Problem 1511

Substituting $(x, y) = (0, 0)$ in (8) yields $f(0)^2 + f(0) = 0$; that is, $f(0) = 0$ or $f(0) = -1$. If $f(0) = 0$, then the substitution $y = 0$ in (8) yields $f(x) = 0$ for all $x \in \mathbb{R}$. However, the zero function does not satisfy (8), so we must have $f(0) = -1$.

We consider two cases regarding the value of $a$.

Case 1. $a > 0$.
Let $x_0 = \frac{1}{\sqrt{a}}$. Substituting $(x, y) = (x_0, -x_0)$ in (8) one obtains $f(x_0)f(-x_0) = 0$. If $f(x_0) = 0$, then the substitution $(x, y) = (x - x_0, x_0)$ in (8) yields
$$f(x) = a x_0 (x - x_0) = \sqrt{a}x - 1, \quad \text{for all } x \in \mathbb{R}.$$

If $f(-x_0) = 0$, then the substitution $(x, y) = (x + x_0, -x_0)$ in (8) yields
$$f(x) = -a x_0 (x + x_0) = -\sqrt{a}x - 1, \quad \text{for all } x \in \mathbb{R}.$$

It is not hard to verify that both functions satisfy (8).

Case 2. $a < 0$.
We are going to show that no function $f$ satisfy (8). A substitution $y = x$ in (8) yields $f(x)^2 + f(2x) = a x^2$, for all $x \in \mathbb{R}$. That is, $f(2x) = a x^2 - f(x)^2$ and $f(-2x) = a x^2 - f(-x)^2$, and so
$$f(2x)f(-2x) = a^2 x^4 - a x^2 (f(x)^2 + f(-x)^2) + (f(x)f(-x))^2.$$

A substitution $y = -x$ in (8) yields $f(x)f(-x) = 1 - a x^2$ for all $x \in \mathbb{R}$. Thus,
$$1 - 4a x^2 = a^2 x^4 - a x^2 (f(x)^2 + f(-x)^2) + (1 - a x^2)^2,$$
or
$$x^2 (f(x)^2 + f(-x)^2) = 2a x^4 + 2x^2 \quad \text{for all } x \in \mathbb{R}.$$
Let $x = \frac{2}{\sqrt{-a}}$. We have $f(x)^2 + f(-x)^2 = 2a \left(\frac{2}{\sqrt{-a}}\right)^2 + 2 = -6$, which is impossible.