CAPS Match 2024

Answer: For every $\alpha \ne 0$, 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 $yf(y)=0$ for all real numbers $y$. Then it is sufficient to realize that $f(x^2+y^2)\ne 0$ iff $x=0\wedge y=0$. Similarly $f(x-y)f(x+y)\ne 0$ iff $x+y=x-y=0\iff x=0\wedge y=0$ 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 $y\ne 0$. Using this equality, $P(y,x)$ shows that $2f(x-y)f(x+y)=\alpha (xf(x)-yf(y))$ for $x\ne y$. Plugging this into the original equation, we obtain $$f(x^2+y^2)=\frac{\alpha }{2}(xf(x)+yf(y))$$ for $x\ne y$. Setting $y=0$ in this equation shows $f(x^2)=\alpha xf(x)/2$ for $x\ne 0$, whereas $P(x,0)$ gives $f(x^2)=f(x{)}^2$ for all $x\in \mathbb{R}$. Hence $\alpha xf(x)/2=f(x{)}^2$, that is, $f(x)=0$ or $f(x)=\alpha x/2$ for $x\ne 0$. In particular, if $f(x)\ne 0$, 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 $x\ne 0$. Then both possible values for $f(0)$ yield functions fulfilling the original equation (if $(x,y)\ne (0,0)$, 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 $z\ne 0$ satisfying $f(z)=\alpha z/2$. Then $f(z^2)=f(z{)}^2=(\alpha /2{)}^2{z}^2\ne 0$, 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 $P(z/2,z/2):f(z^2/2)=z+zf(z/2)$. The left-hand side is 0 or $z^2/2$, the right-hand side $z$ or $z+z^2/2$. Since $z\ne 0$, only $z^2/2=z\iff z=2$ and $0=z+z^2/2\iff z=-2$ 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 $$0\ne b=f(b)=f(x-y)f(x+y)+2f(a)=f(x-y)f(x+y),$$ 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 $x\ne 0$ as possible solution, which we have already found and treated before.