Problem 2335

Answer: For every $\alpha \neq 0$, the zero function and the function with value 1 at 0, but 0 elsewhere are solutions. For $\alpha = 2$, the identity function $x \mapsto x$ is another solution.

Solution-check. The linear function clearly works. Consider the function $f$ such that $f(0) = 1$ and $f(x) = 0$ otherwise. Note that $y f(y) = 0$ for all real numbers $y$. Then it is sufficient to realize that $f(x^2 + y^2) \neq 0$ iff $x = 0 \land y = 0$. Similarly $f(x - y)f(x + y) \neq 0$ iff $x + y = x - y = 0 \Leftrightarrow x = 0 \land y = 0$ which shows that also this function is a solution.

Proof. Denote by $P(x, y)$ the proposition in the problem statement. Comparing $P(x, y)$ with $P(x, -y)$ yields $f(y) = -f(-y)$ for all $y \neq 0$. Using this equality, $P(y, x)$ shows that $2f(x - y)f(x + y) = \alpha(x f(x) - y f(y))$ for $x \neq y$. Plugging this into the original equation, we obtain
$$f(x^2 + y^2) = \frac{\alpha}{2}(x f(x) + y f(y))$$
for $x \neq y$. Setting $y = 0$ in this equation shows $f(x^2) = \alpha x f(x)/2$ for $x \neq 0$, whereas $P(x, 0)$ gives $f(x^2) = f(x)^2$ for all $x \in \mathbb{R}$. Hence $\alpha x f(x)/2 = f(x)^2$, that is, $f(x) = 0$ or $f(x) = \alpha x/2$ for $x \neq 0$. In particular, if $f(x) \neq 0$, then $f(x) = \alpha x/2$. On the other hand, $P(0, 0)$ shows $f(0) = f(0)^2$ and therefore $f(0) = 0$ or $f(0) = 1$.

Consider first the case that $f(x) = 0$ for all $x \neq 0$. Then both possible values for $f(0)$ yield functions fulfilling the original equation (if $(x, y) \neq (0, 0)$, all terms in $P(x, y)$ are zero anyway and $(x, y) = (0, 0)$ was treated before).

Now for the other case: There is a real number $z \neq 0$ satisfying $f(z) = \alpha z/2$. Then $f(z^2) = f(z)^2 = (\alpha/2)^2 z^2 \neq 0$, and hence $f(z^2) = (\alpha/2)z^2$. By comparing the last two statements, we obtain $\alpha = 2$ and then $f(z) = z$.

* $f(0) = 1$. Consider $P(z/2, z/2): f(z^2/2) = z + z f(z/2)$. The left-hand side is 0 or $z^2/2$, the right-hand side $z$ or $z + z^2/2$. Since $z \neq 0$, only $z^2/2 = z \Leftrightarrow z = 2$ and $0 = z + z^2/2 \Leftrightarrow z = -2$ are possible. Either way, $f(2) = 2$ and $f(-2) = -2$, because $f$ is odd. But then $f(4) = f(2^2) = f(2)^2 = 4$, which is impossible, because we just proved that $z = 2$ and $z = -2$ are the only real numbers with $f(z) = z$.

* $f(0) = 0$. We show that $f(x) = 0$ for all positive reals $x$ if $f$ is not the identity function:
(1) There are $0 < a < b$ with $f(a) = 0$, $f(b) = b$. Then $P(x, y)$ for $x = \sqrt{b - a}$ and $y = \sqrt{a}$ yields
$$0 \neq b = f(b) = f(x - y)f(x + y) + 2f(a) = f(x - y)f(x + y),$$
hence $f(x - y) = x - y$ and $f(x + y) = x + y$ and $b = x^2 - y^2 = b - 2a$, forcing the contradiction $a = 0$.
(2) There are $0 < a < b$ with $f(a) = a$, $f(b) = 0$. Analogous to Case 1, we arrive at the contradiction $b = 0$ when investigating $P(\sqrt{b - a}, \sqrt{a})$.
Except for the identity, we only have $f(x) = 0$ for $x > 0$ and thus $f(x) = 0$ for $x \neq 0$ as possible solution, which we have already found and treated before.