Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$: $f(x^2 + f(y)) = y + (f(x))^2$ --- To solve this functional equation, we proceed by evaluating the function at specific values to uncover its structural properties. ### Step 1: Establishing Bijectivity Let us evaluate the original equation at $x = 0$. The equation becomes: $f(f(y)) = y + (f(0))^2$ For clarity, we define the constant $c = f(0)$. This provides our first lemma: **Lemma 1:** $f(f(y)) = y + c^2$ Observe that as $y$ ranges over all real numbers, the expression $y + c^2$ takes on all possible real values. This immediately implies that $f$ is **surjective**. Furthermore, suppose $f(y_1) = f(y_2)$ for some real numbers $y_1$ and $y_2$. Applying $f$ to both sides yields $f(f(y_1)) = f(f(y_2))$, which, by Lemma 1, implies $y_1 + c^2 = y_2 + c^2$, or $y_1 = y_2$. Thus, $f$ is strictly **injective**. Since $f$ is both surjective and injective, it is a bijection on $\mathbb{R}$. ### Step 2: Deriving the Quadratic Property Next, we substitute $y = 0$ into the original equation: $f(x^2 + f(0)) = 0 + (f(x))^2$ Using our defined constant $c = f(0)$, this simplifies to our second lemma: **Lemma 2:** $f(x^2 + c) = (f(x))^2$ ### Step 3: Determining the Value of $f(0)$ We now proceed to show that $c = 0$. By the surjectivity of $f$ established in Step 1, there must exist a real number $a$ such that $f(a) = 0$. Substituting $y = a$ into Lemma 1 gives: $f(f(a)) = a + c^2 \implies f(0) = a + c^2 \implies c = a + c^2$ From this, we obtain the relation:$a = c - c^2$ Next, we substitute $x = a$ into Lemma 2: $f(a^2 + c) = (f(a))^2 = 0^2 = 0$ We have established that $f(a) = 0$. Because $f$ is strictly injective, the equality $f(a^2 + c) = f(a)$ implies that the arguments must be equal: $a^2 + c = a$ With these two equations we get $a^2+c^2=0$ Since $a$ and $c$ are real numbers, $a=0$ and $c=0$. ### Step 4: Reduction to Cauchy's Functional Equation With $c = f(0) = 0$, our previously established lemmas simplify considerably: - From Lemma 1: $f(f(y)) = y$ - From Lemma 2: $f(x^2) = (f(x))^2$ Using them in the original equation: $f(x^2 + f(y)) = f(f(y)) + f(x^2)$ Let $u = x^2$ (noting that $u \ge 0$) and $v = f(y)$ (noting that $v \in \mathbb{R}$ due to surjectivity). The equation transforms into the well-known Cauchy functional equation: $f(u + v) = f(u) + f(v)$ ### Step 5: Establishing Monotonicity The relation $f(u + v) = f(u) + f(v)$ is the well-known Cauchy functional equation. While it possesses infinitely many discontinuous solutions, it yields a unique family of linear solutions, $f(x) = kx$, if the function can be shown to be monotonic. Assume $x_1 > x_2$, which means $x_1 - x_2 > 0$. We can express $x_1 - x_2 = t^2$ for some real non-zero $t$. Applying Cauchy's equation: $f(x_1) = f(x_2 + t^2) = f(x_2) + f(t^2)$ From Lemma 2, $f(t^2)=f(t)^2 \ge 0$, it strictly follows that $f(x_1) \ge f(x_2)$. Therefore, $f$ is monotonically non-decreasing. This monotonicity condition restricts the solutions of Cauchy's equation strictly to linear functions of the form $f(x) = kx$. To find $k$, we substitute this back into our simplified Lemma 1, $f(f(y)) = y$: $k(ky) = y \implies k^2y = y \implies k^2 = 1$ Thus, $k = 1$ or $k = -1$. However, we know from $f(x^2) = (f(x))^2$ that $f(1) \ge 0$. If $k = -1$, then $f(1) = -1$, which yields a contradiction. Therefore, $k$ must be $1$. The unique solution to the functional equation is **$f(x) = x$**. #functional-equations #cauchy-functional-equation #imo