f(x) = x^2 $. Is that true?", "Counterexample: $ f(x) = x^2 + 5 $?\nCheck: $ f(x+y) = (x+y)^2 + 5 = x^2 + 2xy + y^2 + 5 $\n$ f(x) + f(y) + 2xy = x^2 + 5 + y^2 + 5 + 2xy = x^2 + y^2 + 2xy + 10 <br/>\ne x^2 + 2xy + y^2 + 5 $", "Ah! Missing constant term.", "But earlier we had $ c = 0 $, so $ f(0) = 0 $. So $ f(x) = x^2 + bx $, $ f(0) = 0 $ — good.", "But $ f(x) = x^2 + bx $ works — yes.", "But what about $ f(x) = x^2 + bx + c $? We showed $ c = 0 $, $ a = 1 $.", "So only $ c = 0 $, $ a = 1 $, $ b $ free.", "Thus, infinitely many.", "But in the context of olympiad phrasing, if the answer is “infinitely many”, and the solution is a one-parameter family, we state:", "There are infinitely many such functions, one for each real number $ b $, given by $ f(x) = x^2 + bx $.", "But the final answer should be boxed.", "Typically, “infinitely many” is boxed as $ \boxed{\infty} $, though not always standard.", "But in many olympiad problems, when the solution set is infinite, and not finite, they expect “infinitely many”.", "However, look back at similar problems: sometimes symmetry forces $ b = 0 $. But here, no.", "Wait — suppose we set $ y = -x $. Then:\n$ f(0) = f(x) + f(-x) + 2x(-x) \Rightarrow 0 = f(x) + f(-x) - 2x^2 $\nSo $ f(-x) = -f(x) + 2x^2 $", "Now plug $ f(x) = x^2 + bx $:\nLHS: $ f(-x) = x^2 - bx $\nRHS: $ - (x^2 + bx) + 2x^2 = -x^2 - bx + 2x^2 = x^2 - bx $ — matches.", "So consistent.", "No contradiction.", "Thus, all such functions are valid.", "Therefore, the number of real functions satisfying the equation is infinite.", "But to be precise: uncountably infinitely many.", "However, in the format of the competition, and since no option is given, we conclude:", "$$\n\boxed{\ ext{infinitely many}}\n$$", "But the sample answers use mathematical expressions.", "Alternatively, since every solution is determined by a real parameter $ b $, and there is a one-to-one correspondence (binomial), the solution set has cardinality $ \mathfrak{c} $, but for olympiad purposes, infinitely many is acceptable.", "But let us see a standard reference: in functional equations, “find the number” when infinite, answer is “infinitely many” or $ \infty $.", "But in the absence of multiple choice, and to match format, we box the final answer as:", "$$\n\boxed{\ ext{infinitely many}}\n$$", "But to be mathematically rigorous and consistent with olympiad style (where “infinitely many” is accepted), and since no finite number works, we affirm:", "---", "Final Answer:\nAll solutions are of the form $ f(x) = x^2 + bx $ for $ b \in \mathbb{R} $. Since $ b $ can be any real number, there are infinitely many such functions.", "$$\n\boxed{\ ext{infinitely many}}\n$$"]