### Instruction: Let $f(x)$ be a function such that $f(0) = 1$ and
\[f(xy) = f \left( \frac{x^2 + y^2}{2} \right) + (x - y)^2\]for all real numbers $x$ and $y.$  Find $f(x).$.### Response: This problem looks interesting because it involves a functional equation, which means that we have to find a function that satisfies a certain equation for any input.
I also notice that the equation involves $f(0)$, which is given as $1$, so that might be a useful starting point.
Let me try plugging in $x = 0$ and see what happens.
If $x = 0$, then the equation becomes $f(0) = f \left( \frac{0^2 + y^2}{2} \right) + (0 - y)^2$.
Simplifying, I get $1 = f \left( \frac{y^2}{2} \right) + y^2$.
This means that $f \left( \frac{y^2}{2} \right) = 1 - y^2$.
Now, this looks promising, because it gives me a way to express $f$ in terms of a simpler function, namely $1 - y^2$.
But I have to be careful, because this equation only holds for values of the form $\frac{y^2}{2}$, not for any real number.
So I need to find a way to rewrite any real number as $\frac{y^2}{2}$ for some $y$.
One way to do that is to let $y = \sqrt{2x}$, since then $\frac{y^2}{2} = x$.
So, if I plug in $y = \sqrt{2x}$ into the equation I found, I get $f(x) = f \left( \frac{y^2}{2} \right) = 1 - y^2 = 1 - 2x$.
This seems like a possible candidate for $f(x)$, but I have to check if it satisfies the original equation for any $x$ and $y$.
So, let me plug in $f(x) = 1 - 2x$ into the equation and see if it works.
I get $f(xy) = f \left( \frac{x^2 + y^2}{2} \right) + (x - y)^2$.
Substituting $f(x) = 1 - 2x$, I get $1 - 2xy = 1 - 2 \left( \frac{x^2 + y^2}{2} \right) + (x - y)^2$.
Simplifying, I get $-2xy = -x^2 - y^2 + (x - y)^2$.
Expanding, I get $-2xy = -x^2 - y^2 + x^2 - 2xy + y^2$.
Cancelling out, I get $0 = 0$, which is true.
So, $f(x) = 1 - 2x$ does satisfy the original equation for any $x$ and $y$.
Therefore, it is the solution I was looking for.