Let $\mathbb{R}$ be the set of [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number).  Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that


$f(x^2 - y^2) = xf(x) - yf(y)$


for all pairs of real numbers $x$ and $y$.
Solution 1
We first prove that $f$ is [ odd](https://artofproblemsolving.com/wiki/index.php/Odd_function).
Note that $f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0$, and for nonzero $y$, $xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)$, or $yf(-y) = -yf(y)$, which implies $f(-y) = -f(y)$.  Therefore $f$ is odd.  Henceforth, we shall assume that all variables are non-negative.
If we let $y = 0$, then we obtain $f(x^2) = xf(x)$.  Therefore the problem's condition becomes


$f(x^2 - y^2) + f(y^2) = f(x^2)$.


But for any $a,b$, we may set $x = \sqrt{a}$, $y = \sqrt{b}$ to obtain


$f(a-b) + f(b) = f(a)$.


(It is well known that the only [continuous](https://artofproblemsolving.com/wiki/index.php/Continuous) solutions to this functional equation are of the form $f(x) = kx$, but there do exist other solutions to this which are not solutions to the equation of this problem.)
We may let $a = 2t$, $b = t$ to obtain $2f(t) = f(2t)$.
Letting $x = t+1$ and $y = t$ in the original condition yields


$\begin{matrix}f(2t+1) &=& (t+1)f(t+1) - tf(t) \qquad \\ &=& (t+1)[f(t) + f(1) ] - tf(t) \\ &=& f(t) + (t+1)f(1) \qquad \qquad \end{matrix}$


But we know $f(2t + 1) = f(2t) + f(1) = 2f(t) + f(1)$, so we have $2f(t) + f(1) = f(t) + tf(1) + f(1)$, or


$f(t) = tf(1)$.


Hence all solutions to our equation are of the form $f(x) = kx$.  It is easy to see that real value of $k$ will suffice.

Solution 2
As in the first solution, we obtain the result that $f$ satisfies the condition


$f(a) + f(b) = f(a+b)$.


We note that


$f(x) = f\left[ \left(\frac{x+1}{2}\right)^2 - \left( \frac{x-1}{2} \right)^2 \right] = \frac{x+1}{2} f \left( \frac{x+1}{2} \right) - \frac{x-1}{2} f \left( \frac{x-1}{2} \right)$.


Since $f(2t) = 2f(t)$, this is equal to


$\frac{(x+1)[f(x) +f(1)]}{4} - \frac{(x-1)[f(x) - f(1)]}{4} = \frac{xf(1) + f(x)}{2}$


It follows that $f$ must be of the form $f(x) = kx$.

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