Another Functional Equation

Putting $y=0$ gives $f(0) \; = \; f\big(\tfrac12x^2\big) + x^2$ for all $x$ , and hence that $f(u) \,=\, f(0) - 2u$ for $u \ge 0$ . Putting $y=1$ now gives $f(x) \; = \; f\big(\tfrac{x^2+1}{2}\big) + (x-1)^2 \; = \; f(0) - (x^2+1) + (x-1)^2 \; = \; f(0) - 2x$ for all $x$ . It is easy to check that any function of the form $f(x) = \alpha - 2x$ satisfies this functional relation. Since $f(0) = 1$ we have $f(x) = 1 - 2x$ for all $x$ , and so the answer is $\boxed{-4033}$ .

The method I approached this problem was as follows:

I wished to remove the $(x-y)^2$ term somehow. I noted that the difference between the terms inside the functions was exactly half of this annoying term. Thus, I constructed a new function - dependent on the old function - to get rid of the $(x-y)^2$ .

Let $g(x)=f(x)+2x$ . Substituting this in to the statement, we get

$\begin{aligned} g(xy)-2xy &= g\left ( \dfrac{x^2+y^2}{2} \right ) - (x^2+y^2) + (x-y)^2\\ g(xy) &= g \left ( \dfrac{x^2+y^2}{2} \right ) \end{aligned}$

Substituting $y=0$ gives us $g(0)=g \left ( \dfrac{x^2}{2} \right ) = 1$ for all $x\geq 0$ , $g(x)=1$ . If we subtitute $y=-x$ , we get $g(-x^2) = g\left ( \dfrac{x^2}{2} \right ) = g(x^2)$ , so we have $g(x)=1 \forall x \in \mathbb{R}$ . Thus, we must have $f(x)=1-2x$ .

Therefore, $f(2017)=-4033$ .

You just have to plug in y=sqrt(4034), x=0