A functional equation for RMO or IITJEE Mathematics

Ref: Problems for the Mathematical Olympiads by Andrei Negut


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.


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.


More later,

Nalin Pithwa

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