A functional equation for RMO or IITJEE Mathematics

Ref: Problems for the Mathematical Olympiads by Andrei Negut

Problem:

Find all the functions $f: \Re \rightarrow \Re$ with the property that $f(x^{2}-y^{2})=(x-y)(f(x)+f(y))$ holds for all x and $y \in \Re$ .

Solution:

By letting $x=y=0$ , we get $f(0)=0$ and $y=0$ gives us $f(x^{2})=xf(x)$ . Letting $x=0$ yields $f(-y^{2})=yf(y)=-f(y^{2})$ and therefore the given function is odd. Let $x-y=a$ and $x+y=b$ and therefore for all a, $b \in \Re$ , we will have

$f(ab)=a(f(\frac{a+b}{2})+f(\frac{b-a}{2})) \Longrightarrow \frac{f(ab)}{a} - f(\frac{b-a}{2})=f(\frac{a+b}{2})$

But this same relation with a and b switched gives us

$\frac{f(ab)}{a}-f(\frac{b-a}{2})=\frac{f(ab)}{a}-f(\frac{a-b}{2}) \Longrightarrow f(ab)(\frac{1}{a}-\frac{1}{b})$

$= f(\frac{b-a}{2})-f(\frac{a-b}{2}) \Longrightarrow f(ab)\frac{b-a}{ab}=2f(\frac{b-a}{2}) \Longrightarrow \frac{f(ab)}{ab}=\frac{\frac{b-a}{2}}{\frac{b-a}{2}}$

Let us call the above as equation 1, which holds for all $0 \neq a \neq b \neq 0$ (because we divided by a, b and $b-a$ ). But for what positive x and y can we find such a and b with $ab=x$ and $\frac{b-a}{2}=y$ ? We need to have $b=\frac{x}{a}$ and $\frac{x}{a}-a=2y \Longrightarrow a^{2}+2ay-x=0$ and this we always has a real solution. Therefore, for all positive x and y we will have $\frac{f(x)}{x}=\frac{f(y)}{y}$ . So this expression is constant, and therefore, for all positive x, we will have $f(x)=ax$ . Because the function is odd, this will hold for all x and this will be our solution.