Problem 1583

For $y = 0$, the given relation reduces to $f(x) = f(f(x))$, which means that the given relation becomes:
$$f(x + y^{2n}) = f(x) + y^{2n-1}f(y), \quad \forall x, y \in \mathbb{R}. \qquad (1)$$
If we consider $x = 0$ and $y = 1$ in (1), then $f(0) = 0$. Since the equation $f(x) = 0$ has a unique solution, we have the implication:
$$f(x) = 0 \Rightarrow x = 0. \qquad (2)$$
Putting $x = 0$ in (1), we obtain $f(y^{2n}) = y^{2n-1}f(y)$, for all $y \in \mathbb{R}$, which means that relation (1) rewrites as:
$$f(x + y^{2n}) = f(x) + f(y^{2n}), \quad \forall x, y \in \mathbb{R}. \qquad (3)$$
Moreover, $y^{2n-1}f(y) = f(y^{2n}) = f((-y)^{2n}) = -y^{2n-1}f(-y)$, for all $y \in \mathbb{R}$, so $f$ is an odd function. Considering $t = \sqrt[2n]{y} \ge 0$, relation (3) becomes:
$$f(x + t) = f(x) + f(t), \quad \forall x \in \mathbb{R}, t \ge 0. \qquad (4)$$

Using the fact that $f$ is odd, for $t < 0$, we obtain:
$$f(x + t) = -f(-x - t) = -(f(-x) + f(-t)) = f(x) + f(t),$$
which, together with relation (4), implies:
$$f(x + t) = f(x) + f(t), \quad \forall x, t \in \mathbb{R}. \qquad (5)$$
If $f(x_1) = f(x_2)$, according to relation (5), we have $f(x_1 - x_2) = 0$, which, according to relation (2), implies $x_1 = x_2$, meaning $f$ is one-to-one. But since $f(f(x)) = f(x)$, we have $f(x) = x$, which is a solution to the given equation.