## Problem B1

Find all real-valued functions $f$ on the reals such that $f\left(x^{2}-y^{2}\right)=x f(x)-y f(y)$ for all $x, y$.


## Solution

Answer: $\mathrm{f}(\mathrm{x})=\mathrm{kx}$ for some real $\mathrm{k}$.

Putting $\mathrm{y}=0, \mathrm{f}\left(\mathrm{x}^{2}\right)=\mathrm{x} f(\mathrm{x})$. Hence $\mathrm{f}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=\mathrm{f}\left(\mathrm{x}^{2}\right)-\mathrm{f}\left(\mathrm{y}^{2}\right)$. So for any non-negative $\mathrm{x}, \mathrm{y}$, we have $\mathrm{f}(\mathrm{x}-\mathrm{y})=$ $f(x)-f(y)$. Hence also $f(x)=f(x+y-y)=f(x+y)-f(y)$, so $f(x+y)=f(x)+f(y)$ for non-negative $x, y$. Also $f(0)=f\left(0^{2}\right)=0 f(0)=0$, and for non-negative $y, f(-y)=f(0-y)=f(0)-f(y)=-f(y)$. Hence also $f(-y)=-f(y)$ for negative $y$. So we have $f(x+y)=f(x)+f(y)$ for non-negative $x$ and any $y$. But now if $x$ is negative, $f(x+$ $y)=-f(-x-y)=-(f(-x)-f(y))=f(x)+f(y)$. So $f(x+y)=f(x)+f(y)$ for all $x$ and $y$.

Now for any $x$ we have $f(x)+f(x-1)=f(2 x-1)=f\left(x^{2}-(x-1)^{2}\right)=x f(x)-(x-1) f(x-1)=x f(x-1)+x$ $f(1)-(x-1) f(x-1)=x f(1)+f(x-1)$, so $f(x)=x f(1)$. So if $f(1)=k$, then $f(x)=k x$. It is trivial to check that this does indeed satsify the equation given for any $\mathrm{k}$.

